NEET Physics

Wave Optics Electromagnetic Waves

Question 1. The intensity ratio of the maxima and minima in an interference pattern produced by two identical coherent sources of light is 9: 1. The intensities of the used light sources are in the ratio

1. 3:1
2. 4:1
3. 9:1
4. 10:1

Answer: 2. 4:1

For maximum intensity, phase difference = Φ = 0 and for minimum intensity, Φ =180°.

Resultant intensity,

⇒ \(I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi\)

⇒ \(I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\),

and \(I_{\min }=I_1+I_2-2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)

Given that \(\frac{I_{\max }}{I_{\min }}=\frac{9}{1}=\left[\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right]^2\)

⇒ \(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}=\frac{3}{1} \Rightarrow \sqrt{\frac{I_1}{I_2}}=\frac{2}{1}\)

Hence, \(\frac{I_1}{I_2}=4: 1\).

Question 2. In an interference pattern, the intensities of two interfering waves are I and 4I respectively. They produce intensities at two points A and B with phase differences of TC/2 and π respectively. The difference in their intensities is

1. I
2. 2I
3. 4I
4. 5I

Answer: 3. 4I

Given that I₁ = 1 and I₂ = 4I.

Intensity at A is

⇒ \(I_{\mathrm{A}}=I_1+I_2+2 \sqrt{I_1 I_2} \cos \frac{\pi}{2}=I+4 I=5 I\)

Intensity at B is

⇒ \(I_{\mathrm{B}}=I_1+I_2+2 \sqrt{I_1 I_2} \cos \pi=I+4 I-2 \sqrt{I \cdot 4 I}=I\)

Difference in intensities = I_A – I_B = 5I – I = 4I.

Question 3. In Young’s double-slit experiment, the two slits act as coherent sources of equal amplitude a and of wavelength λ. In another experiment with the same set-up, the two slits are sources of equal amplitude a and wavelength λ but are incoherent. The ratio of intensities of light at the midpoint of the screen in the first case to that in the second case is

1. 2:1
2. 1:2
3. 3:4
4. 4:3

Answer: 1. 2:1

When the sources are coherent, the intensity at the central maxima (where phase difference Φ = 0), I₁ = k(a + a)²– k4a².

When the sources are incoherent, there is no phase relationship, hence no sustained interference. The net intensity, I₂ = k(a² + a² ) = k2a².

∴ \(\text { ratio }=\frac{I_1}{I_2}=\frac{k\left(4 a^2\right)}{k\left(2 a^2\right)}=2: 1\)

Question 4. What is the path difference for destructive interference?

1. n
2. \(\frac{(n+1)}{2} \lambda\)
3. n(λ + 1)
4. \(\left(\frac{2 n+1}{2}\right) \lambda\)

Answer: 4. \(\left(\frac{2 n+1}{2}\right) \lambda\)

For destructive interference,

∴ path difference A = odd multiple of \(\frac{\lambda}{2}=(2 n+1) \frac{\lambda}{2}\)

Question 5. A monochromatic beam of light is used for the formation of fringes on the screen by illuminating the two slits in Young’s double-slit experiment. When a thin film of mica is interposed in the path of one of the interfering beams then

1. Die fringe width increases
2. The fringe width decreases
3. The fringe width remains the same but the pattern shifts
4. The fringe pattern disappears

Answer: 3. The fringe width remains the same but the pattern shifts

When a thin film of mica sheet covers one of the two slits, a path difference Δ = (μ – l)f is introduced in one path of interfering waves. So, there will be no change in the fringe width but the central maximum with the whole fringe pattern will shift.

Question 6. A double-slit experiment is performed with a light of wavelength 500 nm. A thin film of thickness 2 μm and a refractive index of 1.5 is introduced in the path of the upper beam. The location of the central maximum will

1. Remain unshifted
2. Shift downward by nearly two fringes
3. Shift upward by nearly two fringes
4. Shift downward by five fringes

Answer: 3. Shift upward by nearly two fringes

Given that λ = 500 nm, thickness t = 2 pm and μ =1.5.

The optical path introduced for wave originating from the upper slit is Δ = (μ-l)t.

Central maximum P0 will shift to P where the path difference is zero.

Thus, at P, S₂P – S₁P – (μ-l)t = 0

=> dsinθ =(μ-l)t

⇒ \(d \frac{y}{D}=(\mu-1) t\)

⇒ \(y=\frac{D}{d}(\mu+1) t=N \beta=N \frac{D \lambda}{d}\)

The Number of fringes shifted is

∴ \(N=\frac{(\mu-1) t}{\lambda}=\frac{(1.5-1)\left(2 \times 10^{-6} \mathrm{~m}\right)}{\left(500 \times 10^{-9} \mathrm{~m}\right)}=2\)

Two fringes will shift upward.

Question 7. Young’s double-slit experiment is first performed in air and then in a medium other than air. It is found that the 8th bright fringe in the medium lies where the 5th dark fringe lies in the air. The refractive index of the medium is nearly

1. 1.25
2. 1.50
3. 1.69
4. 1.78

Answer: 4. 1.78

When in the air, the position of the 5th dark fringe from the centre is

⇒ \(y=\left(4+\frac{1}{2}\right) \beta_{\mathrm{a}}=\frac{9}{2} \beta_{\mathrm{a}}=\frac{9}{2} \frac{D \lambda_{\mathrm{a}}}{d}\)

In the medium, the position of the 8th bright fringe is

⇒ \(y=8 \beta_m=\frac{8 D \lambda_m}{d}\)

⇒ \(\frac{9}{2} \frac{D}{d} \lambda_{\mathrm{a}}=\frac{8 D}{d} \lambda_{\mathrm{m}}\)

The refractive index of the medium is

∴ \(\mu_{\mathrm{m}}=\frac{\lambda_{\mathrm{a}}}{\lambda_{\mathrm{m}}}=\frac{8}{4.5}=\frac{16}{9}=1.78\)

Question 8. Two coherent monochromatic light beams of amplitudes 3 and 5 units are superposed. The ratio of the maximum and minimum intensities in the resulting interference pattern will be

1. 4: 1
2. 16:1
3. 1:4
4. 2:1

Answer: 2. 16:1

Given that the ratio of amplitudes of interfering waves is

⇒ \(\frac{A_1}{A_2}=\frac{5}{3}\)

Now, \(\frac{A_1+A_2}{A_1-A_2}=\frac{5+3}{5-3}=\frac{8}{2}=\frac{4}{1}\)

Hence, \(\frac{I_{\max }}{I_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}=\left(\frac{4}{1}\right)^2=16: 1\)

Question 9. The two coherent sources with intensity ratio β produce interference. The fringe visibility will be

1. \(\frac{2 \sqrt{\beta}}{1+\beta}\)
2. 2β
3. \(\frac{2}{1+\beta}\)
4. \(\frac{\sqrt{\beta}}{1+\beta}\)

Answer: 1. \(\frac{2 \sqrt{\beta}}{1+\beta}\)

Given that the intensity ratio of two coherent waves is

⇒ \(\frac{I_1}{I_2}=\beta\)

Fringe visibility is defined as

\(V=\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}\) → (1)

Now, \(I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2 ; I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)

Substituting these in (1),

∴ \(V=\frac{4 \sqrt{I_1 I_2}}{2\left(I_1+I_2\right)}=\frac{2 I_2 \sqrt{\beta}}{I_2(\beta+1)}=\frac{2 \sqrt{\beta}}{1+\beta}\).

Question 10. Two slits are separated by a distance of 0.5 mm and illuminated with a light of wavelength λ = 600 nm. If the screen is placed at 2.5 m from the slits, the distance of the third bright fringe from the centre will be

1. 1.5 mm
2. 3 mm
3. 6 mm
4. 9 mm

Answer: 4. 9 mm

Given that slit separation = d = 0.5 mm;

wavelength = λ, = 600 nm; D = 2.5 m.

Fringe width is

⇒ \(\beta=\frac{D \lambda}{d}=\frac{(2.5 \mathrm{~m})\left(600 \times 10^{-9} \mathrm{~m}\right)}{\left(0.5 \times 10^{-3} \mathrm{~m}\right)}=3000 \times 10^{-6} \mathrm{~m}=3 \mathrm{~mm}\)

Distance of the third bright fringe = 3β = 3 x 3 mm = 9 mm.

Question 11. In a double-slit experimental arrangement, interference fringes of width 1 mm each are observed when light of wavelength 500 nm is used. Keeping the set-up unaltered, if the light source is replaced by another of wavelength 600 nm, the fringe width will be

1. 1.2 mm
2. 0.5 mm
3. 1 mm
4. 1.5 mm

Answer: 1. 1.2 mm

Fringe width = \(\beta=\frac{D \lambda}{d} \Rightarrow \beta \propto \lambda\)

Ratio \(\frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2}=\frac{500 \mathrm{~nm}}{600 \mathrm{~nm}}=\frac{5}{6}\)

Given that \(\beta_1=1 \mathrm{~mm}, \text { hence } \beta_2=\frac{6}{5} \beta_1=\frac{6}{5}(1 \mathrm{~mm})=1.2 \mathrm{~mm}\)

Question 12. In Young’s double-slit experiment, the spacing between the slits is 0.3 mm and the screen is kept at a distance of 1.5 m. The second bright fringe is found at 6 mm from the central fringe. The wavelength of the light used in the experiment is

1. 625 nm
2. 600 nm
3. 550 nm
4. 500 nm

Answer: 2. 600 nm

Given that d = 0.3 mm, D =1.5 m, 2β = 6 mm.

Now, \(\beta=3 \mathrm{~mm}=\frac{D \lambda}{d}=\frac{(1.5 \mathrm{~m})(\lambda)}{(0.3 \mathrm{~mm})}=5 \times 10^3 \lambda\)

Wavelength \(\lambda=\frac{3 \mathrm{~mm}}{5 \times 10^3}=0.6 \times 10^{-6} \mathrm{~m}=600 \mathrm{~nm}\)

Question 13. In Young’s double-slit experiment if the fringe order is represented by m then the fringe width is

1. Independent of m
2. Directly proportional to m
3. Directly proportional to (2m +1)
4. Inversely proportional to (2m +1)

Answer: 1. Independent of m

In the double-slit experiment, all the interference fringes are of the same width \(\left(\beta=\frac{D \lambda}{d}\right)\) and are independent of the fringe order m.

Question 14. Which among the following cannot produce coherent sources?

1. Fresnel biprism
2. Lloyd’s mirror
3. Young’s double-slit
4. Prism

Answer: 4. Prism

A single prism can produce dispersion, deviation, spectrum, etc., but cannot produce coherent sources.

Question 15. In Young’s double-slit experiment if the separation between the two slits is made three times then the fringe width will become

1. 9 times
2. \(\frac{1}{9} \text { times }\)times
3. 3 times
4. \(\frac{1}{3} \text { times }\)times

Answer: 4. \(\frac{1}{3} \text { times }\)times

Fringe width = \(\beta=\frac{D \lambda}{d}\)

If D and X are constants, \(\beta \propto \frac{1}{d}\)

⇒ \(\frac{\beta_1}{\beta_2}=\frac{d_2}{d_1}\)

Given that d₂ = 3d₁

Hence, \(\beta_2=\frac{\beta_1}{3}\)

Question 16. When exposed to sunlight, a thin film of oil on water exhibits brilliant colours due to the phenomenon of

1. Diffraction
2. Interference
3. Polarization
4. Dispersion

Answer: 2. Interference

Sunlight incident on a thin film of oil spread over the surface of water undergoes multiple reflections and interference. Bright colours are obtained which satisfy the condition for maxima.

Question 17. When a compact disc (CD) is illuminated by a source of white light, coloured lines are observed. This is due to

1. Interference
2. Dispersion
3. Refraction
4. Diffraction

Answer: 4. Diffraction

A compact disc (CD) consists of thousands of pits arranged in the form of spiral tracks. Visible light incident on the pits gets diffracted and constituent colours of the polychromatic light get separated and rainbow-like colours are observed.

Question 18. Young’s double-slit experiment is performed first in the air and then in a medium with a refractive index of 1.78. It is observed that the position of the nth bright fringe in the medium coincides with the 5th dark fringe in the air. The fringe order n has the value

1. 5
2. 7
3. 8
4. 10

Answer: 3. 8

Let y be the position of the 5th dark fringe in the air, so

⇒ \(y=\left(n+\frac{1}{2}\right) \beta=\left(4+\frac{1}{2}\right) \beta=\frac{9}{2} \frac{D \lambda_{\mathrm{a}}}{d}\)

In a medium of refractive index JI =1.78, wavelength = \(\lambda^{\prime}=\frac{\lambda_{\mathrm{a}}}{\mu}=\frac{\lambda_{\mathrm{a}}}{1.78}\)

the position of the nth bright fringe is

⇒ \(y^{\prime}=n \beta^{\prime}=n \frac{D \lambda^{\prime}}{d}=\frac{n D}{d} \frac{\lambda_{\mathrm{a}}}{1.78}\)

Given that \(y=y^{\prime}, \text { hence } \frac{9}{2} \frac{D \lambda_{\mathrm{a}}}{d}=\frac{n}{1.78} \frac{D \lambda_{\mathrm{a}}}{d}\)

∴ \(\frac{9}{2}=\frac{n}{1.78} \Rightarrow n=\frac{9 \times 1.78}{2}=8.01 \approx 8\)

Question 19. The intensity of the maximum in Young’s double-slit experiment is I₀. The distance between the two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D =10d?

1. \(\frac{I_0}{4}\)
2. \(\frac{3}{4} I_0\)
3. \(\frac{I_0}{2}\)
4. I₀

Answer: 3. \(\frac{I_0}{2}\)

Let I be the intensity of the two slits. At the maximum intensity,

⇒ \(I_0=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi\)

⇒  I + I + 2Icos 0° = 4I

\(I=\frac{I_0}{4}\) → (1)

Path difference at point P (in front of slit S₁) is

⇒ \(\Delta=S_2 P-S_1 P=\sqrt{D^2+d^2}-D\)

⇒ \(D\left(1+\frac{d^2}{D^2}\right)^{1 / 2}-D=D+\frac{d^2}{2 D}-D=\frac{d^2}{2 D}\)

Given that d = 5λ and D = l0d =10(5λ) = 50λ.

path difference = \(\Delta=\frac{25 \lambda^2}{2(50 \lambda)}=\frac{\lambda}{4}\)

and phase difference = \(\phi=\frac{2 \pi}{\lambda} \Delta=\frac{\pi}{2}\)

Hence, the intensity at P is

∴ \(I_{\mathrm{p}}=I+I+2 \sqrt{I I} \cos \frac{\pi}{2}=2 I=2\left(\frac{I_0}{4}\right)=\frac{I_0}{2}\) [From (1)]

Question 20. An interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio \(\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}\) will be

1. \(\frac{\sqrt{n}}{n+1}\)
2. \(\frac{2 \sqrt{n}}{n+1}\)
3. \(\frac{\sqrt{n}}{(n+1)^2}\)
4. \(\frac{2 \sqrt{n}}{(n+1)^2}\)

Answer: 2. \(\frac{2 \sqrt{n}}{n+1}\)

Given that = \(\frac{I_1}{I_2}=n=\frac{a_1^2}{a_2^2}\), where a₁ and a₂ are the amplitudes of the coherent sources.

Hence, \(\frac{a_1}{a_2}=\frac{\sqrt{n}}{1}\).

⇒ \(\frac{\left(a_1+a_2\right)}{\left(a_1-a_2\right)}=\frac{\sqrt{n}+1}{\sqrt{n}-1} \Rightarrow \frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{(\sqrt{n}+1)^2}{(\sqrt{n}-1)^2} \Rightarrow \frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{n}+1)^2}{(\sqrt{n}-1)^2}\)

∴ \(\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{(\sqrt{n}+1)^2-(\sqrt{n}-1)^2}{(\sqrt{n}+1)^2+(\sqrt{n}-1)^2}=\frac{2 \sqrt{n}}{n+1}\)

Question 21. In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 run is used. What will be the width of each slit for obtaining ten maxima of double-slit within the principal maxima of a single-slit pattern?

1. 0.2 mm
2. 0.1mm
3. 0.5 mm
4. 0.02 mm

Answer: 1. 0.2 mm

In the given figure, intensity distribution due to a single-slit diffraction pattern is shown. The angular width of the principal maxima is \(\frac{2 \lambda}{a}\), where a = width of each slit.

The linear width is PQ = D (angular width), where D is the distance between the screen and the slits.

\(P Q=\frac{2 \lambda}{a} \cdot D\) → (1)

Given that 10 double-slit fringes are within PQ, we have

\(P Q=10 \beta=10 \frac{D \lambda}{d}\) → (2)

Equating (1) and (2)

⇒ \(\frac{2 \lambda}{a} D=\frac{10 D \lambda}{d}\)

∴ \(a=\frac{d}{5}=\frac{1 \mathrm{~mm}}{5}=0.2 \mathrm{~mm}\)

Question 22. Two slits in Young’s experiment have widths in the ratio 1: 25. The ratio of intensity at the maxima to the minima in the interference pattern \(\left(\frac{I_{\max }}{I_{\min }}\right)\) is

1. \(\frac{9}{4}\)
2. \(\frac{121}{49}\)
3. \(\frac{49}{121}\)
4. \(\frac{4}{9}\)

Answer: 1. \(\frac{9}{4}\)

Assuming that slit width is proportional to the intensity of light, we have

⇒ \(\frac{I_1}{I_2}=\frac{25}{1}=\left(\frac{a_1}{a_2}\right)^2, \text { where } \frac{a_1}{a_2}\) is the ratio of amplitudes.

⇒ \(\frac{a_1}{a_2}=\frac{5}{1} \Rightarrow \frac{a_1+a_2}{a_1-a_2}=\frac{5+1}{5-1}=\frac{3}{2}\)

Hence, \(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{9}{4}\)

Question 23. In Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is λ (λ being the wavelength of the light used) is k. The intensity at a point where the path difference is λ/A will be

1. k
2. \(\frac{k}{2}\)
3. \(\frac{k}{4}\)
4. Zero

Answer: 2. \(\frac{k}{2}\)

Let I be the intensity of the component of the interfering sources.

For path difference Δ = λ, phase difference = 2π

and intensity k = I + I + 2√II cos 2π = 4I

⇒ \(I=\frac{k}{4}\)

For path difference = \(\Delta=\frac{\lambda}{4}, \text { phase difference }=\phi=\frac{2 \pi}{\lambda} \frac{\lambda}{4}=\frac{\pi}{2}\).

∴ intensity = \(I^{\prime}=I+I+2 \sqrt{I I} \cos \frac{\pi}{2}=2 I=2\left(\frac{k}{4}\right)=\frac{k}{2}\).

Question 24. In Young’s double-slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ₁ = 1200 run and λ₂ = 1000 nm. At what minimum distance from the common central bright fringe on the screen, 2 m away from the slits will a bright fringe from one interference pattern coincide with a bright fringe from the other?

1. 8 mm
2. 6 mm
3. 4 mm
4. 3 mm

Answer: 2. 6 mm

Given that slit separation d = 2 mm, λ₁ =1200 run, λ₂ =1000 run and D = 2 m.

Fringe width = \(\beta=\frac{D \lambda}{d}\)

Since λ₁ > λ₂, so β₁ > β₂.

Let nth bright fringe of λ₁ coincide with (n + 1)th bright fringe of λ₂.

So, \(n \frac{D \lambda_1}{d}=(n+1) \frac{D \lambda_2}{d}\)

⇒ n(1200 run) = (n +1)(1000 run)

⇒ n = 5.

Hence, the minimum distance from the common centre is

∴ \(5 \beta_1=\frac{5 D \lambda_1}{d}=\frac{5(2 \mathrm{~m})\left(1200 \times 10^{-9} \mathrm{~m}\right)}{\left(2 \times 10^{-3} \mathrm{~m}\right)}=6 \mathrm{~mm}\).

Question 25. Interference is possible in

1. Light waves only
2. Sound waves only
3. Both light and sound waves
4. Neither light nor sound waves

Answer: 3. Both light and sound waves

Interference is a wave phenomenon, which is observed in longitudinal (sound waves) as well as in transverse waves (light waves).

Question 26. If a yellow light emitted by a sodium lamp in Young’s double-slit experiment is replaced by a monochromatic blue light of the same intensity,

1. Fringe width will decrease
2. Fringe width will increase
3. Fringe width will remain unchanged
4. Fringes will become less intense

Answer: 1. Fringe width will decrease

Fringe width = \(\beta=\frac{D \lambda}{d}\)

λ_yellow > λblue.

Hence, β_yellow > β_blue.

Fringe width will decrease.

Question 27. In Young’s experiment, two coherent sources are placed 0.90 mm apart and a fringe pattern is observed one metre away. If it produces a second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used is

1. 60 x 10^-4 cm
2. l0 x 10^-4 cm
3. 10 x 10^-5 cm
4. 6 x 10^-5 cm

Answer: 4. 6 x 10^-5 cm

Given that d = 0.90 mm and D =1.0 m.

The distance of the 2nd dark fringe from the central fringe is

⇒ \(\beta+\frac{\beta}{2}=\frac{3}{2} \beta=1 \mathrm{~mm}\)

∴ \(\frac{3}{2} \frac{D \lambda}{d}=1 \mathrm{~mm}\)

⇒ \(\lambda=\frac{2 d}{3 D} \times 1 \mathrm{~mm}=\frac{2}{3}\left(\frac{9 \times 10^{-4} \mathrm{~m}}{1.0 \mathrm{~m}}\right)(1 \mathrm{~mm})=6 \times 10^{-7} \mathrm{~m}=6 \times 10^{-5} \mathrm{~cm}\).

Question 28. In Young’s double-slit experiment, the fringe width is found to be 0.4 mm. If the whole set-up is immersed in water of refractive index \(\frac{4}{3}\) without any change in the set-up, the new fringe width would be

1. 0.30 mm
2. 0.40 mm
3. 0.53 mm
4. 450 microns

Answer: 1. 0.30 mm

Given that β = 0.4 mm (in air). Let it be β’ in water.

∴ \(\frac{\beta^{\prime}}{\beta}=\frac{D \lambda^{\prime} / d}{D \lambda / d}=\frac{\lambda^{\prime}}{\lambda}=\frac{f \lambda^{\prime}}{f \lambda}=\frac{c}{c_0}=\frac{1}{\mu}\)

∴ \(\beta^{\prime}=\frac{\beta}{\mu}=\frac{0.4 \mathrm{~mm}}{4 / 3}=0.3 \mathrm{~mm}\)

Question 29. A Young’s double-slit experiment is performed with blue and green lights of wavelengths 4360 A and 5460 A respectively. If X be the distance of the 4th maxima from the central one then

1. X (blue) = X (green)
2. X (blue) > X (green)
3. X (blue) < X (green)
4. \(\frac{X \text { (blue) }}{X \text { (green) }}=\frac{5460}{4360}\)

Answer: 3. X (blue) < X (green)

Given that λ_blue = 4360 Å and λ_green = 5460 Å.

Since fringe width = \(\beta=\frac{D \lambda}{d}\)

⇒ \(\beta_{\text {blue }}<\beta_{\text {green }}\)

⇒ \(4 \beta_{\text {blue }}<4 \beta_{\text {green }}\)

∴ \(X_{\text {blue }}<X_{\text {green }}\)

Question 30. An achromatic combination of lenses consists of

1. Two convex lenses in contact
2. Two concave lenses in contact
3. One convex and one concave lens in contact
4. One convex and one plane mirror in contact

Answer: 3. One convex and one concave lens in contact

An achromatic combination of lenses constitutes a combination of two lenses of different materials satisfying the condition \(\frac{\omega_1}{f_1}+\frac{\omega_2}{f_2}=0\).

As f₁ and f₂ are of opposite signs, one is convex and the other concave.

Question 31. The focal length of a convex lens will be maximum for

1. Blue light
2. Yellow light
3. Greenlight
4. Red light

Answer: 4. Red light

Sunlight emerging from a convex lens undergoes dispersion, and blue light converges most and red light the least as shown in the given.

Hence, f_red is the maximum.

This can also be seen in the relation

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

Now, \(\mu_{\text {blue }}>\mu_{\text {green }}>\mu_{\text {yellow }}>\mu_{\text {red }}\)

∵ μ is least for red, \(\frac{1}{f_{\text {red }}}\) is least, hence is maximum.

Question 32. Cauchy’s dispersion formula is

1. \(\mu=A+B \lambda^{-2}+C \lambda^{-4}\)
2. \(\mu=A+B \lambda^{-2}+C \lambda^4\)
3. \(\mu=A+B \lambda^2+C \lambda^{-4}\)
4. \(\mu=A+B \lambda^2+C \lambda^4\)

Answer: 1. \(\mu=A+B \lambda^{-2}+C \lambda^{-4}\)

Cauchy’s dispersion formula is an empirical relationship between the refractive index and the wavelength of light for a particular transparent material. It is expressed as

⇒ \(\mu=A+\frac{B}{\lambda^2}+\frac{C}{\lambda^4}\)

where μ = refractive index and A, B, and C are constants.

Question 33. If the red light is replaced by a white light then

1. The width of the diffraction pattern will increase
2. The width of the diffraction pattern will decrease
3. A central white band will be obtained
4. No effect will be observed

Answer: 3. A central white band will be obtained

White light is a combination of seven colours. When diffracted through a single slit, the principal maxima will be white where all the components superpose in the same phase

Question 34. Greater accuracy in the determination of the position of a particle with an optical microscope can be had if the beam of light used

1. Has higher wavelength
2. Is polarized
3. Has higher frequency
4. Has greater intensity

Answer: 3. Has higher frequency

For a microscope, resolving power = \(\frac{2 \mu \sin \theta}{1.22 \lambda}\)

For higher resolution, λ should be small which means that a light of higher frequency should be used.

Question 35. The ratio of the resolving power of an optical microscope for two wavelengths λ₁ = 400 nm and λ₂ = 600 nm is

1. 9:4
2. 3:2
3. 16: 81
4. 8: 27

Answer: 2. 3:2

Resolving power of an optical instrument \(\propto \frac{1}{\lambda}\)

⇒ \(\frac{R_1}{R_2}=\frac{\lambda_2}{\lambda_1}=\frac{600 \mathrm{~nm}}{400 \mathrm{~nm}}=\frac{3}{2}\).

Question 36. A linear aperture whose width is 0.20 mm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of monochromatic light of wavelength 500 nm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

1. 0.10 cm
2. 0.25 cm
3. 0.20 cm
4. 0.15 cm

Answer: 4. 0.15 cm

Given that slit width a = 0.2 mm, λ = 500 nm and f = D = 60 cm.

Diffracted rays will be focused by a convex lens at its focal plane, so D = f.

The angular position of the first dark band is \(\theta_1=\frac{\lambda}{a}\) and its distance from the centre of the screen is

∴ \(y=D \theta=D \frac{\lambda}{a}=\frac{(60 \mathrm{~cm})(500 \mathrm{~nm})}{(0.2 \mathrm{~mm})}=0.15 \mathrm{~cm}\)

Question 37. In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle of 30° when light of wavelength 500 nm is incident on the slit. The first secondary maximum is observed at an angle of

1. \(\sin ^{-1}\left(\frac{1}{2}\right)\)
2. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
3. \(\sin ^{-1}\left(\frac{2}{3}\right)\)
4. \(\sin ^{-1}\left(\frac{1}{4}\right)\)

Answer: 2. \(\sin ^{-1}\left(\frac{3}{4}\right)\)

The angular position of the first minimum = θ₁

where sin \(\theta_1=\frac{\lambda}{a}=\sin 30^{\circ} \Rightarrow \frac{\lambda}{a}=\frac{1}{2}\)

Let the angular position of the first maximum be 01:

⇒ \(\sin \theta_1^{\prime}=\frac{3}{2} \frac{\lambda}{a}=\frac{3}{2}\left(\frac{1}{2}\right)=\frac{3}{4}\)

∴ \(\theta_1^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)\).

Question 38. At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

1. π radian
2. \(\frac{\pi}{2}\) radian
3. \(\frac{\pi}{4}\) radian
4. \(\frac{\pi}{8}\) radian

Answer: 1. π radian

While explaining the diffraction pattern due to a single slit, the unobstructed wavefront through the slit divided into 2, 3, and 4, parts take the path difference \(\Delta=a \sin \theta \text { as } \lambda, \frac{3}{2} \lambda, 2 \lambda\),…, etc. For the first minimum, the Huygens wavelets originating from the edge of the slit and from the midpoint of the slit are in opposite phases (phase difference = π) so that all cancel out leading to minimum intensity.

Question 39. For a parallel beam of monochromatic light of wavelength λ, diffraction is produced by a single slit whose width a is of the order of the wavelength of light. If D is the distance of the screen from the slit, the width of the central maxima will be

1. \(\frac{D a}{\lambda}\)
2. \(\frac{2 D a}{\lambda}\)
3. \(\frac{2 D \lambda}{a}\)
4. \(\frac{D \lambda}{a}\)

Answer: 3. \(\frac{2 D \lambda}{a}\)

The angular width of the central maxima is

⇒ \(2 \theta_1=2\left(\frac{\lambda}{a}\right)\)

So, the linear width of the central maxima is

∴ \(y=D\left(2 \theta_1\right)=2 \frac{D \lambda}{a}\)

Question 40. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide. The resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is

1. 1.2 cm
2. 2.4 cm
3. 1.2 mm
4. 2.4 mm

Answer: 4. 2.4 mm

Given that λ, = 600 nm, slit width = a = 1 mm and D = 2 m.

The distance between the first minima on either side of the central bright is the width of the principal maxima which is

⇒ \(y=D\left(2 \theta_1\right)=2 \frac{D \lambda}{a}\)

⇒ \(\frac{2(2 \mathrm{~m})\left(600 \times 10^{-9} \mathrm{~m}\right)}{\left(1 \times 10^{-3} \mathrm{~m}\right)}=2.4 \mathrm{~mm}\).

Question 41. A parallel beam of fast-moving electrons is incident normally in a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is true?

1. The angular width of the central maximum will decrease.
2. The angular width of the central maximum of the diffraction pattern will increase.
3. The angular width of the central maximum will remain unaffected.
4. Diffraction patterns will not be observed on the screen in the case of electrons.

Answer: 1. The angular width of the central maximum will decrease.

The de Broglie wavelength of fast-moving electrons is \(\lambda=\frac{h}{m v}\), which decreases with the increase in speed v. Hence, the angular width of the central maximum, \(2 \theta=\frac{2 \lambda}{a}\) will also decrease.

Question 42. A beam of monochromatic light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is

1. 2π
2. 3π
3. 4π
4. πλ

Answer: 3. 4π

In order to describe the pattern of diffraction for the 2nd minima, we divide the unobstructed wavefront through the slit into four equal segments so that the 1st and 2nd as well as the 3rd and 4th cancel out. The path difference for wavelets originating from the extreme edges is 2A, and the equivalent phase difference is \(\phi=\frac{2 \pi}{\lambda}(2 \lambda)=4 \pi\).

Question 43. The angular resolution of a 10-cm-diameter telescope at a wavelength of 600 nm is of the order of

1. 10⁶ rad
2. 10^-2 rad
3. 10^-4 rad
4. l0^-6 rad

Answer: 4. l0^-6 rad

Angular resolution = \(\Delta \theta=1.22 \frac{\lambda}{D}\),

Where = wave length = 600 nm and D = diameter = 10cm.

∴ \(\Delta \theta=\frac{1.22\left(600 \times 10^{-9} \mathrm{~m}\right)}{\left(10 \times 10^{-2} \mathrm{~m}\right)}=7.32 \times 10^{-6} \mathrm{rad} \approx 10^{-6} \mathrm{rad}\).

Question 44. A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects which can be resolved by the telescope when the mean wavelength of light is 500 nm, is of the order of

1. 0.5 m
2. 5 m
3. 5 mm
4. 5 cm

Answer: 3. 5 mm

According to the Rayleigh criterion, two objects can be resolved if the principal maxima of one falls on the first minima of the other in the diffraction pattern.

Hence, the limit of resolution is

⇒ \(\theta=1.22 \frac{\lambda}{d}=\tan \theta=\frac{y}{D}\)

where y is the minimum distance and D = distance of the objects from the lens.

⇒ \(y=1.22 \frac{\lambda}{d} \times D=1.22 \times \frac{\left(500 \times 10^{-9} \mathrm{~m}\right)(1000 \mathrm{~m})}{\left(10 \times 10^{-2} \mathrm{~m}\right)}\)

∴ 6.10 x 10^-3 m = 6 mm = 5 mm.

Question 45. The diameter of the human eye lens is 2 mm. What will be the minimum separation between two points to resolve them, which are situated at a distance of 50 m from the eye? The wavelength of light is 500 nm.

1. 2.32 m
2. 4.28 m
3. 1.52 cm
4. 12.48 cm

Answer: 3. 1.52 cm

The required minimum separation is

⇒ \(y=D \tan \theta=D \theta=D\left(1.22 \frac{\lambda}{d}\right)\)

Given that d = 2 mm, D = 50 m and A = 500 nm.

⇒ \(y=\frac{50 \mathrm{~m}(1.22)\left(500 \times 10^{-9} \mathrm{~m}\right)}{\left(2 \times 10^{-3} \mathrm{~m}\right)}=15.25 \times 10^{-3} \mathrm{~m}=1.52 \mathrm{~cm}\)

Question 46. An astronaut is looking down on the earth’s surface from a space shuttle at an altitude of 400 km. Assuming that the astronaut’s pupil diameter is 5 mm and the wavelength of visible light is 500 nm, the astronaut will be able to resolve linear objects of size about

1. 0.5 m
2. 5 m
3. 50 m
4. 500 m

Answer: 3. 50 m

As done in the preceding question, the minimum size to be resolved,

∴ \(y=1.22 \frac{D \lambda}{d}=\frac{1.22(400 \mathrm{~km})(500 \mathrm{~nm})}{(5 \mathrm{~mm})}\)

⇒ \(\frac{1.22\left(400 \times 10^3 \mathrm{~m}\right)\left(500 \times 10^{-9} \mathrm{~m}\right)}{\left(5 \times 10^{-3} \mathrm{~m}\right)}=4.88 \times 10 \mathrm{~m} \approx 50 \mathrm{~m}\).

Question 47. A beam of light is incident normally on a diffraction grating through which the first diffraction is seen at 32°. In this case, there will be the 2nd-order diffraction? 

1. At 80°
2. At 64°
3. At 48°
4. There will be no second-order diffraction

Answer: 4. There will be no second-order diffraction

In a diffraction grating, the nth-order diffraction angle (θ_n) is given by dsin 0n = nλ.

For the lst-order diffraction (n = 1),

d sin θ₁ = X

⇒  d sin 32° = X.

For the 2nd-order diffraction (n = 2),

d sin θ₂ = 2X = 2(d sin 32°)

⇒ sin θ₂ = 2 x sin 32° = 2(0.53) =.1.06.

Since sin θ₂ cannot be greater than one, there will be no second-order diffraction.

Question 48. At what angle of incidence will the light reflected from glass (μ = 1.5) be completely polarized?

1. 56.3°
2. 40.3°
3. 72.8°
4. 51.6°

Answer: 1. 56.3°

According to Brewster’s law, refractive index μ = tan i_p, where i_p = angle of polarization.

∴ i_p = tan^-1(1.5) = 56.3°

Question 49. Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarized light of intensity I₀ is incident on P₂. A third polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 45° with that of P₁. The intensity of transmitted light through P₂ is

1. \(\frac{I_0}{4}\)
2. \(\frac{I_0}{8}\)
3. \(\frac{I_0}{16}\)
4. \(\frac{I_0}{2}\)

Answer: 2. \(\frac{I_0}{8}\)

The intensity of the light transmitted through polaroid P₁ is I₀/2 which is plane polarized. Since the pass axes of P₁ and P₂ are inclined at 45°, the intensity of the light transmitted through P₃ (according to Malus’s law)

⇒ \(\left(\frac{I_0}{2}\right) \cos ^2 45^{\circ}=\frac{I_0}{4}\)

Pass axes of P₃ and P₂ are also inclined at 45°, so the final intensity of light transmitted through P₂ will be

∴ \(\frac{I_0}{4} \cos ^2 45^{\circ}=\frac{I_0}{8}\)

Question 50. In the case of linearly polarized light, the magnitude of the electric field vector

1. Does not change with time
2. Varies periodically with time
3. Increases and decreases linearly with time
4. Is parallel to the direction of propagation

Answer: 2. Varies periodically with time

Light is an electromagnetic wave in which electric and magnetic fields are mutually perpendicular and perpendicular to the direction of propagation. In the case of linearly (or plane-) polarized light, \(\vec{E}\) -vector varies periodically with time

Question 51. Which of the following diagrams represents the variation of the electric field vector with time for a circularly polarized light?

Answer: 2.

Circular polarization of an electromagnetic wave describes a polarization state in which at each point, the electric field of the wave has a constant magnitude but its direction rotates with time at a steady rate in a plane perpendicular to the direction of the wave. The tip of the electric field vector at any point in space describes a circle as time passes.

Question 52. In a YDSE, a light of wavelength 500 nm is used. The distance between the slits and the screen is 1.8 m and the slit separation is 0.4 mm. If the screen is moved parallel to itself with a speed of 4 m s^-1, with what speed will the first maxima move?

1. 2 mm s^-1
2. 3 mm s^-1
3. l mm s^-1
4. 5 mm s^-1

Answer: 4. 5 mm s^-1

When the screen is displaced parallel to itself, the position of the central maxima remains unchanged but the fringe width increases.

Now, \(\beta=\frac{D \lambda}{d}\)

⇒ \(\frac{d \beta}{d t}=\frac{\lambda}{d} \frac{d D}{d t}=\frac{500 \times 10^{-9} \mathrm{~m}}{0.4 \times 10^{-3} \mathrm{~m}} \times v_{\text {screen }}\)

⇒ \(\left(\frac{5}{4} \times 10^{-3}\right)\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)=5 \mathrm{~mm} \mathrm{~s}^{-1}\)

This is the speed with which the first maxima will move.

Question 53. Find the resolving power of a microscope whose objective lens has focal length f₀ = 5 cm, and diameter = 2 cm. (Wavelength of light used, λ = 6000 A.)

1. 11.9 x l0⁵ m^-1
2. 10.9 x 10⁵ m^-1
3. 10.9 x l0⁴ m^-1
4. 10.9 x l0³ m^-1

Answer: 2. 10.9 x 10⁵ m^-1

Given that f₀ = 5 cm and sin \(\theta \approx \tan \theta=\frac{D}{f}=\frac{2}{5}\)

The resolving power of the microscope is

∴ \(\frac{2 \mu \sin \theta}{1.22 \lambda}=\frac{2(1)(2 / 5)}{1.22\left(600 \times 10^{-9} \mathrm{~m}\right)}=10.9 \times 10^5 \mathrm{~m}^{-1}\)

Question 54. In a double-slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away was found to be 0.2°. What will be the angular width of the first minima if the entire set-up is immersed in water? \(\left(\mu_{\text {water }}=\frac{4}{3}\right)\).

1. 0.26°
2. 0.15°
3. 0.05°
4. 0.1°

Answer: 2. 0.15°

The angular width of a fringe is

⇒ \(\theta=\frac{\beta}{D}=\frac{D \lambda}{d D}=\frac{\lambda}{d}\)

Now, \(\frac{\theta_{\text {water }}}{\theta_{\text {air }}}=\frac{\lambda_\omega}{\lambda_{\mathrm{a}}}=\frac{C_\omega / v}{C_{\mathrm{a}} / v}=\frac{1}{\mu_\omega}=\frac{3}{4}\)

Hence, \(\theta_{\text {water }}=\frac{3}{4}\left(\theta_{\text {air }}\right)=(0.75)\left(0.2^{\circ}\right)=0.15^{\circ}\).

Question 55. In a YDSE, the ratio of the slit widths is 4: 1. The ratio of the intensity of the maxima to that of the minima in the fringe pattern close to the central fringe on the screen will be

1. 3: 16
2. 2: 9
3. 9: 1
4. 4: 1

Answer: 3. 9: 1

Assuming that the slit width is proportional to the intensity of light transmitted,

⇒ \(\frac{I_1}{I_2}=\frac{4}{1}=\frac{a_1^2}{a_2^2}\),

where a1 and a2 represent the corresponding amplitudes.

∴ \(\frac{a_1}{a_2}=\frac{2}{1} \Rightarrow \frac{a_1+a_2}{a_1-a_2}=\frac{3}{1}\)

Hence, \(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{9}{1}\).

Question 56. In a YDSE, when a thin film of thickness t having refractive index μ is introduced in front of one of the slits, the maximum at the centre of the fringe system shifts by one fringe width. The value of t in terms of λ (the wavelength used) is

1. \(\frac{\lambda}{(2 \mu-1)}\)
2. \(\frac{2 \lambda}{(\mu-1)}\)
3. \(\frac{\lambda}{(\mu-1)}\)
4. \(\frac{\lambda}{2(\mu-1)}\)

Answer: 3. \(\frac{\lambda}{(\mu-1)}\)

Insertion of a thin film in the path of rays coming from one slit will introduce an additional optical path Δ = (μ- l)t, where 4 is the refractive index of the film and t is its thickness. Since one fringe shift corresponds to an additional path μ, therefore

∴ \((\mu-1) t=\lambda \Rightarrow t=\frac{\lambda}{(\mu-1)}\).

Question 57. In an interference experiment, the ratio of amplitudes of coherent waves is \(\frac{a_1}{a_2}=\frac{1}{3}\). The ratio of maximum and minimum intensities of fringes will be

1. 4
2. 2
3. 9
4. 18

Answer: 1. 4

Given that \(\frac{a_1}{a_2}=\frac{1}{3} \Rightarrow \frac{a_2+a_1}{a_2-a_1}=\frac{4}{2}=\frac{2}{1}\)

∴ \(\frac{I_{\max }}{I_{\min }}=\left(\frac{a_2+a_1}{a_2-a_1}\right)^2=\left(\frac{2}{1}\right)^2=4\).

Question 58. Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensities of the component waves are in the ratio

1. 16:9
2. 5:3
3. 25:9
4. 4:1

Answer: 3. 25:9

Given that \(\frac{I_{\max }}{I_{\min }}=\frac{16}{1}=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2\)

⇒ \(\left(\frac{a_1+a_2}{a_1-a_2}\right)=\frac{4}{1} \Rightarrow \frac{a_1}{a_2}=\frac{5}{3} \Rightarrow \frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\frac{25}{9}\).

Question 59. In Young’s double-slit experiment with slit separation 0.1 mm, one observes a bright fringe at an angle (1/40) radian by using a light of wavelength λ₁. When a light of wavelength λ₂ is used, a bright fringe is seen at the same angle in the same set-up. Given that λ₁ and λ₂ are in the visible range (380 nm-740 nm), their values are

1. 380 nm and 525 nm
2. 400 nm and 500 nm
3. 380 nm and 500 nm
4. 625 nm and 500 nm

Answer: 4. 625 nm and 500 nm

The angular position of the nth bright fringe at P„ is

⇒ \(\theta=\frac{P_0 P_n}{O P_0}=\frac{y}{D}\)

If this region contains n bright fringes,

⇒ \(y=n \beta=n \frac{D \lambda}{d}\); hence

⇒ \(\theta=\frac{1}{D}\left(\frac{n D \lambda}{d}\right)=\frac{n \lambda}{d}\)

⇒ \(n \lambda=\theta \cdot d=\frac{1}{40} \times(0.1 \mathrm{~mm})=\frac{10^{-4} \mathrm{~m}}{40}=\frac{10000}{4} \mathrm{~nm}=2500 \mathrm{~nm}\)

Thus, \(\lambda=\frac{2500}{n} \mathrm{~nm} \text {, where } n=1,2,3\),….

For k to lie within the visible range (380 nm-740 nm),

⇒ \(\lambda=\frac{2500 \mathrm{~nm}}{5}, \frac{2500 \mathrm{~nm}}{4}\),…

Hence, λ₁= 500 nm and λ₂ = 625 nm.

Question 60. In a YDSE, the slit separation is 0.320 mm and light of wavelength λ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range -30° < 0 < 30° is

1. 320
2. 321
3. 640
4. 641

Answer: 4. 641

Given that slit separation = d = 0.32 mm and wavelength = λ = 500 nm.

If N bright fringes are contained within the angular range 0 then the path difference is

A = dsinθ = Nλ,

⇒ \(N=\frac{d \sin \theta}{\lambda}=\frac{\left(0.32 \times 10^{-3} \mathrm{~m}\right)(0.5)}{\left(500 \times 10^{-9} \mathrm{~m}\right)}=320\).

Same number of fringes (N = 320) will be contained in the range θ = 0 to θ = -30°.

Thus, the total number of bright fringes contained will be 2N + central bright fringe = 641.

Question 61. Consider a YDSE set-up as shown in the figure. What should be the slit-separation d in terms of wavelength X such that the first minima occurs directly in front of the slit S₁?

1. \(\frac{\lambda}{2(5-\sqrt{2})}\)
2. \(\frac{\lambda}{(5-\sqrt{2})}\)
3. \(\frac{\lambda}{2(\sqrt{5}-2)}\)
4. \(\frac{\lambda}{(\sqrt{5}-2)}\)

Answer: 3. \(\frac{\lambda}{2(\sqrt{5}-2)}\)

Since the position of the first dark fringe P is directly in front of slit Sl, the path difference

at P will be \(\Delta=\frac{\lambda}{2}\)

or \(\frac{\lambda}{2}=\Delta=\mathrm{S}_2 \mathrm{P}-\mathrm{S}_1 \mathrm{P}=\sqrt{4 d^2+d^2}-2 d\)

⇒ (√5 – 2)d

⇒ \(d=\frac{\lambda}{2(\sqrt{5}-2)}\)

Question 62. In a YDSE, the path difference at a certain point on the screen between two interfering waves is \(\frac{1}{8}\) of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe will be close to

1. 0.94
2. 0.74
3. 0.85
4. 0.80

Answer: 3. 0.85

Let I be the intensity of each of the coherent waves. The phase difference is

⇒ \(\phi=\frac{2 \pi}{\lambda} \Delta=\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{8}=\frac{\pi}{4}\)

Intensity at point P is

⇒ \(I_{\mathrm{P}}=I+I+2 \sqrt{I} \sqrt{I} \cos \frac{\pi}{4}=2 I\left(1+\frac{1}{\sqrt{2}}\right)\)

Intensity at the central bright fringe,

⇒ \(I_0=I+I+2 \sqrt{I} \sqrt{I} \cos 0^{\circ}=4 I\)

∴ \(\frac{I_{\mathrm{P}}}{I_0}=\frac{\sqrt{2}+1}{2 \sqrt{2}}=0.85\)

Question 63. In a YDSE, green light (A. = 5303 A) falls on the slits having a slit separation of 19.44 |j,m and a width of 4.05 |im. The number of bright fringes between the first and the second diffraction minima is

1. 4
2. 10
3. 9
4. 5

Answer: 4. 5

Angular positions of the first and the second minima are \(\theta_1=\frac{\lambda}{a} \text { and } \theta_2=\frac{2 \lambda}{a}\)

where a is the width of each slit. The linear separation between the two minima is

⇒ \(D \theta=\left(\theta_2-\theta_1\right) D=\frac{D \lambda}{a}=N \beta\)

where N is the number of bright fringes and j} = fringe width = \(\frac{D \lambda}{d}\)

⇒ \(N\left(\frac{D \lambda}{d}\right)=\frac{D \lambda}{a}\)

∴ \(N=\frac{d}{a}=\frac{19.44 \mu \mathrm{m}}{4.05 \mu \mathrm{m}}=4.8 \approx 5\)

Question 64. The value of the numerical aperture of the objective lens of a microscope is 1.25. If light of wavelength 500 nm is used, the minimum separation between two points to be seen distinctly will be

1. 0.38 μm
2. 0.48 μm
3. 0.24 μm
4. 0.12 μm

Answer: 3. 0.24 μm

The minimum separation between two objects that can be resolved by a microscope is given by

⇒ \(d_{\min }=\frac{1.22 \lambda}{2 \mu \sin \theta}=\frac{1.22 \lambda}{2(N A)}\)

where numerical aperture = 1.25 and λ = 500 run.

∴ \(d_{\min }=\frac{1.22\left(500 \times 10^{-9} \mathrm{~m}\right)}{2(1.25)}=2.44 \times 10^{-7} \mathrm{~m}=0.24 \mu \mathrm{m}\)

Question 65. Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect a light of wavelength 500 nm coming from a star.

1. 305 x 10^-9 radian
2. 152.5 x l0^-9 radian
3. 610 x 10^-9 radian
4. 457.5 x 10^-9 radian

Answer: 3. 610 x 10^-9 radian

The limit of resolution (angular) for a telescope is

⇒ \(\Delta \theta=1.22 \frac{\lambda}{\mathrm{D}}=\frac{1.22\left(500 \times 10^{-9} \mathrm{~m}\right)}{2 \mathrm{~m}} \mathrm{rad}\)

∴ 610 x 10^-9 rad.

Question 66. The diameter of the objective lens of a telescope is 250 cm. For a light of wavelength 600 nm coming from a distant object, the limit of resolution of the telescope is close to

1. 3.0 x 10^-7 rad
2. 2.0 x 10^-7 rad
3. 1.5 x 10^-7 rad
4. 4.5 x 10^-7 rad

Answer: 1. 3.0 x 10^-7 rad

Limit of resolution = \(1.22 \frac{\lambda}{D}=1.22 \frac{\left(600 \times 10^{-9} \mathrm{~m}\right)}{250 \times 10^{-2} \mathrm{~m}}=2.9 \times 10^{-7} \mathrm{rad}\)

∴ 3.0 x 10^-7 rad.

Question 67. Consider a tank made of glass (refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index μ. A student finds that irrespective of what the incident angle i is (as shown in the figure), a beam of light entering the liquid-glass interface is never completely polarized. For this to happen, the tire minimum value of p should be

1. \(\frac{3}{\sqrt{5}}\)
2. \(\frac{5}{\sqrt{3}}\)
3. \(\frac{4}{3}\)
4. \(\sqrt{\frac{5}{3}}\)

Answer: 1. \(\frac{3}{\sqrt{5}}\)

Under the given situation, light will be transmitted from liquid to air if r ≤ θ_C, where θ_C is the critical angle for the liquid.

Further, transmitted light in the air will never be completely polarized if the polarizing angle 0_p>θ_C.

Thus, tan 0_p > tan θ_C.

⇒ \(\frac{\mu_{\mathrm{g}}}{\mu_l}>\frac{\sin \theta_{\mathrm{c}}}{\cos \theta_{\mathrm{c}}} \Rightarrow \frac{\mu_{\mathrm{g}}}{\mu_l}>\frac{1 / \mu_l}{\sqrt{1-\frac{1}{\mu_l^2}}}\)

⇒ \(2 \mu_c->\frac{1}{\sqrt{\mu_l^2-1}} \Rightarrow 9\left(\mu_l^2-1\right)>4 \mu_l^2\)

∴ Hence, \(\mu_l>\frac{3}{\sqrt{5}}\).

Question 68. A system of three polaroids P₁, P₂ and P₃ is set up such that the pass axis of P₃ is crossed with respect to that of P_1. The pass axis of P₂ is inclined at 30° to that of P₁. When a beam of unpolarized light of intensity I₀ is incident on P₁ the intensity of light transmitted by the three polaroids is I. The ratio I0/I is equal to (nearly)

1. 5.33
2. 16.00
3. 10.67
4. 1.80

Answer: 3.

Pass axes of I₁ and I₂ are inclined at 30°, so according to Malus’s law,

⇒ \(I_2=I_1 \cos ^2 30^{\circ}=\frac{3 I_1}{4}\)

But \(I_1=\frac{I_0}{2}, \text { so } I_2=\frac{3}{4}\left(\frac{I_0}{2}\right)=\frac{3 I_0}{8}\)

Similarly, P₂ and P₃ are inclined at 60°, so

⇒ \(I=I_2 \cos ^2 60^{\circ}=\frac{3}{8} I_0 \times \frac{1}{4}=\frac{3}{32} I_0\)

∴ \(\frac{I_0}{I}=\frac{32}{3}=10.67\)

Question 69. An electromagnetic wave is represented by the electric field \(\vec{E}=E_0 \sin (\omega t+6 y-8 z) \hat{n}\). Then, the direction of propagation of the wave is

1. \(\frac{-3 \hat{j}-4 \hat{k}}{5}\)
2. \(\frac{-3 \hat{j}+4 \hat{k}}{5}\)
3. \(\frac{-3 \hat{i}-4 \hat{k}}{5}\)
4. \(\frac{3 \hat{i}-4 \hat{j}}{5}\)

Answer: 2. \(\frac{-3 \hat{j}+4 \hat{k}}{5}\)

From the given wave equation, it is clear that the direction of propagation of the wave is along the negative direction of \(6 \hat{j}-8 \hat{k}\). The unit vector along the direction of propagation will be

∴ \(-\frac{6 \hat{j}-8 \hat{k}}{\sqrt{36+64}}=\frac{-6 \hat{j}+8 \hat{k}}{10}=\frac{-3 \hat{j}+4 \hat{k}}{5}\).

Question 70. A plane electromagnetic wave travels in free space along the electric field component of the wave at a particular point of space and time is E = 6 V m^-1 along the y-direction. Its corresponding magnetic field component B would be

1. 6 x 10^-8 T along the z-direction
2. 6 x 10^-8 T along
3. 2 x 10^-8 T along z-direction
4. 2 x 10^-8 T along y-direction

Answer: 3. 2 x 10^-8 T along z-direction

Since \(\frac{E}{B}=c\), hence magnitude of the magnetic field is

⇒ \(B=\frac{E}{c}=\frac{6 \mathrm{Vm}^{-1}}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}=2 \times 10^{-8} \mathrm{~T}\)

But \(\vec{E}=E_0 \hat{i}\) (given) and the wave travels along the x-direction, \(\hat{E} \times \hat{B}=\hat{i}\), gives the direction along the z-direction, \(\vec{B}=\left(2 \times 10^{-8} \mathrm{~T}\right) \hat{k}\).

Question 71. A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space at time t, \(\vec{E} \text { is }\left(6.3 \mathrm{Vm}^{-1}\right) \hat{j}\). The corresponding magnetic field \(\vec{B}\) a that point will be

1. \(\left(18.9 \times 10^8 \mathrm{~T}\right) \hat{k}\)
2. \(\left(6.3 \times 10^{-8} \mathrm{~T}\right) \hat{k}\)
3. \(\left(18.9 \times 10^{-8} \mathrm{~T}\right) \hat{k}\)
4. \(\left(2.1 \times 10^{-8} \mathrm{~T}\right) \hat{k}\)

Answer: 4. \(\left(2.1 \times 10^{-8} \mathrm{~T}\right) \hat{k}\)

Given that \(\vec{E}=\left(6.3 \cdot \mathrm{Vm}^{-1}\right) \hat{j}\). The direction of propagation of the electromagnetic wave is along the x-direction. Thus,

⇒ \(\frac{E}{B}=c \text { and } \hat{E} \times \hat{B}=\hat{i} \Rightarrow \hat{j} \times \hat{B}=\hat{i}\)

Thus, \(B=\frac{E}{c}=\frac{6.3}{3 \times 10^8} \mathrm{~T}=2.1 \times 10^{-8} \mathrm{~T}\)

and \(\hat{B}=\hat{k}\)

∴ \(\vec{B}=B \hat{B}=\left(2.1 \times 10^{-8} \mathrm{~T}\right) \hat{k}\).

Question 72. If the magnetic field of a plane electromagnetic wave is given by \(B=\left(100 \times 10^{-6} \mathrm{~T}\right) \sin \left[2 \pi \times 2 \times 10^{15}\left(t-\frac{x}{c}\right)\right]\) then the maximum electric field associated with the wave is

1. 6 x l0⁴ N C^-1
2. 4 x l0⁴ N C^-1
3. 3 x l0⁴ N C^-1
4. 4.5 x l0⁴ N C^-1

Answer: 3. 3 x l0⁴ N C^-1

The amplitude of the magnetic field B₀ = 100 x10^-6 T, hence the corresponding amplitude of the electric field is

E₀ = B₀ x c = (100 x10^-6 T)(3 x10⁸ m s^-1 ) = 3 x10⁴ N C^-1.

Question 73. The intensity of a plane-polarized light is 3.3 W m^-2. A polarizer of area 3 x 10^-4 m² rotates at a constant angular frequency of 10jc rad s^-1. The total amount of energy transmitted in one complete revolution is

1. 6.95 x l0^-4 J
2. 1.0 x 10^-4 J
3. 3.95 x 10^-4J
4. 2.95 x l0^-4J

Answer: 2. 1.0 x 10^-4 J

According to Malus’s law, the intensity of a transmitted light is

I = I₀ Cos²θ.

The total energy transmitted in one revolution is

⇒ \(E=\int I A d t=\int I_0 \cos ^2 \theta A d t\)

⇒ \(I_0 A \int \cos ^2 \omega t d t=\frac{I_0 A^{2 \pi}}{\omega} \int_0^{2 \pi} \cos ^2 \omega t d(\omega t)\)

∴ \(\frac{I_0 A \pi}{\omega}=\frac{\left(3.3 \mathrm{~W} \mathrm{~m}^{-2}\right)\left(3 \times 10^{-4} \mathrm{~m}^2\right) \pi}{(10 \pi) \mathrm{rad} \mathrm{s}^{-1}}\)

= 0.99 x 10^-4 J ≈1.0 x 10^-4 J

Question 74. A single slit of width 0.6 x10^-4 m is illuminated by a light of wavelength 600 nm. The highest order of minima on both sides of the central maxima is

1. 200
2. 100
3. 20
4. 10

Answer: 2. 100

For a single-slit diffraction pattern, the position for the minima is d sin0 = nX

⇒ \(\sin \theta=\frac{n \lambda}{d}<1\)

Hence, \(n \leq \frac{d}{\lambda}\)

Here, d = slit width = 0.6 x 10^-4m and λ = 600 x 10^-9 m.

∴ \(n \leq \frac{6 \times 10^{-5} \mathrm{~m}}{6 \times 10^{-7} \mathrm{~m}}=100\)

Question 75. In the given figure, P and Q are two equally intense coherent sources emitting radiation of the same wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of Q by 90°. A, B and C are three points each at the same distance from the midpoint O of PQ. The intensities observed at A, B and C will be in the ratio

1. 0:1:4
2. 0:1:2
3. 4:1:0
4. 2:1:0

Answer: 2. 0:1:2

Resultant intensity due to superposition is

I = 2I₀(1 + cos π), where Φ= phase difference.

At point A: path difference = \(\Delta=P A-Q A=P Q=5 \mathrm{~m}=\frac{20 \mathrm{~m}}{4}=\frac{\lambda}{4}\)

and phase difference = \(\frac{2 \pi}{4}=\frac{\pi}{2}\)

Since P is ahead of Q by \(\frac{\pi}{2}\), hence net phase difference Φ = π.

I_A = 2I₀(l + cos π) = 0.

At point B: B is equidistant from both P and Q, hence path difference = 0.

But P is ahead in phase by \(\frac{\pi}{2}, \text { so } \phi=\frac{\pi}{2}\)

⇒ \(I_{\mathrm{B}}=2 I_0\left(1+\cos \frac{\pi}{2}\right)=2 I_0\)

At point C: path difference = \(\Delta=Q C-P C=\frac{\lambda}{4}\)

But P leads by \(\frac{\pi}{2}\left(\equiv \frac{\lambda}{4}\right)\), hence Φ = 0.

∴ I_c = 2I₀(1 + cos 0°) = 4I₀.

∴ Hence, I_A : I_B : I_c = 0 : 2I₀ : 4I₀ = 0 : 1 : 2.

Question 76. In a YDSE with a slit separation of 1 mm and distance of the screen from the slit 1 m, the wavelength of the light used is 650 nm. At a distance y =1.270 mm from the central fringe where a bright fringe is observed, the path difference is

1. 1.27 μm
2. 2.45 μm
3. 2.27 μm
4. 0.27 μm

Answer: 1. 1.27 μm

Path difference at P due to sources S₁ and S₂ is

Δ = S₂P-S₁P ≈ S₂N = S₁S₂sin θ.

For θ to be small, \(\sin \theta \rightarrow \tan \theta=\frac{y}{D}\)

⇒ \(\Delta=d\left(\frac{y}{D}\right)=(1 \mathrm{~mm}) \frac{(1.27 \mathrm{~mm})}{1 \mathrm{~m}}\)

= 1.27 μm.

Question 77. In a YDSE, the total number of fringes observed between points O and P with λ = 700 nm was 16. With λ’ = 400 nm, the total number of fringes in the same region OP will be

1. 7
2. 12
3. 28
4. 14

Answer: 3. 28

y = N₁p₁ = N₂P₂

⇒ \(16 \cdot \frac{D(700 \mathrm{~nm})}{d}=N_2 \cdot \frac{D(400 \mathrm{~nm})}{d}\)

∴ Hence, \(N_2=\frac{16 \times 700}{400}=28\).

Question 78. In a YDSE, the intensity at a point where the path difference is λ is k units. At another point where the path difference is \(\frac{\lambda}{6}\), the intensity is\(\frac{n}{12}\) K. The value of n is

1. 7
2. 12
3. 6
4. 9

Answer: 4. 9

Intensity, \(I=2 I_0(1+\cos \phi)=2 I_0\left(1+\cos \frac{2 \pi \Delta}{\lambda}\right)\).

For \(\Delta=\lambda, I=4 I_0=k ; \text { for } \Delta=\frac{\lambda}{6} ; \phi=\frac{\pi}{3}\), so

⇒ \(I^{\prime}=2 I_0\left(1+\cos \frac{\pi}{3}\right)=2 I_0 \cdot \frac{3}{2}=3 I_0=\frac{3 k}{4}=9\left(\frac{k}{12}\right)\)

∴ n = 9.

Question 79. Two light waves having the same wavelength λ in vacuum are in phase initially. If the first wave travels a path L₁ through a medium of refractive index μ₁ while the second wave travels a path of length L₂ through a medium of refractive index μ₂ then the phase difference between the two is

1. \(\frac{2 \pi}{\lambda}\left(\frac{L_1}{\mu_1}-\frac{L_2}{\mu_2}\right)\)
2. \(\frac{2 \pi}{\lambda}\left(\mu_1 L_1-\mu_2 L_2\right)\)
3. \(\frac{2 \pi}{\lambda}\left(\frac{L_1}{\mu_2}-\frac{L_2}{\mu_1}\right)\)
4. 0

Answer: 2.

Optical path = μL.

Path difference = Δ = (μ₁L₁ – μ₂L₂).

Phase difference = \(\phi=\frac{2 \pi}{\lambda} \Delta=\frac{2 \pi}{\lambda}\left(\mu_1 L_1-\mu_2 L_2\right)\)

Question 80. In a YDSE, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled then the fringe width becomes

1. Half
2. Four times
3. One fourth
4. Double

Answer: 2. Four times

Fringe width = \(\beta=\frac{D \lambda}{d}\)

Given that d is changed to \(\frac{d}{2}\) and D is changed to 2D

∴ Hence, \(\beta^{\prime}=\frac{D^{\prime} \lambda}{d^{\prime}}=\frac{2 D \lambda}{d / 2}=4 \frac{D \lambda}{d}=4 \beta\)

Question 81. The Brewster angle ip for an interface should be

1. \(30^{\circ}<i_p<45^{\circ}\)
2. \(45^{\circ}<i_{\mathrm{p}}<90^{\circ}\)
3. \(i_p=90^{\circ}\)
4. \(0^{\circ}<i_{\mathrm{p}}<30^{\circ}\)

Answer: 2. \(45^{\circ}<i_{\mathrm{p}}<90^{\circ}\)

Polarization from reflection is

i_p + r = 90° ⇒ = 90°- r.

Thus, i_p <90°.

For i_p = r; 2i_p = 90° ⇒ i_p = 45°

⇒ 45° < i_p <90°.

Question 82. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of a telescope whose objective lens has a diameter of 2 m is

1. 1.83 x l0^-7 rad
2. 7.32 x l0^-7 rad
3. 6.00 x l0^-7 rad
4. 3.66 x l0^-7 rad

Answer: 4. 3.66 x l0^-7 rad

According to the Rayleigh criterion, the angular limit of resolution is

⇒\(\theta=1.22 \frac{\lambda}{D}=1.22 \times \frac{600 \times 10^{-9} \mathrm{~m}}{(2 \mathrm{~m})}\)

= 1.22 x 300 x l0^-9 rad

= 3.66 x l0^-7 rad.

Modem Physics Synopsis

Photon’s energy, \(E=h f=\frac{h c}{\lambda}=\frac{1240 \mathrm{eV} \mathrm{nm}}{\lambda}\)

Photon’s momentum, \(p=\frac{E}{c}=\frac{h}{\lambda}\).

Einstein’s photoelectric equation:

hf = Φ + KE_max,

where f = frequency of the incident light

and Φ = photoelectric work function = \(h f_0=\frac{h c}{\lambda_0}\)

For photoemissions, hf ≥ Φ ⇒ \(\lambda \leq \lambda_0=\frac{h c}{\phi}\)

Stopping potential (V_s): KE_max = eVs => hf = Φ₀ + eV_s.

The de Broglie wavelength of a matter wave is given by

⇒ \(\lambda_{\mathrm{dB}}=\frac{h}{p}=\frac{h}{m v}\)

∵ \(\mathrm{KE}=\frac{p^2}{2 m} \Rightarrow p=\sqrt{2 m E_{\mathrm{k}}}\),

∴ \(\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{2 m E_{\mathrm{k}}}}\)

The wavelength of an electron accelerated by a voltage V: This is given by

∴ \(\lambda=\frac{h}{\sqrt{2 m e V}}=\frac{1.227}{\sqrt{V}} \mathrm{~nm}=\sqrt{\frac{150}{V}}\) Å.

Bohr’s quantum conditions:

Angular momentum, L = mvr_n = \(n\left(\frac{h}{2 \pi}\right)\)

Radius of the nth orbit, r_n = \(\left(\frac{a_0}{Z}\right)\) n²,

where a₀ = Bohr radius = 0.53 Å.

Speed of an electron in the nth orbit, \(v_n=\frac{Z e^2}{2 \varepsilon_0 h} \cdot \frac{1}{n}\)

So, \(v_n \propto \frac{1}{n}\).

KE in the nth orbit, \(\mathrm{KE}_n=\frac{Z e^2}{8 \pi \varepsilon_0 r_n}=Z^2\left(\frac{13.6}{n^2}\right) \mathrm{eV}\).

PE in the nth orbit, \(\mathrm{PE}_n=-\frac{Z e^2}{4 \pi \varepsilon_0 r_n}=-2 \mathrm{KE}_n\).

The total energy in the nth orbit, E_n = KE_n + PE_n

⇒ \(-\frac{\mathrm{Z} e^2}{8 \pi \varepsilon_0 r_n}\)

⇒ \(-\left(\frac{13.6}{n^2}\right) Z^2 \mathrm{eV}\)

Note that | PE| = 2KE and| E_tot| = KE.

Photon energy during emission, hv = E₂-E₁.

image

Photon energy during absorption, hv = E₂-E₁.

image

The energy of the emitted photons, ΔE = E_i-E_f

⇒ \(h v=13.6 \mathrm{Z}^2\left(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{n_{\mathrm{i}}^2}\right)\)

Wave number, \(\frac{1}{\lambda}=\frac{13.6 \mathrm{Z}^2}{h c}\left(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{n_{\mathrm{i}}^2}\right)=R_{\infty} Z^2\left(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{n_{\mathrm{i}}^2}\right)\),

where R_∞ = Rydberg constant = 1.0973 x 10⁷ m^-1

and R_∞hc = 13.6 eV.

Ionization energy = energy required to detach an electron.

∴ \(E_{\text {ion }}=\frac{13.6}{n^2} \mathrm{eV}\)

Ionization potential = \(\frac{13.6}{n^2} \mathrm{~V}\).

Maximum number of possible transitions from the nth state to the ground state, \(N=(n-1)+(n-2)+\ldots+1=\frac{n(n-1)}{2}\).

Constitution of a nucleus: Any nucleus ^AX_Z has the following:

• • Z = number of protons = atomic number
  • A-Z = number of neutrons
  • A- mass number

Isotopes: Nuclei having the same number of protons (Z) but different mass numbers (A) are called isotopes (Example., \({ }_1^2 \mathrm{H},{ }_1^3 \mathrm{H} ;{ }_8^{17} \mathrm{O}, { }_8^{18} \mathrm{O} ;{ }_{92}^{235} \mathrm{U},{ }_{92}^{238} \mathrm{U}\)).

Isobars: Nuclei having the same mass number (A) but different numbers of protons (Z) are called isobars (Example., \({ }_1^3 \mathrm{H},{ }_2^3 \mathrm{He} ;{ }_3^7 \mathrm{Li}, { }_4^7 \mathrm{Be} ;{ }_{18}^{40} \mathrm{Ar},{ }_{20}^{40} \mathrm{Ca}\)).

Isotones: Nuclei having the same number of neutrons (A- Z) called isotones (Example., \({ }_1^3 \mathrm{H},{ }_2^4 \mathrm{He} ;{ }_7^{17} \mathrm{~N},{ }_8^{18} \mathrm{O},{ }_9^{19} \mathrm{~F}\)).

Nuclear radius, R = R₀A^1/3, where R₀ = 1.1 x 10^-15m and A = mass number.

The density of nuclear matter is independent of the mass number A, and its value is around 2.3 x 10¹⁷ kg m^-3.

Atomic mass unit (in short, amu) (symbol: u)

= \(\frac{1}{12} \text { (mass of a }{ }_6^{12} \mathrm{C} \text { atom) }\).

Thus, \(1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{~kg} \approx \frac{931.5 \mathrm{MeV}}{c^2}\).

The mass defect is the amount ΔM by which the mass of an atomic nucleus differs from the sum of the masses of its constituent particles. Thus,

ΔM = Zm_p+(A-Z)m_n-M.

Binding energy = ΔMc².

Q-value of a nuclear process: Q = U_i -U_f = (m_i– m_f )c² = Amc².

Rate of disintegration (decay), \(\frac{d N}{d t}=-\lambda N\).

An instantaneous number of active nuclei, N = N₀e^-λt.

Activity: \(A=\left|\frac{d N}{d t}\right|=\lambda N \text { and } A(t)=A_0 \mathrm{e}^{-\lambda t}\)

Half-life, \(T_{1 / 2}=\frac{1}{\lambda} \ln 2=\frac{0.693}{\lambda}\)

Average life, \(T_{\mathrm{av}}=\frac{1}{\lambda}\)

Number of active (undecayed) nuclei after n half-lives, \(N=N_0\left(\frac{1}{2}\right)^n\).

Photoelectric Effect

Question 1. The number of photoelectrons emitted for a light of frequency v (higher than the threshold frequency v₀) is proportional to

1. The intensity of the light
2. The threshold frequency (v₀)
3. v – v₀ (d) the frequency of the light
4. The frequency of the light

Answer: 1. The intensity of the light

The intensity (I) of light is proportional to the number of photons emitted per second. The greater the intensity, the greater will be the number of photoelectrons emitted (N). Hence, N ∝ I.

Question 2. When a photosensitive surface is illuminated with a radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with a radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelength for the given surface is

1. 5λ
2. 4λ
3. 3λ
4. \(\frac{5 \lambda}{2}\)

Answer: 3. 3λ

According to the equation of the photoelectric effect,

hv = Φ₀ + eV_s.

For the wavelength λ,

⇒ \(\frac{h c}{\lambda}=\phi_0+e V\) → (1)

For the wavelength 2λ,

⇒ \(\frac{h c}{2 \lambda}=\phi_0+\frac{e V}{4}\)

or \(\frac{2 h c}{\lambda}=4 \phi_0+e V\) → (2)

Subtracting (1) from (2),

⇒ \(\frac{h c}{\lambda}=3 \phi_0\)

But \(\phi_0=\frac{h c}{\lambda_0}\), where λ₀ = threshold wavelength.

∴ \(\frac{h c}{\lambda}=3\left(\frac{h c}{\lambda_0}\right) \text { or } \lambda_0=3 \lambda\).

Question 3. The number of ejected photoelectrons increases with an increase in

1. The intensity of light
2. The wavelength of light
3. The frequency of light
4. None of these

Answer: 1. The intensity of light

The emission of photoelectrons takes place due to the absorption of the incident photons. Hence, the number of photoelectrons ejected will be proportional to the intensity of the incident light.

Question 4. The threshold frequency for the photoelectric effect on sodium corresponds to a wavelength of 500 nm. Its work function is

1. 4 x l0^-19 J
2. 3 x 10^-19 J
3. 1J
4. 2 x 10^-19 J

Answer: 1. 4 x l0^-19 J

Photoelectric work function = \(\phi_0=h v_0=\frac{h c}{\lambda_0}\)

⇒ \(\frac{\left(6.67 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{500 \times 10^{-9} \mathrm{~m}}\)

∴ 4 x 10^-19 J.

Question 5. A light of wavelength 500 nm falls on a photosensitive surface with a photoelectric work function of 1.9 eV. The kinetic energy of the fastest photoelectrons emitted will be

1. 0.58 eV
2. 2.48 eV
3. 1.24 eV
4. 1.16 eV

Answer: 1. 0.58 eV

According to Einstein’s photoelectric equation,

hv = Φ₀ + KE_max

⇒ \(\frac{h c}{\lambda}=\phi_0+\frac{1}{2} m v_{\max }^2\)

Substituting the values,

⇒ \(\frac{h c}{500 \mathrm{~nm}}=1.9 \mathrm{eV}+\mathrm{KE}_{\max }\)

∴ \(\mathrm{KE}_{\max }=\frac{1242 \mathrm{eV} \mathrm{nm}}{500 \mathrm{~nm}}-1.9 \mathrm{eV}=2.48 \mathrm{eV}-1.9 \mathrm{eV}=0.58 \mathrm{eV}\).

Question 6. Einstein’s work on the photoelectric effect gives support to

1. E = mc²
2. \(E=\frac{h}{\lambda}\)
3. E = hv
4. \(h v=\frac{1}{2} m v^2\)

Answer: 3. E = hv

Einstein’s work on the photoelectric effect supports the particle nature of light, which consists of photons with the energy E = hv.

Question 7. If the threshold wavelength for a certain metal is 200 nm, the work function of the metal is

1. 6.2 J
2. 6.2 eV
3. 6.2 MeV
4. 6.2 keV

Answer: 2. 6.2 eV

Work function = \(\phi_0=h v_0=\frac{h c}{\lambda_0}\)

Substituting hc =1242 eV nm and λ₀ = 200 nm in the above expression,

∴ \(\phi_0=\frac{1242 \mathrm{eV} \mathrm{nm}}{200 \mathrm{~nm}} \approx 6.2 \mathrm{eV}\).

Question 8. The photoelectric threshold wavelength of silver is 3250 x 10^-10 m. The maximum velocity of the electron ejected from a silver surface by ultraviolet radiation of wavelength 2536 x 10^-10 m is (given that h = 4.14 x 10^-15 eV s and c = 3 x108 m s^-1 )

1. 0.3 x 10⁶ m s^-1
2. 6 x l0⁵ m s^-1
3. 0.6 x 10⁶ m s^-1
4. 61 x 10³ m s^-1

Answer: 2. 6 x l0⁵ m s^-1

Given that threshold wavelength = λ₀ = 3250 Å and wavelength of the incident radiation = λ = 2536 Å.

From the photoelectric equation,

⇒ \(\frac{h c}{\lambda}=\frac{h c}{\lambda_0}+\frac{1}{2} m v_{\max }^2\)

⇒ \(h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)=\frac{1}{2} m v_{\max }^2\)

∴ the maximum velocity of the ejected photoelectron is

⇒ \(v_{\max }=\sqrt{\frac{2 h c}{m}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}\)

⇒ \(\sqrt{\frac{2\left(6.67 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{9.1 \times 10^{-31} \mathrm{~kg}} \times \frac{(3250-2536) \times 10^{10}}{2536 \times 3250 \mathrm{~m}}}\).

≈ 6 x l0⁵ m s^-1.

Question 9. A metallic surface is illuminated with a monochromatic light of wavelength λ, The stopping potential for the photoelectric current for this light is 3V₀. If the same surface is illuminated with a light of wavelength 2λ, the stopping potential is V₀ The threshold wavelength for this photo-sensitive surface is

1. 6λ
2. 4λ
3. \(\frac{\lambda}{4}\)
4. \(\frac{\lambda}{6}\)

Answer: 2. 4λ

For the wavelength A,

⇒ \(\frac{h c}{\lambda}=\frac{h c}{\lambda_0}+e\left(3 V_0\right)\) → (1)

For the wavelength 2λ,

⇒ \(\frac{h c}{2 \lambda}=\frac{h c}{\lambda_0}+e V_0\) → (2)

Multiplying (2) by 3 and then subtracting it from (1), we obtain

∴ threshold wavelength = λ₀= 4λ.

Question 10. Photons of ultraviolet radiation of 6.2 eV fall on an aluminum surface. The kinetic energy of the fastest electron emitted is (given that work function = 4.2 eV)

1. 3.2 x 10^-21 J
2. 3.2 x 10^-19 J
3. 7 x 10^-25J
4. 9 x l0^-32 J

Answer: 2. 3.2 x 10^-19 J

Given that the energy of a photon = hv = 6: 2 eV and Φ₀ = 4.2 eV.

From hv = Φ₀ + KE_max , we have

KE_max = hv – Φ₀ = 6.2 eV – 4.2 eV

= 2.0 eV = 2.0(1.6 x 10^-19 J)

= 3.2 x 10^-19 J.

Question 11. A monochromatic light of frequency 6.0 x 10¹⁴ Hz is produced by a laser beam. The power emitted is 2 x 10^-3 W. The average number of photons emitted by the source per second is

1. 5 x 10¹⁵
2. 5 x l0¹⁶
3. 5 x l0¹⁷
4. 5 x l0¹⁴

Answer: 1. 5 x 10¹⁵

Power of the light source = 2 x 10^-3 W = 2 x 10^-3 J s^-1.

The energy of each photon = hv.

If N is the number of photons emitted per second,

N(hv) = 2 x 10^-3 J

⇒ \(N=\frac{2 \times 10^{-3} \mathrm{~J}}{h \mathrm{v}}=\frac{2 \times 10^{-3} \mathrm{~J}}{\left(6.67 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(6.0 \times 10^{14} \mathrm{~s}^{-1}\right)}=5 \times 10^{15}\).

Question 12. During a photoelectric emission from a metal surface of work function 1.8 eV, the kinetic energy of the most energetic electron is 0.5 eV. The corresponding stopping potential is

1. 1.3 V
2. 2.3 V
3. 0.5 V
4. 1.8 V

Answer: 3. 0.5 V

Given that KE_max = 0.5 eV.

From the work-energy principle,

work done by the stopping voltage = change in KE

⇒ eV_s = KE_max = 0.5 eV

⇒ \(V_{\mathrm{s}}=\frac{0.5 \mathrm{eV}}{e}=0.5 \mathrm{~V}\).

Question 13. A photoelectric emission occurs only when the incident light has more than a certain minimum

1. Power
2. Intensity
3. Frequency
4. Wavelength

Answer: 3. Frequency

For photoelectric emission, the frequency of the incident light must be greater than the threshold (minimum) frequency.

Question 14. Electrons of mass m with a de Broglie wavelength of λ fall on the target in an X-ray tube. The cutoff wavelength λ₀ of the emitted X-rays is

1. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)
2. \(\lambda_0=\frac{2 h}{m c}\)
3. \(\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}\)
4. \(\lambda_0=\lambda\)

Answer: 1. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)

The cutoff wavelength (λ₀) of the continuous X-rays corresponds to the maximum kinetic energy of an electron in the X-ray tube.

For an electron of de Broglie wavelength λ,

⇒ \(\mathrm{KE}=\frac{p^2}{2 m}=\frac{(h / \lambda)^2}{2 m}=\frac{1}{2 m} \cdot \frac{h^2}{\lambda^2}\)

Corresponding to this KE, the maximum energy of each photon is

⇒ \(h v_{\max }=\frac{h c}{\lambda_{\min }}, \text { where } \lambda_{\min }=\lambda_0=\text { cutoff wavelength }\)

∴ \(\frac{h c}{\lambda_0}=\frac{h^2}{2 m \lambda^2} \Rightarrow \lambda_0=\frac{2 m c \lambda^2}{h}\).

Question 15. A monochromatic radiation emitted during the electron transition in hydrogen from the first excited state to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is

1. 25 x 10¹⁵ Hz
2. 4 x 10¹⁵Hz
3. 1.6 x l0¹⁵ Hz
4. 5 x 10¹⁵ Hz

Answer: 3. 1.6 x l0¹⁵ Hz

The energy of a photon emitted from the hydrogen atom is

⇒ \(\Delta E=h v=13.6\left(\frac{1}{1}-\frac{1}{2^2}\right) \mathrm{eV}=\frac{3}{4} \times 13.6 \mathrm{eV}=10.2 \mathrm{eV}\)

Now, hv = Φ₀ + eVs

⇒ 10.2 eV = Φ₀ + 3.57 eV

⇒ Φ₀ = hv₀ = (10.2- 3.57) eV = 6.63 eV.

Hence, the threshold frequency is

∴ \(\mathrm{v}_0=\frac{6.63 \times\left(1.6 \times 10^{-19} \mathrm{~J}\right)}{6.67 \times 10^{-34} \mathrm{Js}}=1.6 \times 10^{15} \mathrm{~Hz}\).

Question 16. When the energy of the incident radiation is increased by 20%, the maximum kinetic energy of the photoelectrons emitted from a metal surface increases from 0.5 eV to 0.8 eV. The work function of the metal is

1. 0.65 eV
2. 1.5 eV
3. 1.3 eV
4. 1.0 eV

Answer: 4. 1.0 eV

Initially, hv = Φ₀ + = E_max + 0.5 eV. → (1)

With 20% increase in hv, we have

⇒ \(h v+\frac{1}{5} h v=\phi_0+0.8 \mathrm{eV}\)

⇒ \(\frac{6}{5} h v=\phi_0+0.8 \mathrm{eV}\)

⇒ \(h v=\frac{5}{6} \phi_0+\frac{2}{3} e V\) → (2)

Equating (1) and (2),

⇒ \(\phi_0+0.5 \mathrm{eV}=\frac{5}{6} \phi_0+\frac{2}{3} \mathrm{eV}\)

∴ work function = Φ₀ = 1.0 eV.

Question 17. A photocell employs the photoelectric effect to convert

1. A change in the frequency of light into a change in the electric voltage
2. A change in the intensity of illumination into a change in the photoelectric current
3. A change in the intensity of illumination into a change in the work function of the photocathode
4. A change in the frequency of light into a change in the electric current

Answer: 2. A change in the intensity of illumination into a change in the photoelectric current

In a photocell, the radiant energy of the incident light produces a stream of electrons constituting an electric current.

The change in the intensity of light changes the photoelectric current.

Question 18. Photons of energy 5 eV are incident on a cathode (C) in a photoelectric cell. The maximum kinetic energy of the emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode (A) if the stopping potential of A relative to C is

1. +4 V
2. +3 V
3. -3 V
4. -1 V

Answer: 3. -3 V

From Einstein’s photoelectric equation,

hv = Φ₀ + KE_max

In the first case,

hv = 5 eV and KE_max = 2 eV.

∴5 eV = Φ₀ + 2eV ⇒ Φ₀ = 3 eV.

In the second case,

hv = 6 eV.

∴ 6 eV = Φ₀ + KE_max

⇒ (6-3) eV = KE_max = (-e)(V_s), where V_s is the stopping potential

⇒ 3eV = e(3V)= -eV_s.

Hence, V_s =-3 V.

Question 19. A photosensitive surface is illuminated successively by monochromatic light waves of wavelengths λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is thrice that in the first case, the work function of the surface of the material is

1. \(\frac{h c}{2 \lambda}\)
2. \(\frac{h c}{\lambda}\)
3. \(\frac{2 h c}{\lambda}\)
4. \(\frac{h c}{3 \lambda}\)

Answer: 1. \(\frac{h c}{2 \lambda}\)

We know that hv = Φ₀ + E₁, where E₁ = maximum KE of the emitted photoelectrons.

In the first case,

⇒ \(\frac{h c}{\lambda}=\phi_0+E_1\) → (1)

and in the second case,

⇒ \(\frac{h c}{\lambda / 2}=\phi_0+E_2\)

Given that E₂ = 3E₁. Hence,

⇒ \(\frac{2 h c}{\lambda}=\phi_0+3 E_1\)

⇒ \(\frac{2}{3} \cdot \frac{h c}{\lambda}=\frac{\phi_0}{3}+E_1\) → (2)

Subtracting (2) from (1),

⇒ \(\frac{h c}{3 \lambda}=\frac{2 \phi_0}{3}\)

Hence, work function = \(\phi_0=\frac{h c}{2 \lambda}\).

Question 20. A monochromatic light of wavelength 667 nm is produced by a helium-neon laser. The power emitted is 9 mW. The average number of photons reaching per second at a target irradiated by the beam is

1. 9 x 10¹⁷
2. 3 x 10¹⁹
3. 3 x l0¹⁶
4. 9 x l0¹⁵

Answer: 3. 3 x l0¹⁶

Given that λ = 667 run and power = P = 9 mW.

Let N = number of photons reaching per second

Power = \(P=\frac{\text { energy }}{\text { time }}=\frac{N(h v)}{1 \mathrm{~s}}\)

⇒ \(9 \times 10^{-3} \mathrm{~J} \mathrm{~s}^{-1}=N\left(\frac{h c}{\lambda}\right) \mathrm{s}^{-1}\)

⇒ \(N=\frac{9 \times 10^{-3} \mathrm{~J}}{h c / \lambda}=\frac{\left(9 \times 10^{-3}\right)\left(667 \times 10^{-9}\right)}{\left(6.67 \times 10^{-34}\right)\left(3 \times 10^8\right)}=3 \times 10^{16}\).

Question 21. When a light beam of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are just emitted. For another emitter, however, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work functions of the two emitters is

1. 1:2
2. 2:1
3. 1:4
4. 4:1

Answer: 2. 2:1

Just for the emission of photoelectrons, its KE is zero, so that

hv = Φ₀ + 0

or \(\frac{h c}{300 \mathrm{~nm}}=\phi_0\) = work function.

Similarly, for the second surface,

⇒ \(\frac{h c}{600 \mathrm{~nm}}=\phi_0^{\prime}\)

∴ \(\frac{\phi_0}{\phi_0^{\prime}}=\frac{\frac{h c}{300 \mathrm{~nm}}}{\frac{h c}{600 \mathrm{~nm}}}=2 \Rightarrow \phi_0: \phi_0^{\prime}=2: 1\).

Question 22. When a light source is at a distance d from a photoelectric cell, the number of photoelectrons emitted from the cell is n. If the distance of the light source from the cell is reduced to d/2, the number of photoelectrons emitted will become

1. \(\frac{n}{2}\)
2. n
3. 2n
4. 4n

Answer: 4. 4n

We know that according to the inverse-square law, the intensity (I) of the incident radiation is inversely proportional to the square of the distance from the surface. Thus,

⇒ \(I \propto \frac{1}{d^2}\)

⇒ \(\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}\) → (1)

The number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. Thus,

⇒ \(\frac{I_1}{I_2}=\frac{n_1}{n_2}\) → (2)

⇒ \(\frac{n_1}{n_2}=\frac{d_2^2}{d_1^2}\)

∴ \(\frac{n}{n_2}=\frac{(d / 2)^2}{d^2}=\frac{1}{4} \Rightarrow n_2=4 n\) [From (1)]

Question 23. A radio transmitter operates at a frequency of 880 kHz and a power of 10 kW. The number of photons emitted per second is

1. 1.327 x l0³⁷
2. 1.327 x l0²⁵
3. 1.7 x l0³¹
4. 1.327 x l0⁴⁵

Answer: 3. 1.7 x l0³¹

Power P = \(\frac{\text { energy }}{\text { time }}=\frac{N(h v)}{1 \mathrm{~s}}\)

⇒ \(N=\frac{P}{h v}=\frac{10 \times 10^3 \mathrm{Js}^{-1}}{\left(6.67 \times 10^{-34} \mathrm{Js}\right)\left(880 \times 10^3 \mathrm{~s}^{-1}\right)}=1.7 \times 10^{31} \mathrm{~s}^{-1}\).

Question 24. A 200-W lamp emits a monochromatic light of wavelength 0.6 nm. Assuming it to be 25% efficient in converting the electrical energy to light, the number of photons of light emitted per second is

1. 1.55 x l0²⁰
2. 3 x l0¹⁹
3. 62 x l0²⁰
4. 6 x l0¹⁸

Answer: 1. 1.55 x l0²⁰

Power = \(\frac{\text { energy }}{\text { time }}=\frac{N}{t}(h v)\),

where hv = energy of each photon and N/t = number of photons emitted per second.

For 25% efficiency,

⇒ \(P^{\prime}=\frac{P}{4}=\frac{200 \mathrm{~W}}{4}=50 \mathrm{~W}\)

⇒ \(50 \mathrm{~W}=\left(\frac{N}{t}\right)\left(\frac{h c}{\lambda}\right)\)

∴ \(\frac{N}{t}=\frac{\left(50 \mathrm{~J} \mathrm{~s}^{-1}\right)\left(0.6 \times 10^{-6} \mathrm{~m}\right)}{\left(6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}=1.55 \times 10^{20} \mathrm{~s}^{-1}\)

Question 25. A photosensitive metallic surface has a work function of hv₀. If photons of energy 2hv₀ fall on this surface, the electrons come out with a maximum velocity of 4 x 10⁶ m s^-1. When the photon energy is increased to 5hv₀, the maximum velocity of the photoelectrons will be

1. 8 x 10⁵ m s^-1
2. 2 x l0⁷ m s^-1
3. 2 x l0⁶ m s^-1
4. 8 x l0⁶ m s^-1

Answer: 4. 8 x l0⁶ m s^-1

We know that hv = \(\phi_0+\frac{1}{2} m v_{\max }^2\).

For the incident photon of energy 2hv₀,

vmax = 4 x 10⁶ m s^-1

⇒ \(2 h v_0=h v_0+\frac{1}{2} m\left(4 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\) → (1)

With photons of energy 5hv₀,

⇒ \(5 h v_0=h v_0+\frac{1}{2} m v_{\max }^2\) → (2)

From (1) and (2),

⇒ \(h v_0=\frac{1}{2} m\left(4 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\) → (3)

and \(4 h v_0=\frac{1}{2} m v_{\max }^2\) → (4)

Now, dividing (3) by (4),

∴ \(\left(\frac{v_{\max }}{4 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}}\right)^2=4 \Rightarrow v_{\max }=8 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\).

Question 26. Light beams of two different frequencies, whose photons have energies of 1 eV and 2.5 eV respectively, illuminate successively a metallic surface whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electrons will be

1. 1:2
2. 1:5
3. 1 :1
4. 1:4

Answer: 1. 1:2

For photons of energy 1 eV, we have

⇒ \(1 \mathrm{eV}=\phi_0+\frac{1}{2} m v_{\max }^2=0.5 \mathrm{eV}+\frac{1}{2} m v_1^2\)

⇒ \(0.5 \mathrm{eV}=\frac{1}{2} m v_1^2\) → (1)

Similarly, for 2.5-eV photons,

⇒ \(2.5 \mathrm{eV}=0.5 \mathrm{eV}+\frac{1}{2} m v_2^2\)

⇒ \(2.0 \mathrm{eV}=\frac{1}{2} m v_2^2\) → (2)

Dividing (2) by (1)

∴ \(\frac{v_2^2}{v_1^2}=\frac{2}{0.5}=4 \Rightarrow \frac{v_1}{v_2}=\frac{1}{2} \Rightarrow v_1: v_2=1: 2\).

Question 27. When ultraviolet rays are incident on a metal surface, photoelectrons are not emitted. The emission of electrons will occur by the incidence of

1. Infrared rays
2. X-rays
3. Radio waves
4. Lightwaves

Answer: 2. X-rays

For the emission of photoelectrons, the frequency of the incident radiations must be greater than the threshold frequency. No photoemission by the UV rays means that the frequency of the UV rays is less than the threshold frequency. Hence, photoemission will occur when the frequency is greater than that of the ultraviolet rays, which is true only for X-rays.

Question 28. The energy of a photon of a light beam is 3 eV. Then, the wavelength of the photon must be

1. 4125 nm
2. 414 nm
3. 41250 nm
4. 4nm

Answer: 2. 414 nm

The energy of a photon is

⇒ \(h v=\frac{h c}{\lambda}=3 \mathrm{eV}\)

∴ Wavelength = \(\lambda=\frac{h c}{3 \mathrm{eV}}=\frac{1242 \mathrm{eV} \mathrm{nm}}{3 \mathrm{eV}}=414 \mathrm{~nm}\)

Question 29. In a photoemissive cell with an exciting wavelength of λ, the fastest electron has a speed of v. If the exciting wavelength is reduced to 3λ/4, the speed of the fastest electron will be

1. \(\left(\frac{3}{4}\right)^{-1 / 2}\).v
2. \(\left(\frac{4}{3}\right)^{1 / 2}\).v
3. Less than \(\left(\frac{4}{3}\right)^{1 / 2}\).v
4. Greater than \(\left(\frac{4}{3}\right)^{1 / 2}\).v

Answer: 4. Greater than \(\left(\frac{4}{3}\right)^{1 / 2}\).v

For photoemission,

⇒ \(\frac{h c}{\lambda}=h v=\phi_0+\frac{1}{2} m v_{\max }^2=\phi_0+\frac{1}{2} m v^2\).

Since the exciting wavelength is reduced to 3λ/4, let the maximum speed of an electron be vmax

∴ \(\frac{h c}{3 \lambda / 4}=\phi_0+\frac{1}{2} m v_{\max }^{\prime 2}\)

⇒ \(\frac{4 h c}{3 \lambda}=\phi_0+\frac{1}{2} m v_{\max }^{\prime 2}\)

∴ \(\frac{\frac{1}{2} m v_{\max }^{\prime 2}}{\frac{1}{2} m v^2}=\frac{\frac{4}{3} \cdot \frac{h c}{\lambda}-\phi_0}{\frac{h c}{\lambda}-\phi_0}=\frac{\frac{4}{3}\left(\frac{h c}{\lambda}-\phi_0\right)+\frac{\phi_0}{3}}{\frac{h c}{\lambda}-\phi_0}\)

⇒ \(\frac{4}{3}+\frac{\phi_0}{3\left(\frac{h c}{\lambda}-\phi_0\right)}=\frac{4}{3}+(\text { a positive constant })\)

⇒ \(\frac{v_{\max }^{\prime}}{v}>\sqrt{\frac{4}{3}}\)

∴ \(v_{\max }^{\prime}>\left(\frac{4}{3}\right)^{1 / 2} \cdot v\).

Question 30. The work function of the surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the

1. Ultraviolet region
2. Visible region
3. Infrared region
4. X-ray region

Answer: 1. Ultraviolet region

Given that work function = Φ₀ = 6.2 eV.

If stopping potential = V_s = 5V, from the photoelectric equation we have

⇒ \(h v=\phi_0+\frac{1}{2} m v_{\max }^2\)

⇒\(\frac{h c}{\lambda}=\phi_0+e V_s\)

∴ Wave length = \(\lambda=\frac{h c}{\phi_0+e V_s}=\frac{1242 \mathrm{eV} \mathrm{nm}}{6.2 \mathrm{eV}+5 \mathrm{eV}}=\frac{1242}{11.2} \mathrm{~nm}=110.9 \mathrm{~nm}\).

The wavelength range for UV rays is 10-310 nm.

Hence, the required wavelength lies in the ultraviolet region

Question 31. For the photoelectric emission from a certain metal, the cutoff frequency is v. If radiations of frequency 2v be incident on the metal plate, the maximum possible velocity of the emitted electron will be (given that m_e = mass of an electron)

1. \(\sqrt{\frac{h v}{2 m_{\mathrm{e}}}}\)
2. \(\sqrt{\frac{h v}{m_{\mathrm{e}}}}\)
3. \(\sqrt{\frac{2 h v}{m_{\mathrm{e}}}}\)
4. \(2 \sqrt{\frac{h v}{m_e}}\)

Answer: 3. \(\sqrt{\frac{2 h v}{m_{\mathrm{e}}}}\)

Given that cutoff frequency = v₀ = v and frequency of the incident radiation = 2v.

⇒ \(h v=\phi_0+\mathrm{KE}_{\max }=\phi_0+\frac{1}{2} m_e v_{\max }^2\)

⇒ \(h(2 v)=h v+\frac{1}{2} m_e v_{\max }^2\)

∴ \(v_{\max }=\sqrt{\frac{2 h v}{m_e}}\)

Question 32. When the intensity of an incident light beam increases, the

1. Photocurrent increases
2. Photocurrent decreases
3. The kinetic energy of the emitted photoelectrons increases
4. The kinetic energy of the emitted photoelectrons decreases

Answer: 1. Photocurrent increases

The maximum KE of photoelectrons depends on the frequency of the incident radiation and not on the intensity of the incident light. However, the rate of emission of electrons (= photocurrent) depends on its intensity. An increase in the intensity increases the photocurrent.

Question 33. The figure shows the plots of the photocurrent versus the anode potential for a photosensitive surface for three different radiations. Which of the following statements is true?

1. The curves a and b represent the incident radiations of different frequencies and different intensities.
2. The curves a and b represent the incident radiations of the same frequency but of different intensities.
3. The curves b and c represent the incident radiations of different frequencies and different intensities.
4. The curves b and c represent the incident radiations of the same frequency and the same intensity.

Answer: 2. The curves a and b represent the incident radiations of the same frequency but of different intensities.

The curves a and b have the same stopping potential V₁. Hence, the incident radiations have the same frequency but different intensities. The curves b and c have the same intensity as the saturation current for both the stopping potentials. V₁ and V₂ are different Hence, the frequency of the incident radiations must be different.

Question 34. The work functions for the metals A, B, and C are respectively 1.92 eV, 2.0 eV, and 5eV. According to Einstein’s equation, the metal(s) which will emit photoelectrons for a radiation of wavelength 4100 A is/are

1. A and B only
2. All the three
3. An only
4. None of these

Answer: 1. A and B only

The work function of a photosensitive surface is given by

⇒ \(\phi_0=h v_0=\frac{h c}{\lambda_0}\)

∴ \(\lambda_0=\frac{h c}{\phi_0}\)

For photoeniission, λ ≤ λ_0.

For A, \(\lambda_0=\frac{1242 \mathrm{eV} \mathrm{nm}}{1.92 \mathrm{e} V}=646.8 \mathrm{~nm}=6468\) Å

For B, \(\lambda_0=\frac{1242 \mathrm{eV} \mathrm{nm}}{2 \mathrm{eV}}=621 \mathrm{~nm}=6210\) Å

For C ,\(\lambda_0=\frac{1242 \mathrm{eV} \mathrm{nm}}{5 \mathrm{eV}}=248.4 \mathrm{~nm}=2484\) Å

Since the given value of λ (= 4100 Å) is less than that of A and B, the emission will be possible only for A and B.

Question 35. The photoelectric work function for a metal is 4.125 eV. The cut-off wavelength for this surface is

1. 412.5 nm
2. 301 nm
3. 600 nm
4. 206.2 nm

Answer: 2. 301 nm

Work function = \(\phi_0=\frac{h c}{\lambda_0}=4.125 \mathrm{eV}\)

∴ cutoff wavelength = \(\lambda_0=\frac{1242 \mathrm{eV} \mathrm{nm}}{4.125 \mathrm{eV}} \approx 301 \mathrm{~nm}\)

Question 36. In the photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of -1.2 V. Then,

1. The energy of the incident photons is 4.7 eV
2. The energy of the incident photons is 2.3 eV
3. If higher-frequency photons are used, the photoelectric current will rise
4. In case the energy of photons is 3.5 eV, the photoelectric current will be the maximum

Answer: 1. The energy of the incident photons is 4.7 eV

Given that work function = Φ₀ = 3.5 eV

and stopping potential = V_s = -1.2 V.

∴ KEmax=(-e)(-1.2V) =1.2eV.

From the photoelectric equation, the energy of the incident photon is

∴ hv = Φ₀ + KE_max = 3.5 eV + 1.2 eV = 4.7 eV.

Question 37. A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m,

1. Each emitted electron carries half the maximum initial energy
2. The number of electrons emitted is a quarter of the initial number
3. Each emitted electron carries one-quarter of the initial maximum kinetic energy
4. The number of electrons emitted is half the initial number

Answer: 2. The number of electrons emitted is a quarter of the initial number

Varying the distance of the light source from a photoelectric cell will change the intensity but not the frequency. Hence, the KE of the photoelectrons will not change but their number will change according to the inverse-square law. Hence,

⇒ \(N \propto I \propto \frac{1}{d^2}\)

⇒ nd² = n’d^2₂ => n(1 m)² = n'(2 m)²

⇒ \(n^{\prime}=\frac{n}{4}\)

Question 38. According to Einstein’s photoelectric equation, the graph between the maximum kinetic energy of the photoelectrons ejected and the frequency of the incident radiation is

Answer: 2.

According to Einstein’s-photoelectric equation,

hv = Φ₀ + KE_max

KE_max = hV – Φ_0.

This represents a straight-line graph of the form y = mx-c, where the slope is h and the intercept on the x-axis is v₀ (the threshold frequency) as in option (2).

Question 39. A beam of light is incident normally on a completely absorbing surface with an energy flux of 25 W cm^-2. If the surface area by 25 cm², the momentum transferred to the surface in a duration of 40 minutes will be

1. 6.3 x 10^-4 Ns
2. 3.5 x 10^-6 N s
3. 5.0 x 10^-3 N s
4. 1.4 x l0^-6 N s

Answer: 3. 5.0 x 10^-3 N s

The intensity of the incident light = I = 25 W cm^-2.

The rate at which energy is absorbed by the surface (of area = A = 25 cm^-2) will be 625 W.

∴ momentum transferred = \(\Delta p=\frac{E}{c}=\frac{(625 \mathrm{~W})(40 \times 60 \mathrm{~s})}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}=5 \times 10^{-3} \mathrm{~N} \mathrm{~s}^2\).

Question 40. A 2-mW laser operates at a wavelength of 500 nm. The number of photons emitted per second is

1. 5 x 10¹⁵
2. 2 x 10¹⁶
3. l x l0¹⁶
4. 1.5 x l0¹⁶

Answer: 1. 5 x 10¹⁵

The power of the laser is

P = 2 mW = 2 x l0^-3 J s^-1.

This is the energy emitted per second associated with N photons each of energy hv.

∴ \(N\left(\frac{h c}{\lambda}\right)=2 \times 10^{-3} \mathrm{~J}\)

Hence, \(N=\frac{\left(2 \times 10^{-3} \mathrm{~J}\right)\left(500 \times 10^{-9} \mathrm{~m}\right)}{\left(6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}=5 \times 10^{15}\)

Question 41. The magnetic field associated with an electromagnetic wave at the origin is given by B = \(B_0\left[\sin \left(3.14 \times 10^7\right) c t+\sin \left(6.28 \times 10^7\right) c t\right]\). If the wave is incident on a silver plate having a photoelectric work function of 4.7 eV, what will be the maximum kinetic energy of the emitted photoelectrons?

1. 7.7 eV
2. 6.82 eV
3. 8.52 eV
4. 12.5 eV

Answer: 1. 7.7 eV

Angular frequency = ω = (6.28 x10⁷ )c

⇒ 2πv = 2π x 10⁷ x (3 x10⁸) Hz ⇒ v = 3 x l0¹⁵Hz.

From the photoelectric equation,

hv = Φ₀ + KE_max

⇒ KE_max = hv – Φ = (6.6 x10^-34 J s)(3 x10¹⁵ s^-1) – 4.7 eV

⇒ \(\frac{6.6 \times 3 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}-4.7 \mathrm{eV}=7.7 \mathrm{eV}\)

Question 42. In a photoelectric-effect experiment, the threshold wavelength of the photosensitive surface is 380 nm. If the wavelength of the incident light is 260 nm, the maximum kinetic energy of an emitted photoelectron will be [given that E (in eV) = 1237/λ, (in nm)]

1. 1.5 eV
2. 15.0 eV
3. 3.0 eV
4. 4.5 eV

Answer: 1. 1.5 eV

From the equation,

hv = hv₀ + KE_max

⇒ \(\frac{h c}{\lambda}=\frac{h c}{\lambda_0}+\mathrm{KE}_{\max }\)

⇒ \(\mathrm{KE}_{\max }=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)=(1237 \mathrm{eV} \mathrm{nm})\left(\frac{1}{260 \mathrm{~nm}}-\frac{1}{380 \mathrm{~nm}}\right)\)

∴ \((1237 \mathrm{eV})\left(\frac{120}{260 \times 380}\right)=1.5 \mathrm{eV}\)

Question 43. The stopping potential corresponding to the incident radiation of wavelength λ, is V. If the wavelength of the incident light on the same photosensitive surface is increased to 3λ, the stopping potential becomes V/4. If the threshold wavelength be kλ then k is

1. 6
2. 3
3. 9
4. 2

Answer: 3. 9

According to Einstein’s photoelectric equation,

⇒ \(\frac{h c}{\lambda}=\phi+e V \text { and } \frac{h c}{3 \lambda}=\phi+\frac{e V}{4}\)

Eliminating V,

⇒ \(\frac{h c}{12 \lambda}=\frac{3}{4} \phi=\frac{3}{4}\left(\frac{h c}{\lambda_0}\right)\)

∴ threshold wavelength = X0 = 9X.

Hence, k = 9.

Question 44. The kinetic energy of the most energetic photoelectrons increases from K to 3K when the wavelength of the incident radiation changes from 500 nm to 200 nm. The photoelectric work function of the metallic surface is

1. 0.50 eV
2. 0.25 eV
3. 0.62 eV
4. 0.70 eV

Answer: 3. 0.62 eV

Einstein’s photoelectric equation is

⇒ \(\frac{h c}{\lambda}=\phi-\mathrm{KE}_{\max }\)

For λ = λ₁ = 500 nm,

⇒ \(\frac{h c}{500 \mathrm{~nm}}=\phi+K\)

For λ = λ₂ = 200 nm,

⇒ \(\frac{h c}{200 \mathrm{~nm}}=\phi+3 K\)

⇒ \(2 \phi=h c\left(\frac{3}{500 \mathrm{~nm}}-\frac{1}{200 \mathrm{~nm}}\right)=\frac{1240 \mathrm{eV} \mathrm{nm}}{1000 \mathrm{~nm}}=1.24 \mathrm{eV}\).

Hence, work function = Φ = 0.62 eV.

Question 45. The given graph shows the variation of the stopping potential (V_s) with the frequency (f) of the incident radiation. The photoelectric work function of the metallic surface is

1. 4.01 eV
2. 2.07 eV
3. 5.01 eV
4. 3.01 eV

Answer: 2. 2.07 eV

hf = Φ + eV_s,

∴ \(V_{\mathrm{s}}=\left(\frac{h}{e}\right) f-\frac{\phi}{e}\)

This represents the given straight-line graph. When V_s = 0,

⇒ \(\left(\frac{h}{e}\right) f-\frac{\phi}{e}=0\)

Hence, the work function is

∴ \(\phi=h f=\left(\frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} \mathrm{eV} \mathrm{s}\right)\left(5 \times 10^{14} \mathrm{~s}^{-1}\right)=2.07 \mathrm{eV}\)

Question 46. In the case of the photoelectric effect, the graph showing the variation of the stopping potential (V_s) against the reciprocal of the wavelength (1/λ) of the incident radiation is given. What will happen to the graph on increasing the intensity of light?

1. The graph will not change.
2. The intercept on the y-axis will change.
3. The slope of the graph will increase.
4. The graph will shift to the right remaining parallel to the given graph.

Answer: 1. The graph will not change.

The stopping potential depends on the frequency (and wavelength) of the incident radiation and is independent of the intensity.

Hence, the V_s – λ^-1 graph does not change.

Question 47. In a photoelectric experiment, the potential difference between the plates increases while keeping the incident light on the photocathode unchanged. Which of the following is the correct statement about the saturation current?

1. It increases.
2. It decreases.
3. It remains unchanged.
4. It first increases and then decreases.

Answer: 3. It remains unchanged.

The saturation current depends on the intensity of the incident light only. An increase in the accelerating voltage will not affect the saturation current and it will remain unchanged.

Question 48. Two monochromatic sources emitting photons of λ₁ = 500 nm and λ₂ =1 nm have the same power of 200 W. What is the ratio of the densities of the photons emitted from the two sources?

1. 200
2. 500
3. 300
4. 0.8

Answer: 2. 500

Power is defined as the energy radiated per unit of time.

nhf = \(n\left(\frac{h c}{\lambda}\right)\) where n is the number of photons emitted per unit time.

∴ photon density = n = \(\frac{P \lambda}{h c} \propto \lambda\)

∴ \(\frac{n_1}{n_2}=\frac{\lambda_1}{\lambda_2}=\frac{500 \mathrm{~nm}}{1 \mathrm{~nm}}=500\).

Question 49. A beam of light with an average flux of 20 W cm^-2 falls normally on a non-reflecting surface having a total surface area of 20 cm². The energy received by the surface during a time span of 1 min is

1. 12 kJ
2. 24 kJ
3. 48 kJ
4. 10 kJ

Answer: 2. 24 kJ

Average flux = 20 J s^-1 cm^-2.

∴ the energy received in1 min on 20 cm² will be

(20 J s^-1 cm^-2 X 60 s)(20 cm²) = 20 x 60 x 20 J = 24 kJ.

Question 50. A light beam of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and the intensity is doubled?

1. Four times
2. One fourth
3. Zero
4. Doubled

Answer: 3. Zero

Let the threshold frequency be v₀.

Initially, frequency = v = 1.5v₀.

Finally, when the frequency is halved, it becomes

⇒ \(v^{\prime}=\frac{v}{2}=\frac{1.5 v_0}{2}=0.75 v_0<v_0\)

For frequencies v’ less than the threshold value, no photoemission is possible, so the photocurrent will be zero.

Matter Waves

Question 1. An electron beam has a kinetic energy equal to 100 eV. The wavelength associated with the beam is

1. 24.6 Å
2. 0.12 Å
3. 1.2 Å
4. 6.3 Å

Answer: 3. 1.2 Å

The kinetic energy of an electron is E = \(E=\frac{p^2}{2 m}\)

∴ its momentum = p = \(\sqrt{2 m E}\)

The associated wavelength is

⇒ \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

Substituting the values,

∴ \(\lambda=\frac{6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\sqrt{2\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(100 \times 1.6 \times 10^{-19} \mathrm{~J}\right)}}=1.2 \times 10^{-10} \mathrm{~m}=1.2\) A.

Alternative method

KE =100 eV means accelerating voltage = V =100 V, i.e.,

∴ \(\lambda=\frac{1.2 \mathrm{~nm}}{\sqrt{V}}=1.2\) A.

Question 2. Radiant energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (given that c = velocity of light)

1. \(\frac{E}{c}\)
2. \(\frac{2 E}{c^2}\)
3. \(\frac{E}{c^2}\)
4. \(\frac{2 E}{c}\)

Answer: 4. \(\frac{2 E}{c}\)

Energy of a photon = E = hv = \(\frac{h c}{\lambda}\) and de Broglie wavelength = \(\lambda=\frac{h}{p}\).

∴ momentum = \(p=\frac{h}{\lambda}=\frac{h c}{\lambda c}=\frac{E}{c}\)

The momentum transferred to the surface is the change in momentum, i.e.,

⇒ \(\overrightarrow{\Delta p}=\vec{p}-(-\vec{p})=2 \vec{p}\)

∴ \(|\overrightarrow{\Delta p}|=2 p=\frac{2 E}{c}\)

Question 3. The wavelength associated with an electron accelerated through a potential difference of 100 V is of the order

1. 1000 Å
2. 100 AÅ
3. 10.5 Å
4. 1.2 Å

Answer: 4. 1.2 Å

The KE gained by the electron is

⇒ \(E=\frac{p^2}{2 m}=e(100 \mathrm{~V})=100 \cdot \mathrm{eV}=100\left(1.6 \times 10^{-19} \mathrm{~J}\right)\)

∴ associated wavelength = \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

Substituting the values,

∴ \(\lambda=\frac{6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\sqrt{2\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(1.6 \times 10^{-17} \mathrm{~J}\right)}}=1.2 \times 10^{-10} \mathrm{~m}=1.2\) A

Alternative method

∴ \(\lambda=\frac{1.227 \mathrm{~nm}}{\sqrt{V}}=\frac{1.227 \mathrm{~nm}}{\sqrt{100}}=1.2\) A.

Question 4. A particle of mass 1 mg has the same wavelength as that of an electron moving with a velocity of 3 x 10⁶ m s-1. The velocity of the particle is (given that the mass of an electron = 9.1 x 10^-31 kg)

1. 2.7 x10^-18 m s^-1
2. 2.7 x10^-21 m s^-1
3. 3 xl0^-31 m s^-1
4. 9 x10^-2 m s^-1

Answer: 1. 2.7 x10^-18 m s^-1

The wavelength associated with the moving particle of mass m is

⇒ \(\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h}{(1 \mathrm{mg}) v}\)

The wavelength of an electron is

⇒ \(\lambda_{\mathrm{e}}=\frac{h}{m_{\mathrm{e}} v_{\mathrm{e}}}=\frac{h}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(3 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\right)}\)

Given that λ = λ_e. Thus

(1 x 10^-6 kg)v = (9.1 x 10^-31 kg)(3 x 10⁶ m s^-1)

⇒ v = 27.3 x 10^-19 m s^-1 = 2.7 x 10¹⁸ m s^-1

Question 5. Electrons, each of mass m and having the de Broglie wavelength λ, fall on the target in an X-ray tube. The cutoff wavelength of the emitted X-rays is

1. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)
2. \(\lambda_0=\frac{2 h}{m c}\)
3. \(\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}\)
4. \(\lambda_0=\lambda\)

Answer: 1. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)

The kinetic energy of the electron incident on the target is

⇒ \(E=\frac{p^2}{2 m}=\frac{1}{2 m}\left(\frac{h}{\lambda}\right)^2=\frac{h^2}{2 m \lambda^2}\)

Its total absorption produces X-ray photons of the maximum energy. Hence,

⇒ \(E=h v_{\max }=\frac{h c}{\lambda_{\min }}=\frac{h c}{\lambda_0}\)

∴ \(\frac{h^2}{2 m \lambda^2}=\frac{h c}{\lambda_0} \Rightarrow \lambda_0=\frac{2 m c \lambda^2}{h}\)

Question 6. Which of the following graphs correctly represents the variation of a particle’s momentum (p) with the associated de Broglie wavelength?

Answer: 4.

The de Broglie wavelength is λ = λ/p, where p is the momentum.

Thus, pλ = constant.

Like Boyle’s law(pV = constant), the p-λ graph represents a rectangular hyperbola, as depicted in (d).

Question 7. If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is

1. 25%
2. 75%
3. 60%
4. 50%

Answer: 2. 75%

Initial wavelength = \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

and final wavelength = A.’ = \(\frac{h}{p^{\prime}}=\frac{h}{\sqrt{(2 m)(16 E)}}=\frac{\lambda}{4}\)

∴ % change in wavelength = \(\frac{\lambda-\lambda^{\prime}}{\lambda} \times 100 \%\)

∴ \(\left(1-\frac{1}{4}\right) \times 100 \%=75 \%\)

Question 8. The wavelengths λ_e of an electron and λ_ph of a photon of the same energy E are related by

1. \(\lambda_{\mathrm{ph}} \propto \sqrt{\lambda_{\mathrm{e}}}\)
2. \(\lambda_{\mathrm{ph}} \propto \frac{1}{\sqrt{\lambda_{\mathrm{e}}}}\)
3. \(\lambda_{\mathrm{ph}} \propto \lambda_{\mathrm{e}}^2\)
4. \(\lambda_{\mathrm{ph}} \propto \lambda_{\mathrm{e}}\)

Answer: 3. \(\lambda_{\mathrm{ph}} \propto \lambda_{\mathrm{e}}^2\)

The wavelength associated with an electron is \(\lambda_{\mathrm{e}}=\frac{h}{\sqrt{2 m_{\mathrm{e}} E_{\mathrm{e}}}}\)

∴ KE of an electron = \(E_{\mathrm{e}}=\frac{h^2}{2 m_{\mathrm{e}} \lambda_{\mathrm{e}}^2}\) → (1)

Energy of a photon = \(E_{\mathrm{ph}}=h v=\frac{h c}{\lambda_{\mathrm{ph}}}\) → (2)

Since E_e = E_ph (given), then, equating (1) and (2), we have

⇒ \(\frac{h c}{\lambda_{\mathrm{ph}}}=\frac{h^2}{2 m_{\mathrm{e}} \lambda_{\mathrm{e}}^2}\)

⇒ \(\lambda_{\mathrm{ph}}=\left(\frac{2 m_{\mathrm{e}} c}{h}\right) \lambda_{\mathrm{e}}^2\)

∴ \(\lambda_{\mathrm{ph}} \propto \lambda_{\mathrm{e}}^2\)

Question 9. The velocity of a particle P is thrice that of an electron, whereas the ratio of the de Broglie wavelength of the particle to that of an electron is 1.824 x 10^-4. Then, the particle will be

1. A neutron
2. Adeuteron
3. An alpha particle
4. A tritium nucleus

Answer: 1. A neutron

For the given, particle, Up = 3ve and wavelength = \(\lambda_P=\left(1.824 \times 10^{-4}\right) \lambda_e\)

∴ \(\frac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{e}}}=\frac{h / m_{\mathrm{P}} v_{\mathrm{P}}}{h / m_{\mathrm{e}} v_e}=\left(\frac{m_{\mathrm{e}}}{m_{\mathrm{P}}}\right)\left(\frac{v_{\mathrm{e}}}{v_{\mathrm{P}}}\right)\)

⇒ \(1.824 \times 10^{-4}=\left(\frac{m_{\mathrm{e}}}{m_{\mathrm{P}}}\right)\left(\frac{1}{3}\right)\)

∴ the mass of the particle is

⇒ \(m_{\mathrm{P}}=\frac{m_{\mathrm{e}}}{3\left(1.824 \times 10^{-4}\right)}=\frac{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times 10^4}{3 \times 1.824}\)

⇒ \(1.663 \times 10^{-27} \mathrm{~kg} \approx 1.67 \times 10^{-27} \mathrm{~kg}\)

This mass corresponds to that of a neutron.

Question 10. An electron of mass m and a photon have the same energy E. The ratio of the de Broglie wavelengths associated with them is (c being the velocity of light)

1. \(c(2 m E)^{1 / 2}\)
2. \(\frac{1}{c}\left(\frac{2 m}{E}\right)^{1 / 2}\)
3. \(\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}\)
4. \(\left(\frac{E}{2 m}\right)^{1 / 2}\)

Answer: 3. \(\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}\)

For an electron, wavelength = \(\lambda_{\mathrm{e}}=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

For a photon,

⇒ \(E=h v=\frac{h c}{\lambda} \Rightarrow \lambda_{\mathrm{ph}}=\frac{h c}{E}\)

∴ \(\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{ph}}}=\frac{h}{\sqrt{2 m E}} \times \frac{E}{h c}=\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}\)

Question 11. A light beam of wavelength 500 run is incident on a metal with a work function of 2.28 eV. The de Broglie wavelength of the emitted electron is

1. λ ≥ 2.8 x 10^-9 m
2. λ ≤ 2.8 x l0^-12 m
3. λ < 2.8 x l0^-10 m
4. λ < 2.8 x 10^-9 m

Answer: 1. λ ≥ 2.8 x 10^-9 m

According to Einstein’s photoelectric equation,

⇒ \(h v=\phi_0+\mathrm{KE}_{\max } \Rightarrow \frac{h c}{\lambda}=\phi_0+\frac{p_{\max }^2}{2 m_{\mathrm{e}}}\)

Substituting the values,

⇒ \(\frac{1242 \mathrm{eV} \mathrm{nm}}{500 \times 10^{-9} \mathrm{~m}}=2.28 \mathrm{eV}+\frac{p_{\max }^2}{2 m_{\mathrm{e}}}\)

⇒ \(p_{\max }^2=(2.484 \mathrm{eV}-2.28 \mathrm{eV})\left(2 m_{\mathrm{e}}\right)\)

⇒ \(p_{\max }=\sqrt{\left(0.204 \times 1.6 \times 10^{-19} \mathrm{~J}\right)\left(2 \times 9.1 \times 10^{-31} \mathrm{~kg}\right)}\)

= 2.4 x 10^-25 kg m s^-1.

∴ the associated wavelength is

⇒ \(\lambda_{\min }=\frac{h}{p_{\max }}=\frac{6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{2.4 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}}=2.75 \times 10^{-9} \mathrm{~m}\)

∴ λ > 2.8 x 10^-9 m

Question 12. The momentum of a photon of energy 1 MeV will be

1. 5 x 10^-22 kg m s^-1
2. 0.33 x l0⁶ kg ms^-1
3. 7 x 10^-24 kg m s^-1
4. 10^-22 kg m s^-1

Answer: 1. 5 x 10^-22 kg m s^-1

The momentum of a photon is

⇒ \(p=\frac{E}{c}=\frac{1 \mathrm{MeV}}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}=\frac{\left(1 \times 10^6\right)\left(1.6 \times 10^{-19} \mathrm{~J}\right)}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}\)

∴  5 x 10^-22 kg m s^-1

Question 13. An electron of mass me, when accelerated through a potential difference V, has a de Broglie wavelength of λ_e. The de Broglie wavelength associated with a proton of mass mp accelerated through the same potential difference will be

1. \(\lambda_{\mathrm{e}}\left(\frac{m_{\mathrm{p}}}{m_{\mathrm{e}}}\right)\)
2. \(\lambda_{\mathrm{e}}\left(\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}\right)\)
3. \(\lambda_{\mathrm{e}} \sqrt{\frac{m_{\mathrm{p}}}{m_{\mathrm{e}}}}\)
4. \(\lambda_{\mathrm{e}} \sqrt{\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}}\)

Answer: 4. \(\lambda_{\mathrm{e}} \sqrt{\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}}\)

KE gained = \(e V=\frac{p^2}{2 m}\) and momentum = p = \(\sqrt{2 m e V}\)

For an electron, \(\lambda_{\mathrm{e}}=\frac{h}{p_{\mathrm{e}}}=\frac{h}{\sqrt{2 m_{\mathrm{e}} e V}}\)

For a proton, \(\lambda_{\mathrm{p}}=\frac{h}{p_{\mathrm{p}}}=\frac{h}{\sqrt{2 m_{\mathrm{p}} e V}}\)

∴ \(\frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}}=\sqrt{\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}} \Rightarrow \lambda_{\mathrm{p}}=\lambda_{\mathrm{e}} \sqrt{\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}}\)

Question 14. What is the kinetic energy of an electron associated with a de Broglie wavelength of 1 nm?

1. 1.5 eV
2. 4.2 eV
3. 2.1 eV
4. 3.1 eV

Answer: 1. 1.5 eV

The wavelength of an electron accelerated through a potential difference of V volts is given by \(\lambda=\frac{1.227}{\sqrt{V}}\).

For λ =1 nm, √V =1.227 ⇒ V =1.5.

∴ kinetic energy = e(1.5 V) =1.5 eV

Question 15. If the following particles move with the same speed, which has the maximum de Broglie wavelength?

1. A proton
2. An α-particle
3. A β-particle
4. A neutron

Answer: 3. A β-particle

The de Broglie wavelength is given by

⇒ \(\lambda=\frac{h}{m v}\)

∴ \(\lambda \propto \frac{1}{m}\)

We know that m_∝ > m_n > m_p > m_e.

The particle with the least mass has the maximum value of the wavelength λ. Thus, β-particles have the maximum de Broglie wavelength.

Question 16. The de Broglie wavelength of a neutron of mass m in thermal equilibrium with heavy water at a temperature T is

1. \(\frac{h}{\sqrt{3 m k T}}\)
2. \(\frac{2}{\sqrt{3 m k T}}\)
3. \(\frac{2 h}{\sqrt{m k T}}\)
4. \(\frac{h}{\sqrt{m k T}}\)

Answer: 1. \(\frac{h}{\sqrt{3 m k T}}\)

A neutron has 3 degrees of freedom.

At die temperature T, its energy is

⇒ \(E=3\left(\frac{1}{2} k T\right)=\frac{3}{2} k T\)

But \(\mathrm{KE}=E=\frac{p^2}{2 m}\)

∴ \(p=\sqrt{2 m E}=\sqrt{2 m\left(\frac{3}{2} k T\right)}\)

⇒ \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{3 m k T}}\)

Question 17. The momentum of a photon of a visible light beam of wavelength 500 nm is about

1. 1.33 x 10^-27 kg m s^-1
2. 2.46 x l0^-27 kg m s^-1
3. 1.84 x l0^-27 kg m s^-1
4. 3.67 x l0^-27 kg m s^-1

Answer: 1. 1.33 x 10^-27 kg m s^-1

The momentum of a photon is

⇒ \(p=\frac{E}{c}=\frac{h c / \lambda}{c}=\frac{h}{\lambda}\)

∴ \(p=\frac{6.67 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{500 \mathrm{~nm}}=1.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\).

Question 18. A monochromatic source of light having a power of 200 W emits 4 x 10²⁰ photons per second. The wavelength of the light is

1. 200 nm
2. 800 run
3. 400 nm
4. 600 nm

Answer: 3. 400 nm

Energy of each photon = \(\frac{200 \mathrm{~J} \mathrm{~s}^{-1}}{4 \times 10^{20} \mathrm{~s}^{-1}}=5 \times 10^{-19} \mathrm{~J}\)

The corresponding wavelength is

∴ \(\lambda=\frac{h}{p}=\frac{h}{E / c}=\frac{h c}{E}=\frac{\left(6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{5 \times 10^{-19} \mathrm{~J}}\)

⇒ 4.0 x 10^-7 m = 400 nm.

Question 19. If a hydrogen atom (of mass m) at rest emits a photon of wavelength λ, the recoil speed of the atom will be

1. \(\frac{m h}{\lambda}\)
2. \(\frac{\lambda h}{m}\)
3. \(m h \lambda\)
4. \(\frac{h}{m \lambda}\)

Answer: 4. \(\frac{h}{m \lambda}\)

Conserving the linear momentum,

| momentum of the photon | = | momentum of the atom |

⇒ \(\frac{h}{\lambda}=m v\)

∴ recoil speed of the atom = v = \(\frac{h}{m \lambda}\).

Question 20. Two particles of masses m and 2m have equal kinetic energies. Their de Broglie wavelengths are in the ratio

1. 1:2
2. 1:1
3. √2:1
4. 1: √2

Answer: 3. √2:1

Kinetic energy = E = \(\frac{p^2}{2 m} \Rightarrow p=\sqrt{2 m E}\)

∴ \(\frac{\lambda_1}{\lambda_2}=\frac{h / p_1}{h / p_2}=\frac{p_2}{p_1}=\sqrt{\frac{2(2 m) E}{2 m E}}=\sqrt{2}\)

λ₁ : λ₂ ⇒ √2:1

Question 21. An electron is accelerated through a potential difference of 10000 V. Its de Broglie wavelength is nearly

1. 12.27 x l0^-12 m
2. 12.2 x 10^-14 m
3. 12.2 x l0^-13 m
4. 12.2 nm

Answer: 1. 12.27 x l0^-12 m

The de Broglie wavelength linked with an electron accelerated through a potential difference of V volts is given by

⇒ \(\lambda=\frac{12.27}{\sqrt{V}} Å=\frac{12.27 \times 10^{-10}}{\sqrt{10000}} \mathrm{~m}=12.27 \times 10^{-12} \mathrm{~m}\)

Question 22. A particle P is formed due to a completely inelastic collision of two particles x and y having de Broglie wavelengths λ_x and λ_y respectively. If x and y are moving in opposite directions, the de Broglie wavelength of P is

1. \(\lambda_x-\lambda_y\)
2. \(\frac{\lambda_{\mathrm{x}} \lambda_{\mathrm{y}}}{\lambda_{\mathrm{x}}-\lambda_{\mathrm{y}}}\)
3. \(\lambda_x+\lambda_y\)
4. \(\frac{\lambda_{\mathrm{x}} \lambda_{\mathrm{y}}}{\lambda_{\mathrm{x}}+\lambda_{\mathrm{y}}}\)

Answer: 2.

The de Broglie wavelengths associated with x and y are λ_x = h/p_x and λ_y = h/p_y respectively.

Conserving the momentum,

P_x – P_y = P [v moving in opposite directions]

⇒ \(\frac{h}{\lambda_x}-\frac{h}{\lambda_y}=\frac{h}{\lambda}\)

⇒ \(\lambda=\frac{\lambda_x \lambda_y}{\lambda_x-\lambda_y}\).

Question 23. Two particles move at a right angle to each other. Their de Broglie wavelengths are λ₁ and λ₂ respectively. The particles suffer a perfectly inelastic collision. The de Broglie wavelength (λ) of the final particle is given by

1. \(\lambda=\frac{\lambda_1+\lambda_2}{2}\)
2. \(\frac{2}{\lambda}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}\)
3. \(\lambda=\sqrt{\lambda_1 \lambda_2}\)
4. \(\frac{1}{\lambda^2}=\frac{1}{\lambda_1^2}+\frac{1}{\lambda_2^2}\)

Answer: 4. \(\frac{1}{\lambda^2}=\frac{1}{\lambda_1^2}+\frac{1}{\lambda_2^2}\)

The de Broglie wavelengths associated with the colliding particles are λ₁ = h/p₁ and λ₂ = h/p₂ respectively. Hence,

⇒ \(p_1=\frac{h}{\lambda_1} \text { and } p_2=\frac{h}{\lambda_2}\)

Conserving the momentum,

⇒ \(\vec{p}=p_1 \hat{i}+p_2 \hat{j}\)

⇒ \(p^2=p_1^2+p_2^2\)

∴ \(\frac{h^2}{\lambda^2}=\frac{h^2}{\lambda_1^2}+\frac{h^2}{\lambda_2^2} \Rightarrow \frac{1}{\lambda^2}=\frac{1}{\lambda_1^2}+\frac{1}{\lambda_2^2}\)

Question 24. If the de Broglie wavelength of an electron is equal to 10 3 times the wavelength of a photon of frequency 6 x10¹⁴ Hz then the speed of the electron is equal to (given that c = 3 x l0⁸ m s^-1, h = 6.64 x 10^-34 Js and me = 9.1 x 10^-31 kg)

1. 1.45 x l0⁶ m s^-1
2. 1.7 x 10⁶ m s^-1
3. 1.1 x l0⁶ m s^-1
4. 1.8 x l0⁶ m s^-1

Answer: 1. 1.45 x l0⁶ m s^-1

Given that \(\lambda_{\mathrm{e}}=10^{-3} \lambda_{\mathrm{ph}}=10^{-3}\left(\frac{c}{6 \times 10^{14} \mathrm{~s}^{-1}}\right)\)

But \(\lambda_{\mathrm{e}}=\frac{h}{m_{\mathrm{e}} v_{\mathrm{e}}} \Rightarrow v_{\mathrm{e}}=\frac{h}{m_{\mathrm{e}} \lambda_{\mathrm{e}}}\)

Substituting the given values, the velocity of an electron becomes

∴ \(v_{\mathrm{e}}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(6 \times 10^{14} \mathrm{~s}^{-1}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(10^{-3} \times 3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}=1.45 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\).

Question 25. A particle A of mass m and charge q is accelerated by a potential difference of 50 V. Another particle B of mass 4m and charge q is accelerated by a potential difference of 2500 V. The ratio of their de Broglie wavelengths λ_A and λ_B is close to

1. 14.14
2. 10.00
3. 0.07
4. 4.47

Answer: 1. 14.14

Kinetic energy = \(\frac{p^2}{2 m}=q V\)

For A, \(p_{\mathrm{A}}=\sqrt{2 m q V}=\sqrt{2 m q(50 \mathrm{~V})}\)

For B, \(p_{\mathrm{B}}=\sqrt{2(4 m) q(2500 \mathrm{~V})}\)

∴ \(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{h / p_{\mathrm{A}}}{h / p_{\mathrm{B}}}=\frac{p_{\mathrm{B}}}{p_{\mathrm{A}}}=\sqrt{\frac{2 \times 4 \times 2500}{2 \times 50}}=10 \sqrt{2}=14.14\).

Question 26. A beam of electrons moving with an energy of E gets scattered from a target having an atomic spacing of 1 Å. The first maximum intensity occurs at θ = 60°. The value of E is

1. 20 eV
2. 40 eV
3. 50 eV
4. 60 eV

Answer: 3. 50 eV

For the first maximum intensity in the diffraction of electrons,

2dsinθ = λ ⇒ 2(1 Å)\(\left(\frac{\sqrt{3}}{2}\right)\) = λ = √3 Å.

The de Broglie wavelength for an electron accelerated through a potential of V volts is

⇒ \(\lambda=\sqrt{\frac{150}{V}} Å=\sqrt{3}\) Å.

∴ V volts = 50 V

Hence, the energy of an electron is E = 50 eV

Question 27. A particle A of mass m_A = m/2 moving along the x-axis with a velocity v₀ collides elastically with another particle B at rest having a mass of m_B = m/3. If the particles continue to move along the x-axis after the collision, the change in the de Broglie wavelength of A in terms of λ₀ is

1. \(\Delta \lambda=2 \lambda_0\)
2. \(\Delta \lambda=\frac{3}{2} \lambda_0\)
3. \(\Delta \lambda=4 \lambda_0\)
4. \(\Delta \lambda=\frac{5}{2} \lambda_0\)

Answer: 3. \(\Delta \lambda=4 \lambda_0\)

Conserving the momentum,

⇒ \(\left(\frac{m}{2}\right) v_0=\left(\frac{m}{2}\right) v_{\mathrm{A}}+\left(\frac{m}{3}\right) v_{\mathrm{B}} \Rightarrow v_0=v_{\mathrm{A}}+\frac{2}{3} v_{\mathrm{B}}\) → (1)

For an elastic collision, e = 1. Hence,

V_B – V_A = V₀ ⇒ V_B = V₀ + V_{A →} (2)

Substituting uB from (2) in (1)

⇒ \(v_0=v_{\mathrm{A}}+\frac{2}{3}\left(v_0+v_{\mathrm{A}}\right) \Rightarrow v_{\mathrm{A}}=\frac{v_0}{5}\)

Hence, after the collision,

⇒ \(\lambda_{\mathrm{A}}=\frac{h}{m_{\mathrm{A}} v_{\mathrm{A}}}=\frac{5 h}{m_{\mathrm{A}} v_0}=5 \lambda\)

∴ Δλ = 5λ₀ – λ₀ = 4λ₀.

Question 28. An electron is accelerated from rest through a potential difference of V volts. If the de Broglie wavelength of the electron is 1.227 x10^-2 nm, the potential difference is

1. 10² V volts
2. 10⁴ V volts
3. 10³ V volts
4. 10V volts

Answer: 3. 10³ V volts

The de Broglie wavelength of the electron is

⇒ \(\lambda=\frac{1.227}{\sqrt{V}} \mathrm{~nm}\)

Given that \(\lambda=1.227 \times 10^{-2} \mathrm{~nm}=\frac{1.227}{\sqrt{V}} \mathrm{~nm}\)

√V = 10² ⇒ V = 10⁴ V.

Bohr Model And Hydrogen Spectrum

Question 1. The energy of a hydrogen atom in the ground state is -13.6 eV. The energy of an He⁺ ion in the first excited state will be

1. -13.6 eV
2. -54.4 eV
3. -27.2 eV
4. -6.8 eV

Answer: 1. -13.6 eV

In the ground state (n = 1) of hydrogen, the total energy is

⇒ \(E_1=-\frac{13.6 \mathrm{eV}}{1^2}=-13.6 \mathrm{eV}\)

In the first excited state (n = 2), the total energy of an He⁺ ion (Z = 2) will be

∴ \(E_1^{\prime}=-\frac{Z^2(13.6 \mathrm{eV})}{2^2}=-\frac{4(13.6 \mathrm{eV})}{4}=-13.6 \mathrm{eV}\).

Question 2. The ground-state energy of a hydrogen atom is -13.6 eV. When the electron is in the first excited state, its excitation energy is

1. 3.4 eV
2. 6.8 eV
3. 10.2 eV
4. Zero

Answer: 3. 10.2 eV

In the ground state (n =1),

energy = E₁ = \(-\frac{13.6 \mathrm{eV}}{1^2}=-13.6 \mathrm{eV}\)

In the first excited state(n = 2),

energy = E₂ = \(-\frac{13.6 \mathrm{eV}}{2^2}=-3.4 \mathrm{eV}\)

The excitation energy is the energy required to raise the electron from the initial state (n_i = 1) to the final state (n_f = 2).

Thus, excitation energy = E₂ – E₁ = -3.4 eV -(-13.6 eV)

= (13.6- 3.4) eV = 10.2 eV.

Question 3. The total energy of an electron in the first excited state of a hydrogen atom is about -3.4 eV. Its kinetic energy in this state is

1. -3.4 eV
2. 3.4 eV
3. 6.8 eV
4. -6.8 eV

Answer: 2. 3.4 eV

In a hydrogen atom, both the total energy and the potential energy are negative such that

E_tot = -KE and PE = -2KE.

∴ total energy = \(E_{\text {tot }}=\mathrm{KE}+\mathrm{PE}=\mathrm{KE}+(-2 \mathrm{KE})=-\mathrm{KE}=\frac{1}{2} \mathrm{PE}\)

Given that E_tot = -3.4 eV.

∴ KE = -E_tot = 3.4 eV.

Question 4. The ground-state energy of an H atom is -13.6 eV. The energy required to ionize an H atom from its second excited state is

1. 13.6 eV
2. 3.4 eV
3. 12.1 eV
4. 1.51 eV

Answer: 4. 1.51 eV

Given that E₁ = -13.6 eV.

The energy in the second excited state (n = 3) is

⇒ \(E_3=-\frac{13.6 \mathrm{eV}}{3^2}=-\frac{13.6}{9} \mathrm{eV}=-1.51 \mathrm{eV}\)

Ionization means the extraction of the electron from the given state to n = ∞.

∴ the energy required to ionize is

E = E_∞ – E₃ = 0- (-1.51 eV) =1.51 eV

Question 5. The ionization energy of a hydrogen atom in its ground state is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between the third and fourth orbits is

1. 3.40 eV
2. 0.66 eV
3. 0.85 eV
4. 1.51 eV

Answer: 2. 0.66 eV

Given that ionization energy = 13.6 eV.

In the third orbit,

⇒ \(E_3=-\frac{13.6 \mathrm{eV}}{n^2}=-\frac{13.6 \mathrm{eV}}{9}\)

In the fourth orbit,

⇒ \(E_4=\frac{-13.6 \mathrm{eV}}{4^2}=\frac{-13.6 \mathrm{eV}}{16}\)

The energy required for the transition from the state n = 3 to the state n = 4 is

⇒ \(\Delta E=E_4-E_3=\frac{-13.6 \mathrm{eV}}{16}-\left(-\frac{13.6 \mathrm{eV}}{9}\right)\)

∴ \(13.6\left(\frac{1}{9}-\frac{1}{16}\right) \mathrm{eV}=\frac{13.6 \times 7}{9 \times 16} \mathrm{eV}=0.66 \mathrm{eV}\)

Question 6. The ratio of the longest wavelengths corresponding to the Lyman series and the Balmer series in the hydrogen spectrum is

1. \(\frac{5}{27}\)
2. \(\frac{7}{29}\)
3. \(\frac{9}{31}\)
4. \(\frac{2}{23}\)

Answer: 1. \(\frac{5}{27}\)

The energy difference during the transition is

⇒ \(\Delta E=h \mathrm{v}=\frac{h c}{\lambda}\)

For the Lyman series,

⇒ \(\frac{h c}{\lambda_1}=E_2-E_1=13.6\left(1-\frac{1}{4}\right)\)

For the Balmer series,

⇒ \(\frac{h c}{\lambda_2}=E_3-E_2=13.6\left(\frac{1}{4}-\frac{1}{9}\right)\)

∴ \(\frac{\lambda_1}{\lambda_2}=\frac{E_3-E_2}{E_2-E_1}=\frac{5 / 36}{3 / 4}=\frac{5}{27}\).

Question 7. The electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths λ₁ and λ₂ emitted in the two cases is

1. \(\frac{27}{20}\)
2. \(\frac{20}{7}\)
3. \(\frac{7}{5}\)
4. \(\frac{27}{5}\)

Answer: 2. \(\frac{20}{7}\)

For the electronic transition from the third excited state (n = 4) to the second excited state (n = 3), the wavelength λ₁ is given by

⇒ \(\frac{h c}{\lambda_1}=E_4-E_3=13.6\left(\frac{1}{9}-\frac{1}{16}\right) \mathrm{eV}\)

For the transition from the second excited state(n = 3) to the first excited state (n = 2), the wavelength λ₂ is given by

⇒ \(\frac{h c}{\lambda_2}=E_3-E_2=13.6\left(\frac{1}{4}-\frac{1}{9}\right) \mathrm{eV}\)

⇒ \(\frac{\lambda_1}{\lambda_2}=\frac{13.6\left(\frac{5}{36}\right) \mathrm{eV}}{13.6\left(\frac{7}{144}\right) \mathrm{eV}}=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7}\).

Question 8. The wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number Z of this hydrogenlike ion is

1. 1
2. 3
3. 2
4. 4

Answer: 3. 2

For the first line of the Lyman series for an H atom, the transition takes place from n = 2 to n = 1. Thus,

⇒ \(\Delta E=\frac{h c}{\lambda_1}=E_2-E_1=13.6\left(1-\frac{1}{4}\right) \mathrm{eV}=13.6 \times \frac{3}{4} \mathrm{eV}\)

For the second line of the Balmer series, the energy needed for the transition for the atomic number Z is

⇒ \(\Delta E=\frac{h c}{\lambda_2}=13.6 \mathrm{Z}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \mathrm{eV}=13.6 \mathrm{Z}^2\left(\frac{3}{16}\right) \mathrm{eV}\)

Since λ₁ = λ₂ (given),

⇒ \(13.6 Z^2\left(\frac{3}{16}\right)=13.6 \times \frac{3}{4}\)

∴ Z² = 4 ⇒ Z = 2.

Question 9. Which of the following transitions gives photons of the maximum energy?

1. From n = 1 to n = 2
2. From n = 2 to n = 1
3. From n = 2 to n = 6
4. From n = 6 to n = 2

Answer: 2. From n = 2 to n = 1

The energy of a photon depends on the change in the principal quantum number (n).

For the transition from n = 2 to n = 1,

⇒ \(\Delta E_1=(13.6 \mathrm{eV})\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4}(13.6 \mathrm{eV})\)

For the transition from n = 6 to n = 2,

⇒ \(\Delta E_2=(13.6 \mathrm{eV})\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{2}{9}(13.6 \mathrm{eV})\)

∵ ΔE₁ > ΔE₂,

∴ the transition from n = 2 to n = 1 will give photons of the maximum energy.

Question 10. The energy of the ground state of a hydrogen atom is -13.6 eV. The energy of the first excited state will be

1. -6.8 eV
2. -3.4 eV
3. -27.2 eV
4. -54.4eV

Answer: 2. -3.4 eV

Given that in the ground state,

⇒ \(E_1=-\frac{13.6}{1^2} \mathrm{eV}=13.6 \mathrm{eV}\)

In the first excited state (n = 2),

⇒ \(E_2=-\frac{13.6 \mathrm{eV}}{2^2}=-3.4 \mathrm{eV}\)

Question 11. Consider the third orbit of an He+ ion. The speed of the electron in this orbit will be (given that 1/4πε₀ = 9 x 10⁹ SI units, Z = 2, and h = 6.6 X 10^-34J S)

1. 2.92 x l0⁶ m s^-1
2. 0.73 x 10⁶ m s^-1
3. 1.45 x l0⁶ m s^-1
4. 3.0 x l0⁸ m s^-1

Answer: 2. 0.73 x 10⁶ m s^-1

The electrostatic attraction provides the centripetal force. Hence,

⇒ \(\frac{m v^2}{r}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(\mathrm{Z} e) e}{r^2}\)

⇒ \(m v^2 r=\frac{1}{4 \pi \varepsilon_0}\left(Z e^2\right)\) → (1)

From the quantization rule,

⇒ \(m v r=\frac{n h}{2 \pi}\) → (2)

Dividing (1) by (2)

⇒ \(v=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(Z e^2\right)(2 \pi)}{n h}\)

Substituting the velocity,

∴ \(v=\left(9 \times 10^9\right) \times \frac{2\left(1.6 \times 10^{-19}\right)^2 \times 2(3.14)}{3\left(6.67 \times 10^{-34}\right)} \mathrm{m} \mathrm{s}^{-1} \approx 1.45 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 12. A hydrogen atom in its ground state is excited by a monochromatic radiation of wavelength 975 Å. The number of spectral lines in the resulting spectrum emitted will be

1. 2
2. 6
3. 10
4. 3

Answer: 2. 6

The energy of the photon absorbed is

⇒ \(E=\frac{h c}{\lambda}=\frac{12420 \mathrm{eV} Å}{975 Å}=12.74 \mathrm{eV}\)

This energy gap corresponds to the transition from n = 4 to n = 1.

∴ the number of spectral lines in the spectrum is

∴ \(E=\frac{h c}{\lambda}=\frac{12420 \mathrm{eV} Å}{975 Å}=12.74 \mathrm{eV}\)

Question 13. The ionization energy of the electron in a hydrogen atom in its ground state is 13.6 eV. The atoms are excited to high energy levels to emit radiations of six wavelengths. The maximum wavelength of the emitted radiations corresponds to the transition between the states

1. n = 3 and n = 2
2. n = 3 and n = l
3. n = 2 and n = 1
4. n = 4 and n = 3

Answer: 4. n = 4 and n = 3

As seen in the preceding question, six spectral lines exist in the spectrum for the transition from n = 4 to n = 3. The maximum wavelength corresponds to the minimum energy gap, which is from n = 4 to n = 3.

Question 14. When an electron jumps from the n = 4 orbit to the n = 2 orbit, we get

1. The second line of the Lyman series
2. The second line of the Balmer series
3. The second line of the Paschen series
4. An absorption line of the Balmer series

Answer: 2. The second line of the Balmer series

From the energy-level diagram, transition I corresponds to the first line of the Balmer series; transition 2 corresponds to the second line of the Balmer series; and transition 3 corresponds to the first line of the Paschen series.

Hence, the option (2) is true.

Question 15. The radius of a hydrogen atom in its ground state is 5.3 x 10^-11 m. After a collision with an electron, it is found to have a radius of 21.2 x l0^-11 m. What is the value of the principal quantum number (n) of the final state of the atom?

1. 3
2. 2
3. 4
4. 16

Answer: 2. 2

The radius of Bohr’s stationary orbit in the nth state (r_n) varies as n².

∴ \(\frac{r_n}{r_1}=\frac{n^2}{1} \Rightarrow \frac{21.2 \times 10^{-11} \mathrm{~m}}{5.3 \times 10^{-11} \mathrm{~m}}=n^2 \Rightarrow n^2=4\)

The principal quantum number in the final state is n = 2

Question 16. The energy levels A, B, and C of a certain atom correspond to increasing values of energy, i.e., E_A < E_B < E_C. If λ₁, λ₂, and λ₃ are the wavelengths of the radiations corresponding to the transitions C→ B, B → A, and C→ A respectively, which of the following relations is correct?

1. λ₁ + λ₂ + λ₃ = 0
2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)
3. \(\lambda_3=\sqrt{\lambda_1^2+\lambda_2^2}\)
4. λ₃ = λ₁+ λ₂

Answer: 2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

From the energy-level diagram,

⇒ \(E_{\mathrm{C}}-E_{\mathrm{B}}=\frac{h c}{\lambda_1}\) → (1)

⇒ \(E_{\mathrm{B}}-E_{\mathrm{A}}=\frac{h c}{\lambda_2}\) → (2)

and \(E_{\mathrm{C}}-E_{\mathrm{A}}=\frac{h c}{\lambda_3}\) → (3)

Adding (1) and (2)

⇒ \(E_{\mathrm{C}}-E_{\mathrm{A}}=h c\left(\frac{1}{\lambda_1}+\frac{1}{\lambda_2}\right)\) (4)

Equating (3) and (4),

∴ \(\frac{h c}{\lambda_3}=h c\left(\frac{\lambda_1+\lambda_2}{\lambda_1 \lambda_2}\right) \Rightarrow \lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

Question 17. If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength λ. When it jumps from the fourth orbit to the third orbit, the corresponding wavelength of the emitted photon will be

1. \(\frac{9 \lambda}{16}\)
2. \(\frac{20 \lambda}{7}\)
3. \(\frac{20 \lambda}{13}\)
4. \(\frac{9 \lambda}{16}\)

Answer: 2. \(\frac{20 \lambda}{7}\)

For the transition from n = 3 to n = 2,

⇒ \(\Delta E=\frac{h c}{\lambda}=(13.6 \mathrm{eV})\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\left(\frac{5}{36}\right) \mathrm{eV}\) → (1)

For the transition from n = 4 to n = 3,

⇒ \(\Delta E=\frac{h c}{\lambda^{\prime}}=(13.6 \mathrm{eV})\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=13.6\left(\frac{7}{144}\right) \mathrm{eV}\) → (2)

Dividing (1) by (2),

∴ \(\frac{\lambda^{\prime}}{\lambda}=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7} \Rightarrow \lambda^{\prime}=\frac{20 \lambda}{7}\).

Question 18. The ratio of the wavelengths of the last line of the Balmer series and the last line of the Lyman series is

1. 1
2. 4
3. 2
4. 0.5

Answer: 2. 4

The last line of any series corresponds to the transition from n = ∞ to any value of n.

Hence, for the last line of the Balmer series,

⇒ \(\Delta E=\frac{h c}{\lambda_1}=(13.6 \mathrm{eV})\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{13.6}{4} \mathrm{eV}\)

For the last line of the Lyman series,

⇒ \(\Delta E=\frac{h c}{\lambda_2}=(13.6 \mathrm{eV})\left(\frac{1}{1^2}-\frac{1}{\infty}\right)=13.6 \mathrm{eV}\)

∴ \(\frac{\lambda_1}{\lambda_2}=\frac{13.6}{13.6 / 4}=4\)

Question 19. The ionization energy of a hydrogen atom is 13.6 eV. The ionization energy of an He⁺ ion would be

1. 13.6 eV
2. 27.2 eV
3. 6.8 eV
4. 54.4 eV

Answer: 4. 54.4 eV

The ionization energy of an H atom in its ground state is the energy required to knock out the electron from n = 1 to n = ∞.

Thus, for the H atom,

⇒ \(E_{\mathrm{H}}=(13.6 \mathrm{eV})\left(\frac{1}{1^2}-\frac{1}{\infty}\right)=13.6 \mathrm{eV}\).

For the He⁺ ion,

⇒ \(E_{\mathrm{He}^{+}}=(13.6 \mathrm{eV}) \mathrm{Z}^2\left(\frac{1}{1^2}-\frac{1}{\infty}\right)=(13.6 \times 4) \mathrm{eV}=54.4 \mathrm{eV}\)

Question 20. Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of the de Broglie wavelength (λ) of that electron as

1. 0.529nλ
2. \(\sqrt{n \lambda}\)
3. nλ
4. 13.6λ

Answer: 3. nλ

According to the de Broglie relation,

wavelength = \(\lambda=\frac{h}{m v}\) → (1)

From Bohr’s quantization condition,

angular momentum = \(\frac{n h}{2 \pi}\)

\(m v r=\frac{n h}{2 \pi} \Rightarrow \frac{h}{m v}=\frac{2 \pi r}{n}\) → (2)

Equating (1) and (2),

⇒ \(\lambda=\frac{2 \pi r}{n}\)

∴ the circumference of the orbit is 2πr = nλ.

Question 21. When a hydrogen atom is raised from the ground state to an excited state,

1. The PE decreases and the KE increases
2. The PE increases and the KE decreases
3. Both PE and KE decrease
4. An absorption spectrum is obtained

Answer: 2. The PE increases and the KE decreases

The total energy of an H atom is negative, as a result of a positive value of the \(\mathrm{KE}\left(=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{2 r}\right)\) and a negative value of the \(\mathrm{PE}\left(=-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}\right)\).

When the atom is raised to the higher excited state, its orbital radius (r) increases, hence the KE decreases; and the PE becomes less negative, i.e., it increases.

Question 22. Some hydrogen atoms are excited from their ground states to the states having the principal quantum number 4. Then, the number of spectral lines observed will be

1. 3
2. 2
3. 5
4. 6

Answer: 4. 6

The total number of spectral lines observed in an H atom for all possible transitions from n = 4 is \(N=\frac{n(n-1)}{2}=\frac{4 \times 3}{2}=6\)

This is also shown in the adjoining energy-level diagram.

Question 23. In terms of the Bohr radius a₀, the value of the second Bohr orbit of a hydrogen atom is

1. 2a₀
2. 4a₀
3. √2a₀
4. 8a₀

Answer: 2. 4a₀

The radius of the nth orbit in an H atom is r_n = n²a₀, where a₀ = Bohr radius (radius of the first orbit).

∴ for the second orbit, r₂ = 2²a₀ = 4a₀

Question 24. The ionization potential of a hydrogen atom is 13.6 V. Some hydrogen atoms in their ground states are excited by a monochromatic radiation of photons of energy 12.1 eV. According to Bohr’s theory, the number of spectral lines emitted by the hydrogen atoms will be

1. 2
2. 4
3. 3
4. 1

Answer: 3. 3

The ionization potential in the ground state is 13.6 V. So, the ionization energy is 13.6 eV, and the energy in the ground state is E₁ = -13.6 eV. Let the absorption of the 12.1 eV energy by the atom raise the electron to the nth state. Thus,

E₁ + ΔE = E_n

⇒ \(-13.6 \mathrm{eV}+12.1 \mathrm{eV}=-\frac{13.6 \mathrm{eV}}{n^2} \Rightarrow-1.5=-\frac{13.6}{n^2} \Rightarrow n=3\)

Hence, the number of spectral lines is

∴ \(N=\frac{n(n-1)}{2}=\frac{3 \times 2}{2}=3\)

Question 25. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of the emission of photons will be (given that m = mass of an electron, R_∞ = Rydberg constant arid h = Planck constant)

1. \(\frac{24 R_{\infty} h}{25 m}\)
2. \(\frac{24 m}{25 R_{\infty} h}\)
3. \(\frac{25 m}{24 R_{\infty} h}\)
4. \(\frac{25 R_{\infty} h}{24 m}\)

Answer: 1. \(\frac{24 R_{\infty} h}{25 m}\)

The energy of an H atom in its nth state in terms of the Rydberg constant (R_∞) is given by

⇒ \(E_n=-\frac{R_{\infty} h c}{n^2}\)

The energy of the photon emitted during the transition 5 → 1 will be

⇒ \(\Delta E=E_5-E_1=R_{\infty} h c\left(\frac{1}{1}-\frac{1}{5^2}\right)=\frac{24}{25} R_{\infty} h c\)

⇒ \(\frac{h c}{\lambda}=\frac{24}{25} R_{\infty} h c \Rightarrow \frac{1}{\lambda}=\frac{24}{25} R_{\infty}\)

∴ \(\frac{p}{h}=\frac{m v}{h}=\frac{24 R_{\infty}}{25} \Rightarrow v=\frac{24 R_{\infty} h}{25 m}\)

Question 26. The maximum wavelength λ which can ionize a hydrogen atom in its ground state is

1. 150.2 nm
2. 100.5 nm
3. 91.3 nm
4. 110 nm

Answer: 3. 91.3 nm

The ionization of an H atom from its ground state results in a transition from n =1 to n = ∞. The required energy is

⇒ \(\Delta E=\frac{h c}{\lambda_0}=(13.6 \mathrm{eV})\left(\frac{1}{1^2}-\frac{1}{\infty}\right)=13.6 \mathrm{eV}\)

∴ \(\lambda_0=\frac{h c}{13.6 \mathrm{eV}}=\frac{1242 \mathrm{eV} \mathrm{nm}}{13.6 \mathrm{eV}}=91.3 \mathrm{~nm}\)

Question 27. Which of the following parameters is the same for an H⁺ ion, an He⁺ ion, and a Li²⁺ ion in their ground states?

1. Energy
2. Speed of an electron
3. The orbital angular momentum of an electron
4. The radius of the orbit

Answer: 3. Orbital angular momentum of an electron

According to Bohr’s quantum condition, the angular momentum of an orbital electron in the nth orbit is quantized as nh/2π. Hence, the orbital angular momentum is the same for all ions in the same state.

Question 28. What can be the maximum photoelectric work function of a metal from which photoelectrons can be ejected by the light of the Balmer series of hydrogen?

1. 13.6 eV
2. 3.4 eV
3. 1.5 eV
4. None of these

Answer: 2. 3.4 eV

In the Balmer series, the maximum energy of a photon corresponds to the transition from n = ∞ to n = 2.

∴ \(\Delta E=\frac{h c}{\lambda}=(13.6 \mathrm{eV})\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{13.6 \mathrm{eV}}{4}=3.4 \mathrm{eV}\)

This maximum value is equal to the maximum work function.

Question 29. Consider the radiations from an H-atom, an He⁺ ion, and a Li²⁺ ion. The ratio of the smallest wavelengths radiated from them will be

1. 1 : 4: 9
2. 9: 4: 1
3. 36: 9: 4
4. 4: 9: 36

Answer: 3. 36: 9: 4

The smallest wavelength corresponds to the maximum energy of radiations, which corresponds to the transition from n = ∞ to n = 1. Thus,

⇒ \(\Delta E=h v=\frac{h c}{\lambda}=13,6\left(\frac{1}{1^2}-\frac{1}{\infty}\right) Z^2 \mathrm{eV}\)

For H – atom, \(\lambda_{\mathrm{H}}=\frac{h c}{13.6 \mathrm{eV}}\)

For He⁺, \(\lambda_{\mathrm{He}^{+}}=\frac{h c}{4(13.6 \mathrm{eV})}\)

For Li²⁺, \(\lambda_{\mathrm{L}^{2+}}=\frac{h c}{9(13.6 \mathrm{eV})}\)

∴ H⁺ : He⁺: Li²⁺ = 36: 9: 4.

Question 30. A hydrogen atom emits ultraviolet radiation of wavelength 102.5 ran. The values of the principal quantum number for the stationary states involved in this transition are

1. n = 2 and n = 1
2. n = 3 and n = 1
3. n = 3 and n = 2
4. n = 4 and n = 2

Answer: 2. n = 3 and n = 1

The possible transitions are shown in the adjoining energy-level diagram in which E₁ = -13.6 eV, E₂ = -3.4 eV, and E₃ = -1.5 eV.

The energy of the photon emitted is

⇒ \(\Delta E=\frac{h c}{\lambda}=\frac{1242 \mathrm{eV} \mathrm{nm}}{102.5 \mathrm{~nm}}=121 \mathrm{eV}\)

This energy corresponds to the transition 3 → 1.

Question 31. A photon of wavelength 50 nm is absorbed by an H atom in its ground state and is ionized. The kinetic energy associated with the ejected electron is

1. 13.6 eV
2. 10.6 eV
3. 11.24 eV
4. 9.12 eV

Answer: 3. 11.24 eV

Given that λ = 50 nm. Hence, the energy of the photon is

⇒ \(E=\frac{h c}{\lambda}=\frac{1242 \mathrm{eV} \mathrm{nm}}{50 \mathrm{~nm}}=24.84 \mathrm{eV}\)

The energy in the ground state of hydrogen is E₁ = -13.6 eV.

∴ ionization energy = (energy spent in the extraction of an electron)

= 13.6 eV.

The energy remaining in the electron is

KE = 24.84 eV – 13.6 eV = 11.24 eV.

Question 32. Find the magnetic field at the center of a hydrogen atom in its ground state. (Given that Bohr radius = a₀ = 5 x 10^-11 m).

1. 20 T
2. 14 T
3. 15.7 T
4. 10.75 T

Answer: 2. 14 T

An electron revolving around the nucleus constitutes an electric current given by

⇒ \(I=\frac{e}{T}=\frac{e v}{2 \pi r}\)

the magnetic field at the center is

⇒ \(B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 e v}{4 \pi r^2}\) →(1)

From Bohr’s quantum condition,

⇒ \(m v r_n=\frac{n h}{2 \pi}\)

For the ground state (n = 1),

⇒ \(m v r=\frac{h}{2 \pi} \Rightarrow v=\frac{h}{2 \pi m r}\)

Substituting v in (1),

⇒ \(B=\frac{\mu_0 e}{4 \pi r^2} \cdot \frac{h}{2 \pi m r}=\frac{\mu_0 e h}{4 \pi\left(2 \pi m r^3\right)}\)

Substituting the standard values,

∴ \(B \approx \frac{10^{-7}\left(1.6 \times 10^{-19}\right)\left(6.62 \times 10^{-34}\right)}{2(3.14)\left(9.1 \times 10^{-31}\right)\left(5 \times 10^{-11}\right)^3} \mathrm{~T} \approx 14 \mathrm{~T}\).

Question 33. How many spectral lines exist in the visible region Of the Balmer series of hydrogen atoms?

1. 5
2. 4
3. 2
4. 3

Answer: 2. 4

Transitions in the Balmer series are given by

⇒ \(\Delta E=\frac{h c}{\lambda}=13.6\left(\frac{1}{2^2}-\frac{1}{n^2}\right) \mathrm{eV}, \text { where } n=3,4,5, \ldots\)

For n = 3, \(\lambda=\frac{h c(4 \times 9)}{13.6 \times 5}=\frac{1242 \mathrm{eV} \mathrm{nm} \times 36}{13.6 \mathrm{eV} \times 5}=657.5 \mathrm{~nm}\)

For n = 4, \(\lambda=\frac{1242 \mathrm{eV} \mathrm{nm} \times 64}{13.6 \mathrm{eV} \times 12}=487 \mathrm{~nm}\)

For n = 5, \(\lambda=\frac{1242 \times 100}{13.6 \times 21} \mathrm{~nm}=434.8 \mathrm{~nm}\)

For n = 6, \(\lambda=\frac{1242 \times 36 \times 4}{13.6 \times 32} \mathrm{~nm}=411 \mathrm{~nm}\)

The line for n = 7 lies in the ultraviolet region. So, only four lines exist in the visible region

Question 34. The energy of the orbital electron in the first excited state of an He⁺ ion is

1. -10.6 eV
2. -25.6 eV
3. 15.6 eV
4. -13.6 eV

Answer: 4. -13.6 eV

For the first excited state, n = 2. So, the energy of an electron is

∴ \(E_2=-\frac{13.6 \mathrm{Z}^2}{n^2} \mathrm{eV}=-\frac{13.6\left(2^2\right)}{2^2} \mathrm{eV}=-13.6 \mathrm{eV}\)

Question 35. If the energy of the electron in the ground state of an H atom is -13.6 eV, find the orbital speed of the electron in the third excited state.

1. 5.45 x l0³ m s^-1
2. 5.45 x l0⁴ m s^-1
3. 5.45 x l0⁶ m s^-1
4. 5.46 x 10⁵ m s^-1

Answer: 4. 5.46 x 10⁵ m s^-1

In the hydrogen atom,|PE| = 2KE and total energy = KE + (-2KE) = -KE.

In the ground state,

⇒ \(-13.6 \mathrm{eV}=\frac{1}{2} m v^2\)

For the third excited state, n = 4. So,

⇒ \(\left|E_4\right|=+\frac{13.6}{4^2} \mathrm{eV}=\frac{1}{2} m v^2\)

∴ \(v=\sqrt{\frac{2\left|E_4\right|}{m}}=\sqrt{\frac{2\left(13.6 \times 1.6 \times 10^{-19} \mathrm{~J}\right)}{16\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}}=5.46 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 36. If an energy of 15 eV is given to a hydrogen atom with the electron in the fourth orbit, the final KE of the electron when it leaves the atom will be

1. 13.6 eV
2. 12.08 eV
3. 14.15 eV
4. 15.85 eV

Answer: 3. 14.15 eV

The energy of the electron in the fourth orbit is

⇒ \(E_4=-\frac{13.6}{4^2} \mathrm{eV}=-0.85 \mathrm{eV}\)

For ionization (extraction of the electron from the atom), an energy of 0.85 eV is now required. Adding 15 eV to the atom will eject the electron with

KE = 15 eV- 0.85 eV = 14.15 eV

Question 37. The total energy of an electron in an atom in an orbit is -3.4 eV. Its kinetic and potential energies are respectively

1. -3.4 eV and -3.4 eV
2. -3.4 eV and -6.8 eV
3. 3.4 eV and -6.8eV
4. 3.4 eV and 3.4 eV

Answer: 3. 3.4 eV and -6.8eV

According to Bohr’s theory, the total energy and the potential energy of an atom are negative.

But KE of an electron = |total energy| = \(\mid \text { total energy }\left|=\frac{1}{2}\right| \mathrm{PE} \mid\)

Given that E_tot = -3.4 eV.

∴ KE =|E_tot| = 3.4eV

and PE = 2(-KE) = -2(3.4 eV) = -6.8 eV.

Question 38. Taking the wavelength of the first Balmer line in the hydrogen spectrum (from n = 3 to n = 2) as 660 runs, the wavelength of the second Balmer line (from n = 4 to n = 2) will be

1. 642.7 nm
2. 889.2 nm
3. 488.9 nm
4. 388.9 run

Answer: 3. 488.9 nm

For the first Balmer line,

⇒ \(\Delta E=\frac{h c}{\lambda_1}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \mathrm{eV}=13.6 \times \frac{5}{4 \times 9} \mathrm{eV}\) → (1)

For the second Balmer line,

⇒ \(\frac{h c}{\lambda_2}=13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \mathrm{eV}=13.6 \times \frac{12}{4 \times 16} \mathrm{eV}\) → (2)

Dividing (1) by (2), we have

⇒ \(\frac{\lambda_2}{\lambda_1}=\frac{5}{36} \times \frac{64}{12}=\frac{20}{27}\)

∴ \(\lambda_2=\frac{20}{27} \lambda_1=\frac{20}{27}(660 \mathrm{~nm})=488.88 \mathrm{~nm} \approx 488.9 \mathrm{~nm}\)

Question 39. A hydrogen atom initially in the ground state is excited by absorbing a photon of wavelength 980 Å. The radius of the atom in the excited state in terms of its Bohr radius a₀ will be (assuming that he = 12500 eV Å)

1. 4a₀
2. 16a₀
3. 9a₀
4. 25a₀

Answer: 2. 16a₀

The energy of the photon of wavelength λ = 980 A is

⇒ \(h c=\frac{12500 \mathrm{eV} Å}{980 Å}=12.76 \mathrm{eV}\)

This energy gap corresponds to the transition from n =1 (E₁ = -13.6 eV) to n = 4 (E₄ = -0.85 eV). Hence, the radius of the orbit of the electron in the state n = 4 is R₄ = a₀(4²) =16a₀.

Question 40. A singly ionized helium ion (He⁺) is in its first excited state. Its ionization energy is

1. 54.40 eV
2. 13.60 eV
3. 48.36 eV
4. 6.04 eV

Answer: 2. 13.60 eV

The ionization energy is the energy required to extract an electron from the atom. It is the positive value of the energy of the electron in the given quantum state.

For a singly ionized helium atom in its first excited state (n = 2), its energy is

⇒ \(E_2=-\frac{13.6 Z^2}{2^2} \mathrm{eV}=-\frac{13.6 \times 4}{4} \mathrm{eV}=-13.6 \mathrm{eV}\)

∴ ionization energy = \(-E_2=-(-13.6 \mathrm{eV})=13.6 \mathrm{eV}\).

Question 41. Consider an electron in a hydrogen atom revolving in its second excited state (having a radius of 4.65 Å). The de Broglie wavelength of this electron is

1. 6.6 Å
2. 9.7 Å
3. 3.5 Å
4. 12.9 Å

Answer: 2. 9.7 Å

According to Planck’s quantum condition,

⇒ \(m v r=\frac{n h}{2 \pi} \Rightarrow p=\frac{n h}{2 \pi r}\)

∴ the de Broglie wavelength is

⇒ \(\lambda=\frac{h}{p}=\frac{2 \pi r}{n}=\frac{2(3.14)(4.65 Å)}{3}=9.7 Å\)

Question 42. For n > > 1, the frequency of a photon emitted from a hydrogen atom 1 during the transition from the (n + 1)th state to the wth state is directly proportional to

1. n
2. \(\frac{1}{n}\)
3. \(\frac{1}{n^2}\)
4. \(\frac{1}{n^3}\)

Answer: 4. \(\frac{1}{n^3}\)

The energy in the nth state of an H atom is

⇒ \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)

Hence, during the transition (n +1)-» n, the energy difference is

⇒ \(\Delta E=h v=13.6\left[\frac{1}{n^2}-\frac{1}{(n+1)^2}\right] \mathrm{eV}\)

∴ \(\mathrm{v} \propto \frac{1}{n^2}-\frac{1}{n^2}\left(1+\frac{1}{n}\right)^{-2}\)

⇒ \(\mathrm{v} \propto \frac{1}{n^2}-\frac{1}{n^2}\left(1-\frac{2}{n}\right)\)

⇒ \(v \propto \frac{1}{n^3}\).

Question 43. If the difference λ_max – λ_min for the Lyman series is 340 Å, the corresponding difference for the Paschen series is

1. 12502 Å
2. 13802 Å
3. 11802 Å
4. 10000 Å

Answer: 3. 11802 Å

For the Lyman series,

⇒ \(\frac{1}{\lambda_{\min }}=R_{\infty}\left(\frac{1}{1^2}-\frac{1}{\infty}\right)=R_{\infty}\)

and \(\frac{1}{\lambda_{\max }}=R_{\infty}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R_{\infty}\)

∴ \(\lambda_{\max }-\lambda_{\min }=\frac{4}{3 R_{\infty}}-\frac{1}{R_{\infty}}=\frac{1}{3 R_{\infty}}\)

⇒ \(\frac{1}{3 R_{\infty}}=340 Å \Rightarrow \frac{1}{R_{\infty}}=3 \times 340 Å\)

For the Paschen series,

⇒ \(\frac{1}{\lambda_{\min }}=R_{\infty}\left(\frac{1}{3^2}-\frac{1}{\infty}\right)=\frac{R_{\infty}}{9}\)

and \(\frac{1}{\lambda_{\max }}=R_{\infty}\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{7 R_{\infty}}{9 \times 16}\)

∴ \(\lambda_{\max }-\lambda_{\min }=\frac{9}{R_{\infty}}\left(\frac{16}{7}-1\right)=\frac{81}{7 R_{\infty}}=\frac{81}{7}(340 \times 3 Å)=11802 Å\).

Question 44. If λ is the maximum wavelength in the Lyman series in the spectrum of a hydrogen atom, the minimum wavelength in the Balmer series of the spectrum of an He⁺ ion is

1. \(\frac{\lambda}{4}\)
2. \(\frac{3 \lambda}{4}\)
3. \(\frac{2 \lambda}{3}\)
4. \(\frac{\lambda}{3}\)

Answer: 2. \(\frac{3 \lambda}{4}\)

For the He⁺ ion, the minimum wavelength corresponds to the maximum energy difference. Hence, for the Balmer series,

⇒ \(\frac{1}{\lambda_{\mathrm{He}^{+}}}=R_{\infty} \mathrm{Z}^2\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{R_{\infty}}{4} \cdot 4=R_{\infty}\)

For the Hatom, the maximum wavelength corresponds to the minimum energy difference. Hence, for the Lyman series,

⇒ \(\frac{1}{\lambda_{\mathrm{H}}}=R_{\infty}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R_{\infty}=\frac{1}{\lambda} \text { (given). }\)

∴ \(\lambda_{\mathrm{He}^{+}}=\frac{1}{R_{\infty}}=\frac{3 \lambda}{4}\).

Question 45. A particle of mass 200 MeV/c² collides with a hydrogen atom at rest. Soon after the collision, the colliding particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle is (N/4) eV, where N is (given that mass of a hydrogen atom =1 GeV/c²)

1. 24
2. 48
3. 51
4. 73

Answer: 3. 51

The mass of the colliding particle is

⇒ \(m=\frac{200 \mathrm{MeV}}{c^2}=200 \times 10^6 \mathrm{u}\)

The mass of the hydrogen atom is

⇒ \(m_{\mathrm{H}}=\frac{1 \mathrm{GeV}}{c^2}=10^9 \mathrm{u}\)

∴ \(\frac{m_{\mathrm{H}}}{m}=\frac{1 \times 10^9 \mathrm{u}}{200 \times 10^6 \mathrm{u}}=\frac{1000}{200}=5\)

⇒ mH = 5m.

Conserving the momentum,

⇒\(m v=m_{\mathrm{H}} v_{\mathrm{H}} \Rightarrow v_{\mathrm{H}}=\left(\frac{m}{m_{\mathrm{H}}}\right) v=\frac{v}{5}\)

∴ loss of energy = \(\frac{1}{2} m v^2-\frac{1}{2} m_{\mathrm{H}} v_{\mathrm{H}}^2\)

⇒ \(\frac{1}{2} m v^2-\frac{1}{2}(5 m)\left(\frac{v}{5}\right)^2=\frac{1}{2} m v^2\left(\frac{4}{5}\right)\)

This energy loss is spent in exciting the H atom in its first excited state, which is

\(\Delta E=E_2-E_1=13.6\left(1-\frac{1}{4}\right) \mathrm{eV}=13.6 \times \frac{3}{4} \mathrm{eV}=10.2 \mathrm{eV}\)

∴ \(\left(\frac{1}{2} m v^2\right)\left(\frac{4}{5}\right)=10.2 \mathrm{eV}\)

∴ initial KE = \(=\frac{1}{2} m v^2=(10.2 \mathrm{eV}) \frac{5}{4}=\frac{51}{4} \mathrm{eV}=\frac{N}{4} \mathrm{eV}\).

Hence, N = 51.

Semiconductors

Question 1. In a semiconductor at room temperature,

1. The valence band is completely filled
2. The valence band is partially empty and the conduction band is partially filled
3. The conduction band is completely empty
4. The valence band is partially filled and the conduction band is partially filled

Answer: 2. The valence band is partially empty and the conduction band is partially filled

At room temperature in semiconductors, the electrons from the covalent bond get sufficient thermal energy so that they get detached from the bond and are able to overcome the forbidden energy gap. Thus, the valence band is partially empty (or ‘holes’ are created) and the conduction band is partially filled. At the absolute zero temperature (0 K), the conduction band in a semiconductor is completely empty and thus behaves as a perfect insulator.

Question 2. In a p-type semiconductor, the majority carriers of current are

1. Protons
2. Holes
3. Neutrons
4. Electrons

Answer: 2. Holes

When a trivalent impurity, such as boron (B) or gallium (Ga), is added to an intrinsic (or pure) semiconductor, such as silicon (Si) or germanium (Ge), it is said to be a p-type semiconductor. Each boron atom has three valence electrons and each silicon atom has four.

The three valence electrons of each. boron atoms form three covalent bonds with the three neighboring atoms. In the fourth covalent bond, one electron from a silicon atom is present and the bond is incomplete with the shortage of one electron.

This missing electron is called a hole. These holes act as the majority charge carriers in the conduction of electricity in a p-type semiconductor.

Question 3. The depletion layer consists of

1. Electrons
2. Protons
3. Mobile ions
4. Immobile ions

Answer: 4. Immobile ions

A depletion layer is formed instantaneously across a p-n junction due to the diffusion of electrons and holes. The n-side has an excess of electrons, which diffuse through the p-side. Similarly, holes diffuse from the p-side to the n-side. During the diffusion, the majority of charge carriers (free electrons in the n-type and holes in the p-type) are gone due to recombinations. This constitutes the depletion layer containing only immobile ions.

Question 4. At absolute zero, Si acts as

1. A conducting metal
2. A nonmetal
3. An insulator
4. None of these

Answer: 3. An insulator

At absolute zero, all the electrons are tied to their atoms in the valence band and no thermal energy is available to detach the electrons. This means that the conduction band is completely empty (no free electrons). Hence, the semiconducting material (Si) acts as an insulator.

Question 5. When an n-type semiconductor is heated,

1. The number of electrons increases, while that of holes decreases
2. The number of holes increases, while that of electrons decreases
3. The number of electrons and holes increases equally
4. The number of electrons and holes remains the same

Answer: 3. The numbers of electrons and holes increase equally

When an n-type extrinsic semiconductor is heated, the thermal energy is sufficient for the electron to break the bond and become free. Simultaneously, a hole is also created. The numbers of electrons and holes remain equal and increase equally with time.

Question 6. The depletion layer in a p-n junction region is caused by

1. Drift of holes
2. Diffusion of charge carriers
3. Migration of impurity ions
4. Drift of electrons

Answer: 2. Diffusion of charge carriers

In a p-n junction, the depletion layer is caused by the diffusion of charge carriers and the consequent recombinations.

Question 7. In a junction diode, the holes are nothing but

1. Protons
2. Neutrons
3. Missing of electrons
4. Extra electrons

Answer: 3. Missing of electrons

Holes and electrons are the two types of charge carriers responsible for currents in a semiconductor. A hole is nothing but the absence of an electron. Although a hole is not a physical particle like an electron, it can ‘shift’ from one atom to another in a semiconductor.

Question 8. To obtain a p-type germanium semiconductor, it must be doped with

1. Indium
2. Arsenic
3. Antimony
4. Phosphorus

Answer: 1. Indium

When an intrinsic tetravalent semiconductor (such as germanium) is doped with a trivalent impurity (such as indium) atom, the impurity atom leaves an empty space in the bond structure and the intrinsic semiconductor becomes a p-type semiconductor.

Question 9. When arsenic is added as an impurity to silicon, the resulting material is

1. An n-type conductor
2. An n-type semiconductor
3. A p-type semiconductor
4. None of these

Answer: 2. An n-type semiconductor

Arsenic is a pentavalent impurity, which when doped with a silicon semiconductor leaves one free electron. Such free electrons (negative particles) constitute currents and thus the semiconductor is called an n-type semiconductor.

Question 10. Which one of the following statements is false?

1. Pure Si doped with trivalent impurities gives a p-type semiconductor.
2. The majority of carriers in an n-type semiconductor are holes.
3. The minority carriers in a p-type semiconductor are electrons.
4. The resistance of an intrinsic semiconductor decreases with an increase in temperature.

Answer: 2. The majority of carriers in an n-type semiconductor are holes.

In a p-type semiconductor, silicon is doped with a trivalent impurity atom leaving behind a hole in the covalent bonds. Here, the majority of carriers are holes and not electrons

Question 11. The device that can act as a complete electronic circuit is an

1. Junction diode
2. Integrated circuit
3. Zener diode
4. Junction transition

Answer: 2. Integrated circuit

An integrated circuit (IC), also called a chip or a microchip, is a set of electronic circuits on one small flat piece of a semiconductor material (normally silicon)

Question 12. The application of a forward bias to a p-n junction

1. Widens the depletion zone
2. Increases the potential difference across the depletion zone
3. Increases the number of donors on the n-side
4. Decreases the electric field in the depletion zone

Answer: 4. Decreases the electric field in the depletion zone

The electric field across a p-n junction is reduced which thus supports the flow of electric current).

Question 13. Which of the following elements produces an n-type semiconductor when added as an impurity into silicon?

1. B
2. P
3. Al
4. Mg

Answer: 2. P

For the growth of an n-type semiconductor, silicon (a tetravalent semiconductor) has to be doped with a pentavalent impurity, such as phosphorus.

Question 14. A semiconductor device is connected in series with a battery and a resistor. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be

1. A p-type semiconductor
2. An intrinsic semiconductor
3. A p-n junction diode
4. An n-type semiconductor

Answer: 3. A p-n junction diode

A p-n junction diode conducts fairly well in its forward bias. Reversing the polarity of the cell brings the diode in the reverse bias, which restricts the current. Hence, the device is a p-n junction diode.

Question 15. In forward-biasing a p-n junction diode,

1. The positive terminal of the battery is connected to the p-side and the depletion region becomes thick
2. The positive terminal of the battery is connected to the n-side and the depletion layer becomes thin
3. The positive terminal of the battery is connected to the n-side and the depletion region becomes thick
4. The positive terminal of the battery is connected to the p-side and the depletion region becomes thin.

Answer: 4. The positive terminal of the battery is connected to the p-side and the depletion region becomes thin.

In a p-n junction diode with forward bias, the depletion layer becomes thin.

Question 16. A p-n junction photodiode is made of a material having a band gap of 2.0 eV. The minimum frequency of radiation that can be absorbed by the material is nearly

1. 1 x 10¹⁴ Hz
2. 20 x 10¹⁴ Hz
3. 5 x 10¹⁴ Hz
4. 10 x 10¹⁴ Hz

Answer: 3. 5 x 10¹⁴ Hz

The incident photon (radiation) must have an energy of 2.0 eV.

Now, E = hv

∴ \(\mathrm{v}=\frac{E}{h}=\frac{2.0 \mathrm{eV}}{6.67 \times 10^{-34} \mathrm{Js}}=\frac{2.0 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.67 \times 10^{-34} \mathrm{~J} \mathrm{~s}}\)

= 4.8 x 10¹⁴ Hz ≈ 5.0 x 10¹⁴ Hz.

Question 17. A p-n junction photodiode is fabricated from a semiconductor having a band gap of 2.5 eV. It can detect a signal of wavelength

1. 6000 Å
2. 6000 nm
3. 4000 Å
4. 4000 nm

Answer: 3. 4000 Å

Energy of photon = \(E=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E}\)

Given that band gap = E = 2.5 eV and he =1242 eV nm.

∴ \(\lambda=\frac{h c}{E}=\frac{1242 \mathrm{eV} \mathrm{nm}}{2.5 \mathrm{eV}}=496.8 \mathrm{~nm}=4968 Å\)

Hence, the required wavelength will be 4000 Å

Question 18. In a p-n junction photocell, the value of the photoelectromotive force produced by a beam of monochromatic light is proportional to the

1. Barrier voltage at the p-n junction
2. The intensity of the light falling on the cell
3. Frequency of the light falling on the cell
4. Voltage applied at the p-n junction

Answer: 2. Intensity of the light falling on the cell

In a p-n junction photocell, the photoelectromotive force produced by incident monochromatic radiation is proportional to the intensity of the incident radiation.

Question 19. Which of the following bonds produces a solid that reflects light in the visible region whose electrical conductivity decreases with temperature and has a high melting point?

1. Metallic bonding
2. Ionic bonding
3. van der Waals bonding
4. Covalent bonding

Answer: 1. Metallic bonding

For solids with metallic bonding, there is a reflection in the visible range. They also have a high melting point, and the electrical conductivity decreases with an increase in temperature.

Question 20. The carbon, silicon, and germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy-band gaps represented by EGC, EGSi, and EGGe respectively. Which of the following relationships is true in their cases?

1. EG_C > EG_Si
2. EG_C – EG_Si
3. EG_C < EG_Ge
4. EG_C < EG_Si

Answer: 1. EG_C > EG_Si

The energy-band gap in carbon is 5.5 eV, that in silicon is 1,14 eV, and that in germanium is 0.67 eV. Thus, EG_C > EG_Si.

Question 21. A reverse bias applied across a p-n junction diode

1. Lowers the potential barrier
2. Raises the potential barrier
3. Increases the majority-carrier current
4. Increases the minority-carrier current

Answer: 2. Raises the potential barrier

A p-n junction diode operating in a reverse bias has its potential barrier raised across the depletion layer.

Question 22. A p-n junction diode can be used as

1. A capacitor
2. An amplifier
3. A rectifier
4. A regulator

Answer: 3. A rectifier

A p-n junction conducts only in a forward bias and offers high resistance in a reverse bias. Hence, it is used as a rectifier.

Question 23. The cause of the potential barrier in a p-n junction diode is the

1. Depletion of negative charges near the junction
2. The concentration of positive charges near the junction
3. Depletion of positive charges near the junction
4. The concentration of positive and negative charges near the junction

Answer: 4. Concentration of positive and negative charges near the junction

During the formation of a junction diode, holes from the p-region diffuse into the n-region, and electrons from the n-region diffuse into the p-region.

In the course of this diffusion process, a thin layer at the junction becomes free from any charge carriers and is called the depletion layer. There is a potential gradient in the depletion layer – negative on the p-side and positive on the n-side. This is called the potential barrier.

Question 24. A piece of copper and another of germanium are cooled from room temperature to 80 K. Then,

1. The resistance of each will increase
2. The resistance of copper will decrease
3. The resistance of copper will decrease, while that of germanium will increase
4. The resistance of copper will increase, while that of germanium will decrease

Answer: 3. The resistance of copper will decrease, while that of germanium will increase

With a decrease in temperature, the resistance of copper (a metallic conductor) decreases due to a decrease in thermal agitation. Germanium (a semiconductor) has its resistance increased, as the thermal energy is insufficient to free the electrons from the covalent bonds.

Question 25. In a forward bias, the width of the potential barrier in a p-n junction diode

1. Increases
2. Decreases
3. Remains constant
4. First increases, then decreases

Answer: 2. Decreases

As more electrons and holes flow in the depletion region in a forward bias, the number of positive and negative ions is reduced, which causes the width of the potential barrier to narrow.

Question 26. The barrier potential of a p-n junction depends on

• The type of the semiconductor material,
• The amount of doping,
• Temperature.

Which of the following combinations is correct?

1. (1) and (2) only
2. (1) only
3. (2) and (3) only
4. (1), (2), and (3)

Answer: 4. (1), (2) and (3)

The potential barrier of a p-n junction depends on

1. The type of the material (Si: 1.14 V but Ge: 0.67 V),
2. The amount of doping, and
3. The temperature.

Question 27. For an n-type semiconductor, which of the following statements is true?

1. Electrons are majority carriers and trivalent atoms are dopants.
2. Electrons are minority carriers and pentavalent atoms are dopants.
3. Holes are minority carriers and pentavalent atoms are dopants.
4. Holes are majority carriers and pentavalent atoms are dopants.

Answer: 3. Holes are minority carriers and pentavalent atoms are dopants.

Doping an intrinsic tetravalent semiconductor (Si or Ge) by a pentavalent impurity brings out a drastic increase in conductivity due to electrons as the majority charge carriers and holes as the minority charge carriers.

Question 28. If a small amount of antimony is added to a germanium crystal,

1. The antimony atoms become acceptor atoms
2. There will be more free electrons than holes in the semiconductor
3. The resistance of the germanium crystal increases
4. The germanium crystal becomes a p-type semiconductor

Answer: 2. There will be more free electrons than holes in the semiconductor

On adding a pentavalent impurity, such as antimony, to germanium, the intrinsic semiconductor becomes an n-type extrinsic semiconductor. Hence, there will be more free electrons than holes.

Question 29. Choose only the false statement from the following.

1. Substances with energy gaps of the order of 10 eV are insulators.
2. The conductivity of a semiconductor increases with an increase in temperature.
3. In conductors, the valence and conduction bands may overlap.
4. The resistivity of a semiconductor increases with an increase in temperature.

Answer: 4. The resistivity of a semiconductor increases with an increase in temperature.

With an increase in the temperature of a semiconductor, the electrons forming covalent bonds get sufficient thermal energy to get detached. Hence, the conductivity increases and the resistivity decreases. Thus, an increase in resistivity is a false consequence.

Question 30. In a semiconducting material, the mobilities of electrons and holes are μ_e and μ_h respectively. Which of the following is true?

1. \(\mu_{\mathrm{e}}>\mu_{\mathrm{h}}\)
2. \(\mu_{\mathrm{e}}<\mu_{\mathrm{h}}\)
3. \(\mu_e=\mu_h\)
4. \(\mu_{\mathrm{e}}<0 \text { but } \mu_{\mathrm{h}}>0\)

Answer: 1. \(\mu_{\mathrm{e}}>\mu_{\mathrm{h}}\)

The mobility (μ) of an electron or a hole is defined as the drift velocity per unit electric field (μ = v_d/E). The effective mass of an electron is smaller than that of a hole. Hence, the mobility of an electron is higher than that of a hole (μ_e > μ_h).

Question 31. When there is a saturation current in a diode, what is its resistance?

1. Zero
2. Infinity
3. Periodic
4. Data insufficient

Answer: 2. Infinity

The dynamic resistance is defined as the slope of the V-I characteristic curve representing the equation r = ΔV/ΔI. At saturation, ΔI = 0 but

∴ \(\Delta V \neq 0, \text { so } r=\frac{\Delta V}{0}=\infty\)

Question 32. The temperature-dependence of resistivity, ρ = f(T), of a semiconductor, is represented by

Answer: 3.

As the temperature of a semiconductor is increased, the electrons in the valence band gain sufficient energy to escape the pull of their atoms. Consequently, a conduction of electrons results. This decreases the resistivity nonlinearly, as shown in the option (3).

Question 33. The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without any filter is 10 V. The DC component of the output voltage is

1. 10√2 V
2. 10 V
3. \(\frac{10}{\pi} \mathrm{V}\)
4. \(\frac{20}{\pi} \mathrm{V}\)

Answer: 3. \(\frac{10}{\pi} \mathrm{V}\)

The input AC voltage is V = (10 V)sin ωt. The output is rectified to be V = (10 V)sinωt (only for the positive half cycle, as shown). The average output is

⇒ \(V_{\mathrm{av}}=\frac{1}{T} \int V_0 \sin \omega t d t\)

⇒ \(\frac{1}{T}\left[V_0 \int_0^{T / 2} \sin \omega t d t\right]\)

⇒ \(\frac{V_0}{T}\left[\frac{\cos \omega t}{\omega}\right]_{T / 2}^0\)

∴ \(\frac{V_0}{\pi}=\frac{10 \mathrm{~V}}{\pi}\)

Question 34. If a full-wave rectifier circuit is operating from 50-Hz mains, the fundamental frequency of the ripple will be

1. 25 Hz
2. 100 Hz
3. 70.7 Hz
4. 50 Hz

Answer: 2. 100 Hz

In a full-wave rectifier circuit, the output has current and voltage signals for both positive and negative halves of the input sinusoidal cycles. Hence, the frequency of the output will be double that of the input, i.e., 2 x 50 Hz = 100 Hz.

Question 35. For conduction through a p-n junction diode, which of the following is correct?

1. A high potential on the n-side and a low potential on the p-side
2. A high potential on the p-side and a low potential on the n-side
3. The same potential on both the p- and n-sides
4. Undetermined

Answer: 2. A high potential on the p-side and a low potential on the n-side

Conduction occurs in a p-n junction when it is in a forward bias, i.e., when the p-side is connected to its positive terminal (higher potential) and the n-side to its negative terminal (lower potential).

Question 36. In the energy-band diagram of a material shown in the adjoining figure., the empty circles and filled circles represent holes and electrons respectively. The material is

1. An insulator
2. A metal
3. An n-type semiconductor
4. A p-type semiconductor

Answer: 4. A p-type semiconductor

In a p-type semiconductor, holes are the majority carriers (in the valence band) and electrons are the minority carriers, as shown in the energy-band diagram. Hence, the given diagram corresponds to a p-type semiconductor.

Question 37. In the given circuit, the voltage across the load is 12 V. The current in the Zener diode varies from 0 A to 50 mA. What is the maximum wattage of the diode?

1. 12 W
2. 6 W
3. 0.6 W
4. 1.2 W

Answer: 3. 0.6 W

The maximum current through the Zener diode is I_z = 50 mA = 0.05 A, and the voltage across the load is V =12 V.

∴ the maximum, wattage of the diode will be

I_zV = (0.05 A)(12 V) = 0.6 W.

Question 38. When a p-n junction diode is reverse-biased, the resistance measured by an ohmmeter will be

1. High
2. Zero
3. Infinity
4. Low

Answer: 1. High

In the reverse bias, the obstruction to the current through a p-n junction diode is very high. So, the ohmmeter will measure a high resistance.

Question 39. If the internal resistance of the cell is negligible then the current flowing through the circuit is

1. 0.06 A
2. 0.02 A
3. 0.08 A
4. 0.1 A

Answer: 4. 0.1 A

The diode D₁ is forward-biased, while D₂ is reverse-biased. Thus, D₂ offers a high resistance and hence it is nonconducting. The current through the circuit is

∴ \(I=\frac{5 \mathrm{~V}}{20 \Omega+30 \Omega}=\frac{5 \mathrm{~V}}{50 \Omega}=0.1 \mathrm{~A}\)

Question 40. In the given figure, a diode D is connected to an external resistor of resistance R = 100 Ω and a battery of emf 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be

1. 20 mA
2. 35 mA
3. 30 mA
4. 40 mA

Answer: 3. 30 mA

The potential drop across the resistor R is

V = 3.5V – 0.5 V = 3.0 V.

∴ the current through the circuit is

⇒ \(I=\frac{V}{R}=\frac{3.0 \mathrm{~V}}{100 \Omega}=0.03 \mathrm{~A}=30 \mathrm{~mA}\)

Question 41. Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is

1. 0.75 A
2. 0.25 A
3. 0.5 A
4. zero

Answer: 3. 0.5 A

In the given circuit, D₁ is in a forward bias (conducting), whereas D₂ is in a reverse bias (nonconducting). Hence, the effective resistance is R =10Ω and the current delivered by the cell is

∴ \(I=\frac{5 \mathrm{~V}}{10 \Omega}=0.5 \mathrm{~A}\)

Question 42. In the given diagram, the input is across the terminals A and C, and the output is across B and D. Then, the output is

1. Zero
2. The same as the input
3. Half-wave-rectified
4. Full-wave-rectified

Answer: 4. Full-wave-rectified

The given circuit containing four p-n junction diodes forms a bridgelike arrangement, as shown in the circuit diagram, which gives a full-wave rectified output, as shown in the graph given below.

An AC voltage source is applied across AC, which gives a DC output across BD. As shown source in the waveforms, in a positive half cycle across AC, the diodes D₁ and D₄ are conducting, and the current through the load is along BD. Similarly, in the negative half cycle, the diodes D₂ and D₃ are conducting, with the current through R from B to D.

Question 43. The given circuit has two ideal diodes connected as shown in the figure. The current flowing through the resistor of resistance R₁ will be

1. 2.5 A
2. 10.0 A
3. 1.43 A
4. 4.13 A

Answer: 1. 2.5 A

In the given diagram, D₁ is nonconducting.

So, the current through the cell is

∴ \(I=\frac{10 \mathrm{~V}}{R_1+R_3}=\frac{10 \mathrm{~V}}{4 \Omega}=2.5 \mathrm{~A}\)

Question 44. Consider the junction diode as ideal. The value of the current flowing through AB is

1. 10^-2 A
2. 10^-3 A
3. 10^-4 A
4. 0 A

Answer: 1. 10^-2 A

Potential difference = V_A– V_B = 4 V- (-6 V) =10 V.

∴ Current = I = \(\frac{V_{\mathrm{A}}-V_{\mathrm{B}}}{R}=\frac{10 \mathrm{~V}}{1000 \Omega}=0.01 \mathrm{~A}=10^{-2} \mathrm{~A}\)

Question 45. Which one of the following represents a forward-biased diode?

Answer: 1.

A p-n junction diode is in its forward bias when the p-side is at a higher potential compared to the n-side, as shown in option (1).

Question 46. The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of the resistance R connected in series L with the diode for obtaining the maximum current?

1. 6:76 Ω
2. 20 Ω.
3. 5 Ω
4. 5.6 Ω

Answer: 3. 5 Ω

For the diode, the voltage drop is V_D = 0.5 V, and its maximum power rating is P_max = 100 mW = 100 x 10^-3 W.

∴ diode resistance = \(R_{\mathrm{D}}=\frac{V_{\mathrm{D}}^2}{P_{\max }}=\frac{(0.5 \mathrm{~V})^2}{10^{-1} \mathrm{~W}}=2.5 \Omega\).

∴ the current through the diode is

⇒ \(I_D=\frac{V_{\mathrm{D}}}{R_{\mathrm{D}}}=\frac{0.5 \mathrm{~V}}{2.5 \Omega}=0.2 \mathrm{~A}\)

Since this 0.2 A current flows through the circuit, the total resistance is

⇒ \(R_{\mathrm{T}}=\frac{V}{I_{\mathrm{D}}}=\frac{1.5 \mathrm{~V}}{0.2 \mathrm{~A}}=7.5 \Omega\)

the resistance of the series resistor is

R = R_T – R_D = 7.5Ω – 2.5Ω = 5.0Ω.

Question 47. If in a p-n junction, a Square input signal of 10 V is applied as shown in the figure, the output signal across R_L will be

Answer: 2.

The p-n junction diode is conducting in a forward bias. In the given case, it is for V = 0 V to V = 5 V only.

So, the output across R will be 0-5 V, as shown in the option (2).

Question 48. In the case of forward-biasing a p-n junction diode, which of the following figures correctly depicts the direction of the flow of carriers?

Answer: 2.

When a p-n junction is formed, a diffusion of charge carriers across the junction takes place. The charge carriers undergo recombinations with their opposite charges. This is indicated by the direction of the flow of carriers, as shown in option (2).

Question 49. Of the diodes shown in the following diagrams, which one is reverse-biased?

Answer: 3.

In option (3), the p-side is at a zero potential (i.e., earth-connected), while the n-side is at +5 V. This indicates a reverse bias.

Question 50. For a given circuit with an ideal p-n junction diode (D), which of the following statements is correct?

1. In a forward bias, the voltage across R is V.
2. In a reverse bias, the voltage across R is V.
3. In a forward bias, the voltage across R is 2V.
4. In a reverse bias, the voltage across R is 2V.

Answer: 1. In a forward bias, the voltage across R is V.

An ideal diode has effectively zero resistance in its forward bias. So, the potential drop across R is V.

Question 51. The application of a forward bias in a p-n junction

1. Increases the number of donors on the n-side
2. Increases the electricity Held in the depletion zone
3. Increases the potential difference across the depletion zone
4. Widens the depletion zone

Answer: 1. Increases the number of donors on the n-side

In a forward-biased p-n junction diode, there is an increase in the number of donors on the n-side and a decrease in the potential barrier.

Question 52. A Zener diode is used for

1. Producing oscillations in an oscillator
2. Amplification
3. Stabilization
4. Rectification

Answer: 3. Stabilization

A Zener diode is a highly doped p-n junction device used for voltage stabilization.

Question 53. A Zener diode is specified as having a breakdown voltage of 9.1 V with a maximum power dissipation of 364 mW. What is the maximum current that the diode can handle?

1. 40 mA
2. 60 mA
3. 50 mA
4. 45 mA

Answer: 1. 40 mA

Power = voltage x current.

∴ \(I_{\max }=\frac{\text { power }}{\text { voltage }}=\frac{364 \times 10^{-3} \mathrm{~W}}{9.1 \mathrm{~V}}=40 \times 10^{-3} \mathrm{~A}=40 \mathrm{~mA}\)

Question 54. What is the direction of the electric field (in the depletion layer) of a p-n junction diode?

1. From the p-side to the n-side
2. From the n-side to the p-side
3. Randomly oriented
4. No electric field

Answer: 2. From the n-side to the p-side

Across the p-n junction, the electric field \(\overrightarrow{E_{\mathrm{b}}}\) is directed from the n-side to the p-side.

Question 55. A Zener diode having a breakdown voltage equal to 15 V is used in a voltage regulator circuit. The current through the diode is

1. 5 mA
2. 10 mA
3. 15 mA
4. 20 mA

Answer: 1. 5 mA

The p.d. across the l-kΩ resistor is V_z =15 V.

∴ Current = \(I^{\prime}=\frac{15 \mathrm{~V}}{1 \mathrm{k} \Omega}=15 \mathrm{~mA}\)

The p.d. across the 25-Ω resistor is

20 V-15 V = 5 V

∴ Current = I = \(\frac{5 \mathrm{~V}}{250 \Omega}=20 \mathrm{~mA} .\)

Hence, the current through the Zener is

I_z = I-I’ = 20 mA – 15 mA = 5 mA.

Question 56. The given graph represents the V-I characteristics of a semiconductor device. Which of the following statements is correct?

1. It is the V-I characteristic of a solar cell where point A represents the open-circuit voltage and point B is the short-circuit current.
2. It is the V-I characteristics of a solar cell, and the points A and B represent the open-circuit voltage and current respectively.
3. It is the V—I characteristics of a photodiode, and the points A and B represent the open-circuit voltage and current respectively.
4. It is the V-I characteristics of an LED, and the points A and B represent the open-circuit voltage and the short-circuit current respectively.

Answer: 1. It is the V-I characteristic of a solar cell where point A represents the open-circuit voltage and point B is the short-circuit current.

The given graph represents the V-I characteristic curve of a solar cell with V_oc as the open-circuit voltage and I_sc as the short-circuit current.

Question 57. In an unbiased p-n junction, holes diffuse from the p-region to the n-region because of the

1. Attraction of the free electrons of the n-region
2. Higher hole concentration in the p-region than the n-region
3. Higher concentration of electrons in the n-region than in the p-region
4. Potential difference across the p-n junction

Answer: 2. Higher hole concentration in the p-region than the n-region

Diffusion takes place from the higher concentration to the lower concentration. Hence, in an unbiased p-n junction, holes diffuse from the p-side to the n-side as the p-region has a higher hole concentration.

Question 58. For a p-type semiconductor, which of the following statements is true?

1. Electrons are majority carriers and trivalent atoms are dopants.
2. Holes are majority carriers and trivalent atoms are dopants.
3. Holes are majority carriers and pentavalent atoms are dopants.
4. Electrons are majority carriers and pentavalent atoms are dopants.

Answer: 2. Holes are majority carriers and trivalent atoms are dopants.

In a p-type semiconductor, an intrinsic (tetravalent Ge or Si) semiconductor is doped with a trivalent impurity. This creates vacancies of valence electrons called holes, which act as the majority charge carriers.

Question 59. The reverse-breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current I_z through the Zener is

1. 7 mA
2. 17 mA
3. 10 mA
4. 15 mA

Answer: 3. 10 mA

Given that I_z = 5.6 V.

Hence, the voltage across R₁ is 9 V- 5.6 V = 3.4 V.

∴ main current = I = \(\frac{3.4 \mathrm{~V}}{200 \Omega}\)

= 17 mA.

The current through R_z is

∴ \(I_2=\frac{V_{\mathrm{Z}}}{R_{\mathrm{Z}}}=\frac{5.6 \mathrm{~V}}{800 \Omega}=7 \mathrm{~mA}\)

∴ I_z =I – I₂ = 17 mA – 7 mA = 10 mA.

Question 60. At 0.3 V and 0.7 V, the Ge and Si diodes become conducting respectively, as shown in the given figure. If the ends of the Ge diode are reversed, the change in the potential V₀ will be

1. 0.2 V
2. 0.6 V
3. 0.4 V
4. 0.8 V

Answer: 3. 0.4 V

When the Ge and Si diodes are both forward-biased, a current passes through Ge with a potential drop of 0.3 V.

Hence, the initial output voltage is V₀ =12 V- 0.3 V =11.7 V.

When Ge is reverse-biased, a current flows through Si alone with a potential drop of 0.7 V. Hence, the output voltage is

V₀ =1 2 V – 0.7 V =11.3 V.

∴ the change in the output voltage is

V₀ – V₀‘ = 11.7 V – 11.3 V = 0.4 V

Question 61. The current through the tire Zener diode in the given circuit will be

1. 14 mA
2. 5 nrA
3. 9 mA
4. Zero

Answer: 3. 9 mA

The potential drop across the 5-kΩ resistor is

V =120 V – V_z =120 V- 50 V = 70 V.

the main current through the battery is \(I=\frac{70 \mathrm{~V}}{5 \mathrm{k} \Omega}=14 \mathrm{~mA}\)

The current through the 10-kΩ resistor is

∴ [latexI_0=\frac{50 \mathrm{~V}}{10 \mathrm{k} \Omega}=5 \mathrm{~mA}][/latex).

the current through the Zener diode is

I_z = I – I₀ = 14 mA – 5 mA = 9 mA

Question 62. In the given circuit, the current through the Zener diode is close to

1. 6.0 mA
2. 6.7 mA
3. 4.0 mA
4. Zero

Answer: 4. Zero

When the Zener voltage (Vz) reaches 10 V, the potential drop across the R₁ resistor is 12 V – 10 V = 2 V.

∴ main current = \(=\frac{2 \mathrm{~V}}{500 \Omega}=4 \mathrm{~mA}\)

The Zener diode with two R₂ resistors makes a parallel combination.

The current through each R₂ resistor (=1500 Ω.) is

∴ \(I_2=\frac{V_{\mathrm{Z}}}{R_2 / 2}=\frac{10 \mathrm{~V}}{750 \Omega}=13.3 \mathrm{~mA}\)

Since l₂ cannot be greater than the main current, the Zener diode will not reach its breakdown voltage. So, I_z = 0.

Question 63. The adjoining figure represents a voltage-regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance of the circuit is R_L= 4 kΩ. The series resistance
of the circuit is R_i = 1 kΩ. If the battery voltage VB varies from 8 V to 16 V, what are the minimum and maximum values of the current through the Zener diode?

1. 0.5 mA and 8.5 mA
2. 1.5 mA and 8.5 mA
3. 0.5 mA and 6 mA
4. 1 mA and 8.5 mA

Answer: 1. 0.5 mA and 8.5 mA

Given that breakdown voltage = V_z = 6 V.

With V_B = 8 V, the potential drop across R_L is 8V – 6V = 2V.

∴ main current = \(I=\frac{2 \mathrm{~V}}{R_{\mathrm{i}}}=\frac{2 \mathrm{~V}}{1 \mathrm{k} \Omega}=2 \mathrm{~mA}\)

The current through R_L is

⇒ \(I_1=\frac{V_{\mathrm{Z}}}{R_{\mathrm{L}}}=\frac{6 \mathrm{~V}}{4 \mathrm{k} \Omega}=1.5 \mathrm{~mA}\)

∴ I_z = I- I₁ = 2 mA-1.5 mA = 0.5 mA.

With V_B = 16 V

⇒ \(I=\frac{16 \mathrm{~V}-6 \mathrm{~V}}{1 \mathrm{k} \Omega}=10 \mathrm{~mA}\)

The current through R_L is I₁=1.5 mA.

So, I_z = I – I₂ = 10 mA – 1.5 mA = 8.5 mA.

Question 64. The adjoining figure shows a DC-voltage-regulator circuit with a Zener diode of breakdown voltage 6 V. If the unregulated input voltage varies between 10 V and 16 V, what is the maximum Zener current?

1. 7.2 mA
2. 3.5 mA
3. 1.5 mA
4. 2.5 mA

Answer: 2. 3.5 mA

Given that Zener breakdown voltage = V_z = 6 V.

The Zener current will be maximum when the supply voltage is maximum (= 16 V).

The voltage across R_s is 16 V- 6 V = 10 V.

The main current (through R_s) is \(I_{\mathrm{s}}=\frac{10 \mathrm{~V}}{2 \mathrm{k} \Omega}=5 \mathrm{~mA}\)

But \(I_{\mathrm{L}}=\frac{V_{\mathrm{Z}}}{R_{\mathrm{L}}}=\frac{6 \mathrm{~V}}{4 \mathrm{k} \Omega}=1.5 \mathrm{~mA}\)

∴ The maximum current through the zener is

I_z = I_s– I_L = 5 mA-1.5 mA = 3.5 mA.

Question 65. A diode has a potential drop of 0.5 V in its forward bias. The maximum current that can flow through the diode is 10 mA. Find the resistance R to be connected in series with the diode so that the setup can be connected across a battery of 1.5 V.

1. 50 Ω
2. 25 Ω
3. 100 Ω
4. 20 Ω

Answer: 3. 100 Ω

The potential drop across the resistor is

V = IR = 1.5 V- 0.5 V = 1.0 V.

Given that I = 10 mA = 10^-2 A.

the required resistance is

⇒ \(R=\frac{1.0 \mathrm{~V}}{10^{-2} \mathrm{~A}}=100 \Omega\)

Question 66. If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm then its band-gap energy is

1. 1.1 eV
2. 3.1 eV
3. 2.0 eV
4. 1.5 eV

Answer: 2. 3.1 eV

The band gap is equal to the energy (hv) of the photon.

∴ bad gap = \(\Delta E=\frac{h c}{\lambda}=\frac{1240 \mathrm{eV} \mathrm{nm}}{400 \mathrm{~nm}}=3.1 \mathrm{eV}\).

Question 67. The increase in the depletion region in a p-n junction diode is due to

1. Its reverse bias
2. Its forward bias
3. Both its forward and reverse biases
4. An increase in the forward current

Answer: 1. Its reverse bias

In a reverse bias, the potential barrier becomes higher, so the width of the depletion layer is increased and the junction offers a large resistance.

Question 68. The solid materials which have negative temperature coefficients of resistance are

1. Insulators only
2. Semiconductors only
3. Insulators and semiconductors
4. Metals only

Answer: 3. Insulators and semiconductors

The resistance and resistivity of metallic conductors increase with increasing temperature, so they have a positive coefficient of resistance. In (intrinsic) semiconductors, the resistivity decreases with increasing temperature. The same is the case with insulators, in which electrons absorb energy at a higher temperature to contribute to conduction. Hence, both semiconductors and insulators have a negative temperature coefficient of resistance.

The Nucleus

Question 1. If the radius of the \({ }_{13}^{27} \mathrm{Al}\) nucleus is taken to be RAl then the radius of the \({ }_{52}^{125} \mathrm{Te}\) nucleus is nearly

1. \(\left(\frac{52}{13}\right)^{1 / 3} R_{\mathrm{Al}}\)
2. \(\frac{5}{3} R_{\mathrm{Al}}\)
3. \(\frac{3}{5} R_{\mathrm{Al}}\)
4. \(\left(\frac{13}{52}\right)^{1 / 3} R_{\mathrm{Al}}\)

Answer: 2. \(\frac{5}{3} R_{\mathrm{Al}}\)

For a nucleus of mass number A, the radius is given by R = \(R_0 A^{1 / 3}\)

∴ \(\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=\left(\frac{125}{27}\right)^{1 / 3}=\frac{5}{3}\)

⇒ \(R_{\mathrm{Te}}=\frac{5}{3} R_{\mathrm{Al}}\)

Question 2. The mass number of helium is 4 and that for sulphur is 32. The radius of the sulfur nucleus is larger than the helium nucleus by a multiple of

1. 4
2. 2√2
3. 2
4. 8

Answer: 3. 2

Given that A_He = 4 and A_s = 32

∴ \(\frac{R_{\mathrm{S}}}{R_{\mathrm{He}}}=\left(\frac{32}{4}\right)^{1 / 3}=(8)^{1 / 3}=2\)

Hence, R_s = 2R_He.

Question 3. The mass density of a nucleus varies with the mass number as

1. A
2. A²
3. \(\frac{1}{A}\)
4. A₀

Answer: 4. A₀

The mass density of a nucleus is given by

⇒ \(\rho=\frac{\text { mass }}{\text { volume }}=\frac{A m}{\frac{4}{3} \pi R^3}=\frac{A m}{\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3}=\frac{3 m}{4 \pi R_0}\)

which is constant and independent of A.

Thus, p ∝ A⁰.

Question 4. What is the radius of the iodine atom? (Given that atomic number = 53 and mass number = 126.)

1. 2.5 x 10^-11 m
2. 2.5 x 10^-9 m
3. 7 x 10^-9 m
4. 7 x 10^-6 m

Answer: 1. 2.5 x 10^-11 m

The electronic configuration of iodine is 2, 8, 18, 18, 7. Hence the principal quantum number of the outermost electron is n = 5.

∴ the radius of the outermost orbit is

⇒ \(R_n=R_0\left(\frac{n^2}{Z}\right)=(0.53 Å)\left(\frac{5^2}{53}\right)=0.25 Å=2.5 \times 10^{-11} \mathrm{~m}\)

Question 5. The nuclei \({ }_6^{13} \mathrm{C} \text { and }{ }_7^{14} \mathrm{~N}\) can be described as

1. Isotopes
2. Isotones
3. Isobars
4. None

Answer: 2. Isotones

In \({ }_6^{13} \mathrm{C}, n_{\mathrm{p}}\) = 6 and nn = 7.

In \({ }_7^{14} \mathrm{~N}, n_{\mathrm{p}}\) = 7 and «n = 7.

Both have the same number of neutrons (= 7).

Hence, they are isotones of each other.

Question 6. When an a-particle of mass m moving at a velocity v bombards on a heavy nucleus of charge Ze, its distance of the closest approach from the nucleus depends on m as

1. m
2. \(\frac{1}{\sqrt{m}}\)
3. \(\frac{1}{m}\)
4. \(\frac{1}{m^2}\)

Answer: 3. \(\frac{1}{m}\)

For an a-particle, charge = 4e.

For a heavy nucleus, charge = Ze.

By the principle of energy conservation, the kinetic energy of the a-particle gets converted into its PE at the distance of the closest approach (d). Thus,

⇒ \(\frac{1}{2} m v^2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Z e)(4 e)}{d}\)

⇒ \(d=\frac{2 Z e^2}{\pi \varepsilon_0 m v^2} \Rightarrow d \propto \frac{1}{m}\)

Question 7. The binding energy of a deuteron is 2.2 MeV and that of a \({ }_2^4 \mathrm{He}\) nucleus is 28 MeV. If two deuterons are fused to form one \({ }_2^4 \mathrm{He}\) nucleus then the energy released is

1. 23.6 MeV
2. 30.2 MeV
3. 25.8 MeV
4. 19.2 MeV

Answer: 1. 23.6 MeV

A fusion reaction releases energy. The reaction is \({ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{Q}\)

Given that BE of a deuteron \(\left({ }_1^2 \mathrm{H}\right)=2.2 \mathrm{MeV}\) and BE of a helium \(\left({ }_2^4 \mathrm{He}\right) \text { nucleus }=28 \mathrm{MeV}\).

∴ the energy released is

Q = (BE of the daughter nuclei)- (BE of the parent nuclei)

= 28 MeV- 2(2.2 MeV) = 23.6 MeV.

Question 8. The binding energy per nucleon of the \({ }_3^7 \mathrm{Li} \text { and }{ }_2^4 \mathrm{He}\) nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}\) the value of the energy Q released is

1. 19.6 MeV
2. 17.3 MeV
3. -2.4 MeV
4. 8.4 MeV

Answer: 2. 17.3 MeV

The given nuclear reaction is \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}\)

Now, BE of the two \({ }_2^4 \mathrm{He} \text { nuclei }\) = 2(7.06 MeV) A_He = 7.06 MeV x 8

= 56.48 MeV.

and BE of the \({ }_3^7 \mathrm{Li} \text { nucleus }\) = 5.60 MeV x 7

= 39.20 MeV.

∴ the energy released is Q = 56.48 Mev = 17.28 Mev = 17.3 Mev.

Question 9. A certain mass of hydrogen is changed into helium by the process of fusion. The mass defect in this fusion reaction is 0.02866 u. The energy liberated per atomic mass unit is nearly (given that 1 u = 931 MeV)

1. 2.67 MeV
2. 6.67 MeV
3. 26.7 MeV
4. 13.35 MeV

Answer: 2. 6.67 MeV

Given that mass defect = m = 0.02866 u.

The energy released = E = mc².

∴ energy released per amu of \({ }_2^4 \mathrm{He}\) is

⇒ \(\frac{E}{4}=\frac{0.02866 \times 931 \mathrm{MeV}}{4 c^2} \cdot c^2=6.67 \mathrm{MeV}\)

Question 10. The mass of the \({ }_3^7 \mathrm{Li}\) nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of the \({ }_3^7 \mathrm{Li}\) nucleus is nearly

1. 46 MeV
2. 3.9 MeV
3. 5.6 MeV
4. 23 MeV

Answer: 3. 5.6 MeV

Mass defect = Δm = 0.042 u.

So, the binding energy is

BE = Amc² = (0.042)(931 MeV).

∴ BE per nucleon = \(\frac{B E}{A}=\frac{0.042 \times 931 \mathrm{MeV}}{7}=5.58 \mathrm{MeV}=5.6 \mathrm{MeV}\).

Question 11. A nucleus \({ }_Z^A \mathrm{X}\) has a mass represented by m(A, Z). If mp and ran denote the masses of a proton and a neutron respectively, the binding energy is given by

1. \(\mathrm{BE}=\left[m(A, Z)-Z m_{\mathrm{p}}-(A-Z) m_{\mathrm{n}}\right] c^2\)
2. \(\mathrm{BE}=\left[\mathrm{Z} m_{\mathrm{p}}+(A-\mathrm{Z}) m_{\mathrm{n}}-m(A, \mathrm{Z})\right] c^2\)
3. \(\mathrm{BE}=\left[\mathrm{Z} m_{\mathrm{p}}+A m_{\mathrm{n}}-m(A, \mathrm{Z})\right] c^2\)
4. \(\mathrm{BE}=m(A, Z)-Z m_{\mathrm{p}}-(A-Z) m_{\mathrm{n}}\)

Answer: 2. \(\mathrm{BE}=\left[\mathrm{Z} m_{\mathrm{p}}+(A-\mathrm{Z}) m_{\mathrm{n}}-m(A, \mathrm{Z})\right] c^2\)

For the nucleus \({ }_Z^A \mathrm{X}\), the number of protons = Z and the number of neutrons = A-Z.

∴ the total mass of its constituents = initial mass

= Mi = Zm_p + (A-Z)m_n,

where wp and mÿ are the masses of a proton and a neutron respectively. When all the neutrons and protons combine to form a nucleus of mass M, the loss of mass, AM appears as the binding energy of the nucleus.

Thus, BE = [Zm_p + (A-Z)m_n – m(A, Z)]c².

Question 12. In the reaction \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n}\), if the binding energies of \({ }_1^2 \mathrm{H},{ }_1^3 \mathrm{H} \text { and }{ }_2^4 \mathrm{He}\) are respectively a, b and c then the energy released in this reaction is

1. c + a – b
2. c – a – b
3. a + b – c
4. a + b + c

Answer: 2. c – a – b

The given reaction is \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n}\)

In this nuclear fusion reaction, the reactants \(\left({ }_1^2 \mathrm{H} \text { and }{ }_1^3 \mathrm{H}\right)\) fuse together to form a more stable nucleus \(\left({ }_2^4 \mathrm{He}\right)\) having more binding energy with the release of an energy Q.

∴ energy released = (BE of the final product)- (BE of the reactants)

= c – (a + b) = c – a – b.

Question 13. If in a nuclear fusion process, the masses of the fusing nuclei are m₁ and m₂ and the mass of the resultant nucleus is m₃ then

1. m₃ = m₁ + m₂
2. m3 > m₁ + m₂
3. m₃ < m₁ + m₂
4. m₃ = m₁ – m₂

Answer: 3. m₃ < m₁ + m₂

In a nuclear fusion reaction, the final mass of the fused nucleus is less than the total mass of the fused nuclei. This loss of mass appears as the energy released.

∴ m₁+ m₂ > m₃ => m₃ < m₁ + m₂.

Question 14. The nuclear reaction \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_{\mathrm{Z}+1}^A \mathrm{Y}+{ }_{-1}^0 \mathrm{e}+\overline{\mathrm{v}}\) represents a

1. Fission
2. Fusion
3. γ-decay
4. β-decay

Answer: 4. β-decay

In the given reaction, an electron \(\left({ }_{-1}^0 \mathrm{e}\right)\) and an antineutrino \((\bar{v})\) are emitted. Such a reaction represents a β-decay.

Question 15. A fission of nuclei is possible because of the binding energy per nucleon in them

1. Increases with the mass number for high mass numbers
2. Increases with the mass number for low mass numbers
3. Decreases with the mass number for high mass numbers
4. Decreases with the mass number for low mass numbers

Answer: 3. Decreases with the mass number for high mass numbers

From the curve showing the BE per nucleon versus the mass number (A), it is observed that the nuclei with higher values of A (like \({ }^{235} \mathrm{U}\)) have lower values of BE/A, and they are thus unstable and can get fissioned easily. Further, for such heavier nuclei, BE/A decreases with an increase in A.

Question 16. Let m_p denote the mass of a proton and m_n that of a neutron. A given nucleus of binding energy BE contains Z protons and N neutrons. The mass of the nucleus is given by

1. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{P}}-(\mathrm{BE}) c^2\)
2. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{P}}+(\mathrm{BE}) c^2\)
3. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{p}}-\frac{\mathrm{BE}}{c^2}\)
4. \(m(N, Z)=N m_{\mathrm{n}}+\mathrm{Z} m_{\mathrm{p}}+\frac{\mathrm{BE}}{c^2}\)

Answer: 3. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{p}}-\frac{\mathrm{BE}}{c^2}\)

The binding energy of a nucleus with Z protons and N neutrons is given by

BE = Amc² – [Zm_p + Nm_n– m(N, Z)]c².

∴ \(\frac{\mathrm{BE}}{c^2}=Z m_{\mathrm{p}}+N m_{\mathrm{n}}-m(N, Z)\)

⇒ \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{p}}-\frac{\mathrm{BE}}{c^2}\)

Question 17. The mass of a proton is 1.0073 u and that of a neutron is 1.0087 u. The binding energy of \({ }_2^4 \mathrm{He}\) is (mass of helium nucleus being 4.0015 u)

1. 28.4 MeV
2. 0.061 keV
3. 0.0305 J
4. 0.0305 erg

Answer: 1. 28.4 MeV

Total mass of the constituents = 2(m_p + m_n)

= 2(1.0073 +1.0087) u = 4.032 u. J

Mass of the fHe nucleus = M = 4.0015 u.

∴ BE of the helium nucleus = (4.032 – 4.0015) x 931 MeV

= 0.0305 x 931 MeV = 28.4 MeV.

Question 18. In the fission reaction \({ }_{92}^{236} \mathrm{U} \rightarrow{ }^{117} \mathrm{X}+{ }^{117} \mathrm{Y}+\mathrm{n}+\mathrm{n}\) the binding energy per nucleon of X and Y is 8.5 MeV, whereas that of \({ }_2^4 \mathrm{He}\) is 7.6 MeV. The total energy liberated will be about

1. 200 MeV
2. 2000 MeV
3. 2 MeV
4. 1 keV

Answer: 1. 200 MeV

The binding energy of the nuclei X and Y taken together is

BE_f = (8.5 + 8.5)(117 MeV) = 1989 MeV.

The binding energy of the \({ }^{236} \mathrm{U}\) nucleus is

BE_i = 236(7.6 MeV) = 1793.6 MeV.

the total energy liberated is

ΔE = BE_i – BE_f = (1989-1793.6) MeV

= 195.4 MeV ≈ 200 MeV.

Question 19. The average binding energy of a nucleon inside an atomic nucleus is about

1. 8 eV
2. 8 erg
3. 8 MeV
4. 8 J

Answer: 3. 8 MeV

The average binding energy of a nucleon (proton or neutron) is the average value of the BE per unit mass number (BE/A). The (BE/A)-versus-A curve has a broad maximum in the range from A = 30 to A =120. This corresponds to an average value of BE/A of about 8 MeV.

Question 20. The energy released in the fission of a single
\({ }_{92}^{235} \mathrm{U}\) nucleus is 200 MeV. The fission rate of a \({ }_{92}^{235} \mathrm{U}\)-filled nuclear reactor operating at a power level of 5 W is

1. 1.56 x l0¹⁰ s^-1
2. 1.56 X 10¹⁶ s^-1
3. 1.56 x l0¹¹ s^-1
4. 1.56 x l0¹⁷ s^-1

Answer: 3. 1.56 x l0¹¹ s^-1

Output power = (number of fissions of \({ }^{235} \mathrm{U}\) per unit time)(200 MeV)

⇒ 5 J s^-1 = N(200MeV)

∴ \(N=\frac{5 \mathrm{~J} \mathrm{~s}^{-1}}{\left(200 \times 10^6\right)\left(1.6 \times 10^{-19} \mathrm{~J}\right)}=\frac{5 \times 10^{11} \mathrm{~J} \mathrm{~s}^{-1}}{3.2 \mathrm{~J}}=1.56 \times 10^{11} \mathrm{~s}^{-1}\).

Question 21. The binding energy per nucleon is maximum in the case of

1. \({ }_2^4 \mathrm{He}\)
2. \({ }_{26}^{56} \mathrm{Fe}\)
3. \({ }_{56}^{141} \mathrm{Ba}\)
4. \({ }_{92}^{235} \mathrm{U}\)

Answer: 2. \({ }_{26}^{56} \mathrm{Fe}\)

In the (BE/A)-versus-A curve, the maximum occurs at (8.8 MeV) /A for \({ }_{26}^{56} \mathrm{Fe}\).

Question 22. If the binding energy per nucleon in the \({ }_3^7 \mathrm{Li} \text { and }{ }_2^4 \mathrm{He}\) nuclei are respectively 5.60 MeV and 7.06 MeV, the energy of each proton in the reaction \({ }_3^7 \mathrm{Li}+\mathrm{p} \rightarrow 2{ }_2^4 \mathrm{He}\) is

1. 2.4 MeV
2. 17.3 MeV
3. 8.4 MeV
4. 19.6 MeV

Answer: 2. 17.3 MeV

The total BE in the \({ }_3^7 \mathrm{Li}\) nucleus is 7(5.60 MeV) = 39.20 MeV.

The total BE in the two \({ }_2^4 \mathrm{He}\) nuclei is 2(4 x 7.06 MeV) = 56.48 MeV.

KE of the proton = difference in BE

= 56.48 MeV – 39.20 MeV

= 17.28 MeV ≈ 17.3 MeV.

Question 23. The stable nucleus that has a radius half that of \({ }^{56} \mathrm{Fe}\) is

1. \({ }^7 \mathrm{Li}\)
2. \({ }^{21} \mathrm{Na}\)
3. \({ }^{16} \mathrm{~S}\)
4. \({ }^{40} \mathrm{~K}\)

Answer: 1. \({ }^7 \mathrm{Li}\)

The radius of a nucleus is R = R₀A ^1/3, where A is the mass number.

Here, \(\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3}\)

⇒ \(\frac{R}{R / 2}=\left(\frac{56}{A_2}\right)^{1 / 3} \Rightarrow 2^3=8=\frac{56}{A_2} \Rightarrow A_2=7\)

Thus, the required nucleus is \({ }^7 \mathrm{Li}\).

Question 24. If nip and mn are the masses of a proton and a neutron respectively, for an element of mass m having Z protons and N neutrons,

1. m > Zm_p + Nmn
2. m = Zm_p + Nm_n
3. m < Zm_p + Nm_n
4. m may be greater than, less than, or equal to Zm_p + Nm_n, depending on the nature of the element

Answer: 3. m < Zm_p + Nm_n

During the formation of a nucleus, the mass of the nucleus is less than that of its constituents, and the mass defect appears as energy. Thus,

m < Zm_p + Nm_n.

Question 25. In a nuclear fission reaction, energy is released because the

1. The total mass of the products is more than the mass of the nucleus
2. The total binding energy of the products formed due to the nuclear fission is more than that of the fissionable parent materials
3. The total binding energy of the products formed due to the nuclear fission is less than that of the fissionable parent materials
4. The combined mass of some particles is converted into energy

Answer: 2. Total binding energy of the products formed due to the nuclear fission is more than that of the fissionable parent materials

In a nuclear fission reaction, the products (or daughter nuclei) are more stable, because their binding energy per nucleon (BE/A) is increased. Thus, the total BE of the products formed is more than that of the reactants.

Question 26. Which of the following are suitable for a fusion process?

1. Light nuclei
2. Heavy nuclei
3. Elements lying in the middle of the periodic table
4. Elements lying in the middle of the binding-energy curve

Answer: 1. Light nuclei

Lighter nuclei have smaller values of BE/A, hence they are unstable. At higher temperatures, light nuclei fuse so as to increase BE/A and become more stable. Hence, light nuclei are suitable for a fusion process.

Question 27. Solar energy is mainly caused due to the

1. Fission of the uranium nuclei present in the sun
2. Gravitational contraction
3. Fusion of protons during the synthesis of heavier elements
4. Burning of hydrogen in oxygen

Answer: 3. Fusion of protons during the synthesis of heavier elements

Solar energy is produced by a thermonuclear process in which lighter nuclei (or protons) fuse together to form heavier elements.

Question 28. In any fission process, the ratio of the total mass of the fission products to that of the parent nuclei

1. Is less than 1
2. Is greater than 1
3. Is equal to 1
4. Depends on the masses of the parent nuclei

Answer: 1. Is less than 1

In nuclear fission, a heavy nucleus breaks up into comparatively lighter nuclei with the release of energy. This energy liberated is due to the loss of mass. Hence, the total mass of the fission products is less than the total mass of the parent nuclei.

∴ \(\frac{\text { total mass of the fission products }}{\text { total mass of the parent nuclei }}<1\)

Question 29. A nucleus of uranium at rest decays into nuclei of thorium and helium. Then,

1. The helium nucleus has more kinetic energy than the thorium nucleus
2. The helium nucleus has less momentum than the thorium nucleus
3. The helium nucleus has more momentum than the thorium nucleus
4. The helium nucleus has less kinetic energy than the thorium nucleus

Answer: 1. The helium nucleus has more kinetic energy than the thorium nucleus

The given reaction is \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{90}^{232} \mathrm{Th}+{ }_2^4 \mathrm{He}\).

The initial momentum before the decay is zero.

By the principle of conservation of linear momentum, \(\left|p_{\mathrm{Th}}\right|=\left|p_{\mathrm{He}}\right|\), and they move in opposite directions. But KE = p2/2m. So,

⇒ \(\mathrm{KE}_{\mathrm{He}}=\frac{p^2}{2 m_{\mathrm{He}}} \text { and } \mathrm{KE}_{\mathrm{Th}}=\frac{p^2}{2 m_{\mathrm{Th}}}\)

∵ \(m_{\mathrm{He}}<m_{\mathrm{Th}}\),

∴ \(\mathrm{KE}_{\mathrm{He}}>\mathrm{KE}_{\mathrm{Th}}\).

Question 30. A nucleus \({ }_n^m \mathrm{X}\) emits an a-particle and two (1-particles. The resulting nucleus is

1. \({ }_n^{m-6} Z\)
2. \({ }_n^{m-4} \mathrm{X}\)
3. \({ }_{n-2}^{n-4} \mathrm{Y}\)
4. \({ }_{n-4}^{m-6} \mathrm{Z}\)

Answer: 2. \({ }_n^{m-4} \mathrm{X}\)

The given nuclear reaction is \({ }_n^m \mathrm{X} \stackrel{\alpha}{\longrightarrow}{ }_{n-2}^{m-4} \mathrm{Y} \stackrel{2 \beta}{\longrightarrow}{ }_n^{m-4} \mathrm{Z}\)

The elements Z and X have the same atomic number (= n), so they are identical.

Hence, the resulting nucleus is \({ }_n^{m-4} \mathrm{X}\).

Question 31. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter nucleus is an

1. Isobar of the parent nucleus
2. The isomer of the parent nucleus
3. Isotone of the parent nucleus
4. Isotope of the parent nucleus

Answer: 4. Isotope of the parent nucleus

For simplicity, consider the decay with one a-particle and two p-particles. Thus, for a given nucleus \({ }_Z^A \mathrm{X}\) we have

⇒ \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_2^4 \alpha+{ }_{\mathrm{Z}-2}^{A-4} \mathrm{Y}\)

Further,

⇒ \({ }_{\mathrm{Z}-2}^{A-4} \mathrm{Y} \rightarrow{ }_2^{-1}{ }_{-1}^0 \beta+{ }_{\mathrm{Z}}^{A-4} \mathrm{Z}\)

Thus, \({ }_Z^A \mathrm{X} \text { and }{ }_Z^{A-4} \mathrm{Z}\) have the same atomic number (= Z) but different mass numbers (respectively A and A- 4). Hence, they are isotopes of each other.

Question 32. The nuclei of which of the following pairs of nuclei are isotones?

1. \({ }_{34}^{74} \mathrm{Se} \text { and }{ }_{31}^{71} \mathrm{Ga}\)
2. \({ }_{42}^{92} \mathrm{Mo} \text { and }{ }_{40}^{92} \mathrm{Zr}\)
3. \({ }_{38}^{84} \mathrm{Sr} \text { and }{ }_{38}^{86} \mathrm{Sr}\)
4. \({ }_{20}^{40} \mathrm{Ca} \text { and }{ }_{16}^{32} \mathrm{~S}\)

Answer: 1. \({ }_{34}^{74} \mathrm{Se} \text { and }{ }_{31}^{71} \mathrm{Ga}\)

Isotones are the nuclides that have the same number of neutrons, which means the same value of A-Z.

For, \({ }_{34}^{74} \mathrm{Se}\) A-Z = 74- 34 = 40.

For, \({ }_{31}^{71} \mathrm{Ga}\) A-Z = 71-31 = 40.

Hence, \({ }_{34}^{74} \mathrm{Se} \text { and }{ }_{31}^{71} \mathrm{Ga}\) are isotones of each other.

Question 33. An unstable nucleus of mass M emits a photon of frequency v, and the nucleus recoils. The recoil energy will be

1. zero
2. hv
3. \(\frac{h^2 v^2}{2 M c^2}\)
4. Mc²– hv

Answer: 3. \(\frac{h^2 v^2}{2 M c^2}\)

Since the nucleus is initially at rest, the total momentum of the photon and the nucleus is zero.

∴ momentum of the photon = \(\frac{h}{\lambda}=\frac{h v}{c}=p\)

∴ the recoil KE of the nucleus is

⇒ \(\frac{p^2}{2 M}=\frac{1}{2 M}\left(\frac{h v}{c}\right)^2=\frac{h^2 v^2}{2 M c^2}\).

Question 34. In the nuclear decay \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_{\mathrm{Z}+1}^A \mathrm{Y} \rightarrow{ }_{\mathrm{Z}-1}^{A-4} \mathrm{~B} \rightarrow{ }_{\mathrm{Z}-1}^{A-4} \mathrm{~B}\), the particles emitted, in the sequence, are

1. β, α, γ
2. γ, β, α
3. β, γ, α
4. α, β, γ

Answer: 1. β, α, γ

Consider the subsequent transitions in the same order below:

⇒ \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_{\mathrm{Z}+1}^A \mathrm{Y}+{ }_{-1}^0 \beta\)  (β-emission),

⇒ \(\stackrel{\mathrm{Z}+1}{A} \mathrm{Y} \rightarrow{ }_{\mathrm{Z}-1}^{A-4} \mathrm{~B}+{ }_2^4 \mathrm{He}\)  (α-emission)

and \({ }_{Z-1}^{A-4} \mathrm{~B} \rightarrow{ }_{Z-1}^{A-4} \mathrm{~B}\)  (γ-emission).

Hence, the sequence is β, α, γ.

Question 35. The radius of a germanium (Ge) nuclide is measured to be twice the radius of a \({ }_4^9 \mathrm{~B}\) nuclide. The number of nucleons in the Ge nuclide is

1. 73
2. 74
3. 75
4. 72

Answer: 4. 72

The number of nucleons is the mass number (A) of the nucleus.

Now, R = R₀A^1/3

∴ \(\frac{R_{\mathrm{Ge}}}{R_{\mathrm{B}}}=\left(\frac{A}{9}\right)^{1 / 3}\)

⇒ \(\frac{2}{1}=\left(\frac{A}{9}\right)^{1 / 3} \Rightarrow 2^3=\frac{A}{9}\)

∴ A = 72 = number of nucleons.

Question 36. The volume occupied by an atom is greater than the volume of the nucleus by a factor of about

1. 10¹⁰
2. 10¹⁵
3. 10²
4. 10⁵

Answer: 2. 10¹⁵

The average atomic size is of the order of 10^-10 m.

The nuclear radius is around1 fm = 10^-15 m

Hence, \(\frac{\text { volume of the atom }}{\text { volume of the nucleus }}=\frac{\frac{4}{3} \pi\left(10^{-10} \mathrm{~m}\right)^3}{\frac{4}{3} \pi\left(10^{-15} \mathrm{~m}\right)^3}=\left(10^5\right)^3=10^{15}\).

Question 37. When a deuterium is bombarded on an \({ }_8^{16} \mathrm{O}\) nucleus, an α-particle is emitted. The product nucleus is.

1. \({ }_7^{13} \mathrm{~N}\)
2. \({ }_5^{10} \mathrm{~B}\)
3. \({ }_4^9 \mathrm{Be}\)
4. \({ }_7^{14} \mathrm{~N}\)

Answer: 4. \({ }_7^{14} \mathrm{~N}\)

The given reaction is \({ }_1^2 \mathrm{H}+{ }_8^{16} \mathrm{O} \rightarrow{ }_2^4 \alpha+{ }_Z^A \mathrm{X}\)

where the product nucleus has the mass number A and the atomic number Z. Since the mass number is conserved,

2 +16 = 4 + A ⇒ A = 14.

The atomic number is conserved, so

l + 8 = 2 + Z

Z = 9 – 2 = 7.

Hence, the product nucleus is \({ }_7^{14} \mathrm{~N}\)

Question 38. In the reaction \(\mathrm{X}(\mathrm{n}, \alpha) \rightarrow{ }_3^7 \mathrm{Li}\), the element X is

1. \({ }_2^4 \mathrm{He}\)
2. \({ }_5^{10} \mathrm{~B}\)
3. \({ }_5^9 \mathrm{~B}\)
4. \({ }_4^{11} \mathrm{Be}\)

Answer: 2. \({ }_5^{10} \mathrm{~B}\)

The given nuclear reaction can be expressed as

⇒ \({ }_{\mathrm{Z}}^A \mathrm{X}+{ }_0^1 \mathrm{n} \rightarrow{ }_3^7 \mathrm{Li}+{ }_2^4 \mathrm{He}\)

Conserving the mass number,

A + l = 7 + 4 ⇒ A = 10.

Conserving the atomic number,

Z + 0 = 3 + 2 ⇒ Z = 5.

Thus, \({ }_Z^A \mathrm{X}={ }_5^{10} \mathrm{X}\)corresponds to the element boron \(\left({ }_5^{10} B\right)\)

Question 39. If a nuclear decay is expressed as \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta^{+}+\mathrm{X}\), the unknown particle is a/an

1. Neutron
2. Antineutrino
3. Proton
4. Neutrino

Answer: 4. Neutrino

The beta-positive decay reaction is given as

⇒ \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+{ }_1^0 \beta+{ }_{\mathrm{Z}}^A \mathrm{X}\)

Conserving the mass number,

11 = 11 + 0 + A ⇒ A = 0.

Conserving the atomic number,

6 = 5 +1 + Z ⇒ Z = 0.

With β⁺ activity, an unknown particle is emitted. This particle has a zero charge (Z = 0) and a zero mass (A = 0). Hence, the unknown particle is a neutrino.

Question 40. Complete the equation for the following fission process: \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{38}^{90} \mathrm{~S}+\ldots\)

1. \({ }_{54}^{143} \mathrm{Xe}+3{ }_0^1 \mathrm{n}\)
2. \({ }_{54}^{145} \mathrm{Xe}\)
3. \({ }_{57}^{142} \mathrm{Xe}\)
4. \({ }_{54}^{142} \mathrm{Xe}+{ }_0^1 \mathrm{n}\)

Answer: 1. \({ }_{54}^{143} \mathrm{Xe}+3{ }_0^1 \mathrm{n}\)

Let the unknown product be \({ }_Z^A \mathrm{X}\). Hence,

⇒ \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{38}^{90} \mathrm{Sr}+{ }_{\mathrm{Z}}^A \mathrm{X}\).

Conserving the mass number,

235 +1 = 90 + A => A = 146.

Conserving the atomic number,

92 + 0 = 38 + Z ⇒ Z = 54

Science \({ }_{54}^{146} \mathrm{Xe}\) is not given as the product, the product must be \({ }_{54}^{143} \mathrm{Xe}\) and \(3{ }_0^1 \mathrm{n}\), which gives A = 143 + 3 = 146 and Z = 54.

Question 41. A nucleus \({ }_n^m X\) emits one a- and two p-particles. The resulting nucleus is

1. \(\underset{n}{m-4} X\)
2. \({ }_{n-2}^{m-4} \mathrm{Y}\)
3. \({ }_{n-4}^{m-4} \mathrm{Z}\)
4. None of these

Answer: 1. \(\underset{n}{m-4} X\)

The given decay process is expressed as

⇒ \({ }_n^m \mathrm{X} \rightarrow{ }_2^4 \mathrm{He}+2{ }_{-1}^0 \beta\)

We can express the decay in the two steps shown below:

⇒ \({ }_n^m \mathrm{X} \rightarrow{ }_2^4 \mathrm{He}+{ }_{n-2}^{m-4} \mathrm{Y}\) (α-decay)

and \({ }_{n-2}^{m-4} \mathrm{Y} \rightarrow 2_{-1}^0 \beta+{ }_n^{m-4} \mathrm{Z}\) (β-decay)

Since \({ }_n^m \mathrm{X}\) and the final product \({ }_n^{m-4} \mathrm{Z}\) have the same atomic number (= n), X and Z may be the same element. Hence, the resulting nucleus is \({ }_n^{m-4} \mathrm{X}\).

Question 42. A nucleus at rest breaks into two fragments, which have their velocities in the ratio 2:1. What will be the ratio of their nuclear sizes (radii)?

1. \(2^{1 / 3}: 1\)
2. \(1: 2^{1 / 3}\)
3. \(3^{1 / 2}: 1\)
4. \(1: 3^{1 / 2}\)

Answer: 2. \(1: 2^{1 / 3}\)

Since the linear momentum is conserved,

⇒ \(\left|m_1 v_1\right|=\left|m_2 v_2\right|\)

Scince \(\frac{v_1}{v_2}=\frac{2}{1}\)

⇒ \(\frac{m_1}{m_2}=\frac{1}{2}=\frac{\frac{4}{3} \pi R_1^3 \rho}{\frac{4}{3} \pi R_2^3 \rho}\)

∴ \(\left(\frac{R_1}{R_2}\right)^3=\frac{1}{2} \Rightarrow R_1: R_2=1: 2^{1 / 3}\).

Question 43. An element A decays into another element C by the following two-step process: \(\mathrm{A} \rightarrow \mathrm{B}+{ }_2^4 \mathrm{He}; \quad \mathrm{B} \rightarrow \mathrm{C}+2 \mathrm{e}\) Then,

1. A and C are isotopes
2. A and C are isobars
3. A and B are isotopes
4. A and B are isobars

Answer: 1. A and C are isotopes

Let the element A have the mass number A and the atomic number Z.

So, in the first case,

⇒ \({ }_{\mathrm{Z}}^A \mathrm{~A} \rightarrow{ }_{\mathrm{Z}-2}^{A-4} \mathrm{~B}+{ }_2^4 \mathrm{He}\)

Similarly, in the second case,

⇒ \({ }_{Z-2}^{A-4} \mathrm{~B} \rightarrow{ }_Z^{A-4} \mathrm{C}+2{ }_{-1}^0 \mathrm{e}\)

Since the elements A and C have the same atomic number (= Z) but different mass numbers, they are isotopes of each other.

Question 44. What is the binding energy per nucleon in the \({ }_{26}^{56} \mathrm{Fe}\) nucleus? [Given that m\(\left({ }^{56} \mathrm{Fe}\right)\) = 55.936 u, mn = 1.00866 u and mp = 1.007274 u.]

1. 8.52 MeV
2. 10.56 MeV
3. 20.52 MeV
4. 50.5 MeV

Answer: 1. 8.52 MeV

The nucleus \({ }_{26}^{56} \mathrm{Fe}\) consists of 26 protons and 30 neutrons.

∴ 26 m_p = 26(1.007274 u) = 26.189124 u

and 30m_n = 30(1.00866 u) = 30.2598 u.

The total mass of the constituents = 56.448924 u.

Mass of the nucleus = m(\({ }_{26}^{56} \mathrm{Fe}\)) = 55.936 u.

∴ mass loss = ΔM = 0.513 u.

∴ the binding energy per nucleon is

∴ \(\frac{\Delta M(931 \mathrm{MeV})}{56}=8.52 \mathrm{MeV}\)

Question 45. Consider the nuclear fission \({ }^{20} \mathrm{Ne} \rightarrow 2^4 \mathrm{He}+{ }^{12} \mathrm{C}\). Given that the binding energy per nucleon of \({ }^{20} \mathrm{Ne},{ }^4 \mathrm{He} \text { and }{ }^{12} \mathrm{C}\) are respectively 8.03 MeV, 7.07 MeV and 7.86 MeV. Identify the correct statement.

1. An energy of 3.6 MeV will be released.
2. An energy of 9.72 MeV will be absorbed.
3. An energy of 8.3 MeV will be released.
4. An energy of 11.9 MeV has to be supplied

Answer: 2. An energy of 9.72 MeV will be absorbed.

The Q-value of a nuclear reaction is given by

ΔQ = (BE of the products)-(BE of the reactants)

Energy is absorbed if ΔQ is negative and released if ΔQ is positive.

Hence, BE of \({ }^{20} \mathrm{Ne}\) = 20(8.03 MeV) =160.6 MeV,

BE of two \({ }^4 \mathrm{He}\) = 2(7.07 MeV)4 = 56.56 MeV

and BE of \({ }^{12} \mathrm{C}\) = 12(7.86 MeV) = 94.32 MeV.

∴ BE of all the products = 150.88 MeV.

∴ ΔQ =.150.88 MeV – 160.6 MeV = -9.72 MeV.

Hence, 9.72 MeV is the amount of energy absorbed.

Question 46. The ratio of the mass densities of the nuclei of \({ }^{40} \mathrm{Ca}\) and \({ }^{16} \mathrm{O}\) is close to

1. 3
2. 5
3. 1
4. 0.1

Answer: 3. 1

The density of the nuclear matter is constant and is independent of the size of the nucleus. Thus,

∴ \(\frac{R_{\mathrm{Ca}}}{R_{\mathrm{O}}}=1\)

Question 47. In a radioactive-decay chain, the initial nucleus is \({ }_{90}^{232} \mathrm{Th}\). In the end, there are six a-particles, and four which have been emitted. If the resulting nucleus is \({ }_{\mathrm{Z}}^A \mathrm{X}\) the values of A and Z are respectively

1. 202 and 80
2. 208 and 82
3. 200 and 81
4. 208 and 80

Answer: 2. 208 and 82

The nuclear reaction can be expressed as

⇒ \({ }_{90}^{232} \mathrm{Th} \rightarrow 6\left({ }_2^4 \mathrm{He}\right)+4{ }_{-1}^0 \beta+{ }_Z^A \mathrm{X}\)

Now, A + 4(0) + 6(4) = 232 ⇒ A = 208.

Also, Z + 4(-1) + 6(2) = 90 ⇒ Z = 82.

Question 48. The masses of various atomic particles are given below:

mp =1.0072 u, mn =1.0087 u, me = 0.00548 u, mv = 0 and md = 2.0141 u,

where p is a proton, n is a neutron, e is an electron, \(\bar{v}\) is an antineutrino and d is a deuteron. Which of the following processes is allowed by the conservation of momentum and energy?

1. n + n → a deuterium atom (with an electron bound to the nucleus)
2. \(\mathrm{p} \rightarrow \mathrm{n}+\overline{\mathrm{e}}+\overline{\mathrm{v}}\)
3. \(\overline{\mathrm{e}}+\mathrm{e} \rightarrow \gamma\)
4. n + p → d + γ

Answer: 4.

A nuclear reaction occurs when the mass of the products is less than the mass of the reactants. Let us consider the given options one by one.

1. n + n → a deuterium atom is energetically not feasible.
2. p → n + \(\overline{\mathrm{e}}+\overline{\mathrm{v}}\) is energetically not feasible.
3. e + \(\overline{\mathrm{e}}\) → γ is incorrect as the momentum is not conserved.
4. n + p → d + γ is energetically possible with the momentum conserved

Question 49. Given that mass of \({ }_3^7 \mathrm{Li}=1.0160 \mathrm{u}\), mass of \({ }_2^4 \mathrm{He}=4.0026 \mathrm{u}\) and mass of \({ }_1^1 \mathrm{H}=1.0070\). When 20 g of \({ }_3^7 \mathrm{Li}\) is converted into \({ }_2^4 \mathrm{He}\) by proton capture, the energy released is

1. 6.82 x 10⁵ kW h
2. 8 x 10⁶ kW h
3. 1.35 x l0⁶ kW h
4. 4.5 x 10⁵ kW h

Answer: 3. 1.35 x l0⁶ kW h

For the reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow 2{ }_2^4 \mathrm{He}\), we obtain.

⇒ \(m\left({ }_3^7 \mathrm{Li}\right)+m\left({ }_1^1 \mathrm{H}\right) \rightarrow 2 m\left({ }_2^4 \mathrm{He}\right)\).

Mass loss = Δm = [7.0160 +1.0070- 2(4.0026)] u = 0.0178 u.

For 20 g of \({ }_3^7 \mathrm{Li}\) the energy released will be

⇒ \(E=\left(\frac{20}{7} N_{\mathrm{A}}\right)(0.0178 \mathrm{u})\), where NA is the Avogadro constant

⇒ \(\frac{20}{7} \times\left(6 \times 10^{23} \mathrm{~mol}^{-1}\right) \times 0.0178 \mathrm{u}\)

⇒ \(\frac{20 \times 6 \times 10^{23} \times 0.0178 \times 10^9 \times 1.6 \times 10^{-19}}{7 \times 1000 \times 3600} \mathrm{~kW} \mathrm{~h}\),

∴ 1.35 x l0⁶ kW h.

Question 50. The radius (R) of a nucleus of mass number A can be estimated by the formula R = (1.3 x l0^-15 m)A^1/3. It follows that the mass density of the nuclear matter is of the order of (given that m_p = m_n ≈ 1.67 x l0^-27 kg)

1. 10³ kg m^-3
2. 10¹⁷ kg m^-3
3. 10²⁴ kg m^-3
4. 10¹⁰ kg m^-3

Answer: 2. 10¹⁷ kg m^-3

Density = \(\frac{\text { mass }}{\text { volume }}=\frac{m_{\mathrm{p}} A}{\frac{4}{3} \pi R^3}=\frac{m_{\mathrm{p}} A}{\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3}\)

⇒ \(\frac{3}{4 \pi} \cdot \frac{m_{\mathrm{p}}}{R_0^3}=\frac{3}{4 \pi} \cdot \frac{1.67 \times 10^{-27} \mathrm{~kg}}{\left(1.3 \times 10^{-15} \mathrm{~m}\right)^3}\)

⇒ \(\frac{3 \times 1.67}{4 \pi(1.3)^3} \times 10^{18} \mathrm{~kg} \mathrm{~m}^{-3}\)

∴ 1.81 x l0¹⁷ kg m^-3

Question 51. When the uranium nuclide \({ }_{92}^{235} \mathrm{U}\) is bombarded with a neutron, it generates the nuclide \({ }_{36}^{89} \mathrm{Kr}\), three neutrons and the nuclide

1. \({ }_{40}^{91} \mathrm{Zr}\)
2. \({ }_{36}^{101} \mathrm{Kr}\)
3. \({ }_{36}^{103} \mathrm{Kr}\)
4. \({ }_{56}^{144} \mathrm{Ba}\)

Answer: 4. \({ }_{40}^{91} \mathrm{Zr}\)

⇒ \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{92}^{236} \mathrm{U} \rightarrow{ }_{36}^{89} \mathrm{Kr}+{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+3{ }_0^1 \mathrm{n}\)

Conserving the mass number,

235 + 1 = 89 + 3 + A ⇒ A =144.

Conserving the atomic number,

92 = 36 + Z ⇒ Z = 56.

Hence, the given nucleus \({ }_Z^A X\) is actually \({ }_{56}^{144} \mathrm{Be}\).

Question 52. The energy equivalent of 0.5 g of matter is

1. 4.5 x l0¹⁶ J
2. 4.5 x 10¹³ J
3. 0.5 x l0¹³ J
4. 1.5 x l0¹³ J

Answer: 2. 4.5 x 10¹³ J

From Einstein’s mass-energy equivalence,

E = mc² = (0.5 x 10^-3 kg)(3 x 108 m s^-1)²

= 4.5 x l0¹³ J

Radioactivity

Question 1. The half-life of a radioactive material is 3 h. If the initial amount is 300 g then after 18 h it will remain

1. 4.68 g
2. 9.375 g
3. 48.8 g
4. 93.75 g

Answer: 1. 4.68 g

Given that total time = t =18 h and half-life = T_1/2 = 3 h.

∴ the number of half-lives is

⇒ \(n=\frac{t}{T_{1 / 2}}=\frac{18 \mathrm{~h}}{3 \mathrm{~h}}=6\)

Now, the amount of active material left after 6 half-lives is

∴ \(N=\left(\frac{1}{2}\right)^6 N_0=\frac{300 \mathrm{~g}}{64}=4.68 \mathrm{~g}\)

Alternative method

⇒ \(\begin{aligned}
300 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 150 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 75 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 37.5 \mathrm{~g} \\
\stackrel{T_{1 / 2}}{\longrightarrow} 18.75 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 9.37 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 4.68 \mathrm{~g}
\end{aligned}\)

After 6 half-lives, 4.68 g of active nuclei remains

Question 2. The half-lives of two radioactive elements A and B are respectively 20 min and 40 min. Initially, the samples of A and B have equal numbers of nuclei. After 80 minutes, the ratio of the remaining numbers of the A and B nuclei will be

1. 1:16
2. 1:4
3. 1 :1
4. 4: 1

Answer: 2. 1:4

For A, half-life = T_A = 20 min, and for B, half-life = T_B = 40 min.

Total time = 80 min = 4T_A = 2T_B

∴ \(N_{\mathrm{A}}=\left(\frac{1}{2}\right)^4 N_0 \text { and } N_{\mathrm{B}}=\left(\frac{1}{2}\right)^2 N_0\)

Hence, the ratio between the numbers of active nuclei of A and B is

∴ \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{\left(\frac{1}{2}\right)^4}{\left(\frac{1}{2}\right)^2}=\frac{4}{16}=\frac{1}{4} \Rightarrow N_{\mathrm{A}}: N_{\mathrm{B}}=1: 4\)

Question 3. The half-life of radium is 1600 y. The fraction of a sample of radium that would remain active after 6400 y is

1. \(\frac{1}{4}\)
2. \(\frac{1}{2}\)
3. \(\frac{1}{8}\)
4. \(\frac{1}{16}\)

Answer: 4. \(\frac{1}{16}\)

For radium, half-life = T_1/2 = 1600 y.

Total time = t = 6400 y = 4T_1/2.

⇒ \(N=\left(\frac{1}{2}\right)^4 N_0\)

Hence, the fraction of nuclei which remain active will be

∴ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^4=\frac{1}{16}\)

Question 4. The activity of a radioactive sample is measured as 9750 counts min^-1 at t = 0 and 975 counts min^-1 at t = 5 min. The decay constant is approximately

1. 0.92 mm^-1
2. 0.69 min^-1
3. 0.46 min^-1
4. 0.23 min^-1

Answer: 3. 0.46 min^-1

Activity (A) is measured as the rate of disintegration.

∵ \(N=N_0 \mathrm{e}^{-\lambda .}\)

∴ \(A=\left|\frac{d N}{d t}\right|=N_0 \lambda \mathrm{e}^{-\lambda t}=A_0 \mathrm{e}^{-\lambda t}\)

where AQ is the activity at t = 0.

Given that A0 = 9750 counts min^-1 at t = 0, and A = 975 counts min^-1 at t = 5 min.

∴ \(A=A_0 \mathrm{e}^{-\lambda t} \Rightarrow \frac{A_0}{A}=\frac{9750}{975}=\mathrm{e}^{\lambda t}\)

∴ In 10 = λ(5 min)

∴ decay constant = \(\lambda=\frac{\ln 10}{5 \min }=\frac{2.3}{5 \mathrm{~min}}=0.46 \mathrm{~min}^{-1}\)

Question 5. The relation between disintegration constant (λ) and half-life (T_1/2) is

1. \(\lambda=\frac{\log _{10} 2}{T}\)
2. \(\lambda=\frac{\log _2 \mathrm{e}}{T}\)
3. \(\lambda=\frac{\log _{\mathrm{e}} 2}{T}\)
4. \(\lambda=\frac{\log _{\mathrm{e}} T_{1 / 2}}{2}\)

Answer: 3. \(\lambda=\frac{\log _{\mathrm{e}} 2}{T}\)

Since the radioactive decay follows the exponential law, N = N₀e^-λt.

After one half-life, N₀ changes to N₀/2

⇒ \(\frac{N_0}{2}=N_0 \mathrm{e}^{-\lambda T} \Rightarrow 2=\mathrm{e}^{\lambda T}\)

∴ \(\lambda T=\ln 2 \Rightarrow \lambda=\frac{\ln 2}{T}=\frac{\log _e^2}{T}\)

Question 6. A sample of a radioactive element contains 4 x 10¹⁰ active nuclei. If the half-life of the element is 10 days, the number of decayed nuclei after 30 days is

1. 0.5 x l0¹⁰
2. 2 x l0¹⁰
3. 3.5 x l0¹⁰
4. 1 x l0¹⁰

Answer: 3. 3.5 x l0¹⁰

Given that N₀ = 4 x l0¹⁰, half-life = T_1/2=10 days, and total time = t = 30 days.

number of half-lives = n = \(\frac{t}{T_{1 / 2}}=\frac{30 \text { days }}{10 \text { days }}=3\)

Hence, the number of active nuclei after 3 half-lives is

⇒ \(N=\left(\frac{1}{2}\right)^3 N_0=\frac{N_0}{8}\)

∴ the number of decayed nuclei is

∴ \(N_0-N=N_0\left(1-\frac{1}{8}\right)=\frac{7}{8} N_0=\frac{7}{8}\left(4 \times 10^{10}\right)=3.5 \times 10^{10}\).

Question 7. The half-life of a radioactive sample is 6 h. After 24 h, its activity is 0.01 μCi. What was the initial activity?

1. 0.04 Ci
2. 0.08 μCi
3. 0.16 μCi
4. 0.24 Ci

Answer: 3. 0.16 μCi

The activity at the time t is

⇒ \(A(t)=A_0 \mathrm{e}^{-\lambda t}\),

where A₀ = initial activity.

Given that half-life = T_1/2 = 6 h and total time = t = 24 h.

The activity at the time t is 0.01 μCi.

∴ \(0.01 \mu \mathrm{Ci}=A_0 \mathrm{e}^{-(\ln 2)(t / T)}\)

∴ initial activity = \(A_0=(0.01 \mu \mathrm{Ci}) \mathrm{e}^{(\ln 2)(24 / 6)}\)

= (0.01 (μCi)e^{41n 2} = (0.01 μCi)(16) = 0.16 μCi.

Question 8. The half-life of radium is 1600 y. What fraction of the same remains undecayed after four half-lives?

1. \(\frac{1}{8}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{16}\)
4. \(\frac{1}{24}\)

Answer: 3. \(\frac{1}{16}\)

Given that half-life = T_1/2 = 1600 y and total time = t = 4T_1/2 = 4 half-lives.

∴ the number of undecayed (active) nuclei after 4 half-lives will be

⇒ \(N=\left(\frac{1}{2}\right)^4 N_0=\frac{N_0}{16}\)

∴ required fraction = \(\frac{N}{N_0}=\frac{1}{16}\)

Question 9. M alpha particles per second are emitted from N atoms of a radioactive element. The half-life of the radioactive element is

1. \(\frac{M}{N} \mathrm{~s}\)
2. \(\frac{N}{M} \mathrm{~s}\)
3. \(\frac{N(0.693)}{M} \mathrm{~s}\)
4. \(\frac{M(0.693)}{N} \mathrm{~s}\)

Answer: 3. \(\frac{N(0.693)}{M} \mathrm{~s}\)

For a nuclear disintegration,

⇒ \(N=N_0 \mathrm{e}^{-\lambda t} \text { and } A=\frac{d N}{d t}=\lambda N_0 \mathrm{e}^{-\lambda t}=\lambda N\)

Given that disintegration rate = A = M s-1

∴\(M=\lambda N \Rightarrow \lambda=\frac{M}{N}\)

∴ half life \(T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{\ln 2}{M / N} \mathrm{~s}=\frac{N}{M}(0.693) \mathrm{s}\)

Question 10. The half-life of a radioactive element is 5 y. What percentage of it will remain active after 25 y?

1. 25%
2. 6.25%
3. 1.25%
4. 3.125%

Answer: 4. 3.125%

Given that half-life = T_1/2 = 5 y and total time = t = 25 y = 5(5 y) = 5T_1/2.

After 5 half-lives, the number of active (undecayed) nuclei will be

⇒ \(N=\left(\frac{1}{2}\right)^5 N_0=\frac{N_0}{32}\)

∴ \(\frac{N}{N_0}=\frac{1}{32}=\frac{100 \%}{32}=3.125 \%\)

Question 11. Two radioactive substances A and B have their decay constants 5λ. and A respectively. At t = 0, they have the same number of nuclei. The ratio of the number of active nuclei of A to those of B will be 1/e² after a time period of

1. 4λ
2. 2λ
3. \(\frac{1}{2 \lambda}\)
4. \(\frac{1}{4 \lambda}\)

Answer: 3. \(\frac{1}{2 \lambda}\)

The number of active nuclei of element A is N_A = N₀e^-5λt and that of element B is N_B = N₀e^-λt.

Given that \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{1}{\mathrm{e}^2}=\frac{\mathrm{e}^{-5 \lambda t}}{\mathrm{e}^{-\lambda t}}=\frac{1}{\mathrm{e}^{4 \lambda t}}\)

4λt = 2.

Hence, the required time is t = \(t=\frac{1}{2 \lambda}\)

Question 12. Nl atoms of a radioactive element emit N₂ beta particles per second. The decay constant of the element is

1. \(\frac{N_1}{N_2}\)
2. \(\frac{N_2}{N_1}\)
3. N₁ In 2
4. N₂ In 2

Answer: 2. \(\frac{N_2}{N_1}\)

Activity = \(A=\left|\frac{d N}{d t}\right|=\lambda N_0 \mathrm{e}^{-\lambda t}=\lambda N\)  → (1)

Since N₂ beta particles are emitted from N₁ nuclei, we have

A = N₂ S^-1 and N = N₁

From (1), N₂= N₁.

Hence, decay constant = \(=\lambda=\frac{N_2}{N_1}\)

Question 13. The half-life of a radioactive substance is 30 min. The time taken between the 40% decay and the 85% decay of this radioactive substance is

1. 45 min
2. 15 min
3. 60 min
4. 30 min

Answer: 3. 60 min

∵ 40% decay = 60% active nuclei,

∴ \(N_1=\frac{60}{100} N_0\)

∵ 85% decay =15% active nuclei,

∴ \(N_2=\frac{15}{100} N_0\)

∴ \(\frac{N_2}{N_1}=\frac{1}{4}=\left(\frac{1}{2}\right)^2\)

Thus, the time duration is 2 half-lives = 2(30 min) = 60 min

Question 14. A mixture consists of two radioactive substances A₁ and A₂ having the half-lives 20 s and 10 s respectively. Initially, the mixture has 40 g of A₁ and 160 g of A₂. The amount of the two in the mixture will become equal after a period of

1. 60 s
2. 80 s
3. 40 s
4. 20 s

Answer: 3.

For A₁, half-life = 20 s.

For A₂, half-life = 10 s.

Let A₁ be reduced to N from N₀ in the time t = (t/T_1/2) half-lives

∴ \(N=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}=(40 \mathrm{~g})\left(\frac{1}{2}\right)^{t / 20 \mathrm{~s}}\)

Similarly, for A₂,

⇒ \(N=(160 \mathrm{~g})\left(\frac{1}{2}\right)^{t / 10 \mathrm{~s}}\)

∴ \(\left(\frac{1}{2}\right)^{t / 20 s}=4\left(\frac{1}{2}\right)^{t / 10 s}\)

⇒ \(\frac{t}{20 \mathrm{~s}}+2=\frac{t}{10 \mathrm{~s}}\)

⇒ \(\frac{t}{20 \mathrm{~s}}=2 \Rightarrow t=40 \mathrm{~s}\)

Question 15. A sample of a radioactive element has a mass of 10 g at the instant t = 0. The approximate mass of the element in the sample after two mean lives is

1. 3.70 g
2. 1.35 g
3. 6.30 g
4. 2.50 g

Answer: 2. 1.35 g

At f = 0, M₀ = 10 g.

Mean life = \(\tau=\frac{1}{\lambda}\)

∴ the active mass after 2 mean lives (= 2/λ) will be

∴ \(M=M_0 \mathrm{e}^{-\lambda(2 / \lambda)}=M_0 \mathrm{e}^{-2}=\frac{10 \mathrm{~g}}{\mathrm{e}^2}=\frac{10 \mathrm{~g}}{7.39}=1.35 \mathrm{~g}\)

Question 16. If the half-lives of a radioactive substance for its α- and β -decays are 4 y and 12 y respectively, the ratio of the initial activity to that after 12 y will be

1. 6.25%
2. 12.5%
3. 25%
4. 50%

Answer: 1. 6.25%

For the α-decay, T_α= 4 y.

For the β-decay, T_β =12 y

∴ net activity = \(\frac{d N_{\mathrm{A}}}{d t}=\frac{d N_{\mathrm{B}}}{d t}+\frac{d N_{\mathrm{C}}}{d t}\)

∴ \(\lambda_{\text {eff }} N_{\mathrm{A}}=\lambda_1 N_{\mathrm{A}}+\lambda_2 N_{\mathrm{A}}\)

⇒ \(\lambda_{\text {eff }}=\lambda_1+\lambda_2\)

⇒ \(\frac{1}{\lambda_{\text {eff }}}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{T_1+T_2}{T_1 T_2}\)

∴ effective half-life = \(T_{\text {eff }}=\frac{T_1 T_2}{T_1+T_2}=\frac{4 \times 12}{16} \mathrm{y}=3 \mathrm{y}\)

∴ number of half-lives = n = \(n=\frac{12 y}{3 y}=4\)

∴ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^4=\frac{1}{16}=6.25 \%\)

Question 17. The half-life of a substance is 20 min. What is the time interval between its 33% decay and 67% decay?

1. 40 min
2. 25 min
3. 20 min
4. 30 min

Answer: 3. 20 min

Given that half-life = T_1/2 = 20 min.

Assume that at a time t₁, undecayed nuclei = (100- 33)%.

∴ N₁ = (67%)N₀.

Similarly, assume that at a time t₂,

N₂ = (100- 67)%. N₀ = (33%)N₀.

∴ \(\frac{N_1}{N_2}=\frac{N_0 \mathrm{e}^{-\lambda t_1}}{N_0 \mathrm{e}^{-\lambda t_2}} \Rightarrow \frac{67}{33}=\mathrm{e}^{\lambda\left(t_2-t_1\right)}\)

Taking \(\frac{67}{33}=2\), we have

⇒ \(2=\mathrm{e}^{\left(t_2-t_1\right) \ln 2 / T}=\mathrm{e}^{\ln 2^{\Delta t / T}}=2^{\Delta t / T}\)

∴ \(\frac{\Delta t}{T}=1\).

Hence, the required time interval is Δt = T_1/2 = one half-life = 20 min

Question 18. A radioisotope X with a half-life of 1.4 x l0⁹ y decays to Y, which is a stable isotope. A sample of a rock from a cave was found to contain X and Y in a ratio 1:7. The age of the rock is

1. 1.96 x l0⁹ y
2. 3.92 x l0⁹ y
3. 8.40 x l0⁹ y
4. 4.20 x l0⁹ y

Answer: 4. 4.20 x l0⁹ y

Let us assume that at t = 0, the total number of active X nuclei = N₀ which decay to Y nuclei such that N_x : N_Y =1: 7.

∴ \(N_{\mathrm{X}}=\frac{N_0}{8} \text { and } N_{\mathrm{Y}}=\frac{7 N_0}{8}\)

Now, \(N_{\mathrm{X}}=\frac{N_0}{8}=\left(\frac{1}{2}\right)^3 N_0\)

This occurs after 3 half-lives. Thus,

required time = t = \(3 T_{1 / 2}=3\left(1.4 \times 10^9 \mathrm{y}\right)=4.2 \times 10^9 \mathrm{y}\).

Alternative method

⇒ \(N_{\mathrm{X}}=\frac{N_0}{8}=N_0 \mathrm{e}^{-\lambda t}\)

⇒ \(8=2^3=\mathrm{e}^{\lambda t}\)

∴ \(\ln 2^3=\ln \mathrm{e}^{\lambda t} \Rightarrow 3 \ln 2=\lambda t\)

⇒ \(t=3\left(\frac{\ln 2}{\lambda}\right)=3 T=3\left(1.4 \times 10^9 \mathrm{y}\right)=4.2 \times 10^9 \mathrm{y}\)

Question 19. A sample of a radioactive material decays at the rate of 500 per second at a certain time. The activity falls to 200 per second after 50 min. What is the decay constant (λ) of the sample? (Given that In 2.5 = 0.916.)

1. 6.10 x l0^-4 s^-1
2. 3.05 x l0^-4 s^-1
3. 5.0 x 10^-4 s^-1
4. 1.0 x l0^-4 s^-1

Answer: 2. 3.05 x l0^-4 s^-1

Initial decay rate = A₀ = 500 s^-1.

⇒ A(t) = A₀e^-λt

⇒ 200 s^-1 = (500 s-1 )e^-λ(5° min)

2.5 = e^{λ(50 min)}

∴ In 2.5 = In e^{λ(50 min)}

⇒ In 2.5 = λ(50 min).

Hence, decay constant = \(\frac{\ln 2.5}{50 \times 60 \mathrm{~s}}=\frac{0.916}{30} \times 10^{-2} \mathrm{~s}^{-1}\)

= \(3.05 \times 10^{-1} \mathrm{~s}^{-1}\).

20. In a radioactive material, the activity at a time t₁ is R₁, and at a later time t₂, it is R₂. If the decay constant of the material is λ then

1. \(R_1=R_2 \mathrm{e}^{-\lambda\left(t_1-t_2\right)}\)
2. \(R_1=R_2 \mathrm{e}^{\lambda\left(t_1-t_2\right)}\)
3. \(R_1=R_2\left(\frac{t_2}{t_1}\right)\)
4. R1 = R2

Answer: 1. \(R_1=R_2 \mathrm{e}^{-\lambda\left(t_1-t_2\right)}\)

We know that A(t) = A₀e^-λt.

∴ at \(t=t_1, R_1=R_0 \mathrm{e}^{-\lambda t_1} \text { and at } t=t_2, R_2=R_0 \mathrm{e}^{-\lambda t_2}\)

Hence, \(\frac{R_1}{R_2}=\mathrm{e}^{-\lambda\left(t_1-t_2\right)}\)

⇒ \(R_1=R_2 \mathrm{e}^{-\lambda\left(t_1-t_2\right)}\).

21. The half-life of a radioactive isotope X is 20 y. It decays to another element Y, which is stable. The two elements X and Y were found to be in the ratio 1: 7 in a sample of a given rock. The age of the rock is estimated to be

1. 40 y
2. 100 y
3. 60 y
4. 80 y

Answer: 3. 60 y

The solution is similar to that of Q.18.

Only the half-life is changed to T_1/2 = 20 y.

Hence, the age of the rock will be 3 half-lives, i.e., T_1/2 = 3(20 y) = 60 y

22. The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and N₀/e counts per minute at t = 5 min. The time period at which the activity reduces to half its value is

1. \(\ln \frac{2}{5} \min\)
2. \(\frac{5}{\ln 2} \min\)
3. \(5 \log 2 \mathrm{~min}\)
4. \(5 \ln 2 \mathrm{~min}\)

Answer: 4. \(5 \ln 2 \mathrm{~min}\)

At t = 0, activity = N₀.

At t = 5 min,

activity = \(\frac{N_0}{\mathrm{e}}=N_0 \mathrm{e}^{-\lambda(5 \mathrm{~min})}\)

⇒ \(\lambda(5 \mathrm{~min})=1 \Rightarrow \lambda=\frac{1}{5 \mathrm{~min}}\)

When the activity reduces to half, the time taken is one half-life. Thus,

∴ \(t=T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{\ln 2}{1 / 5 \mathrm{~min}}=(5 \ln 2) \mathrm{min}\).

Question 23. A radioactive isotope is being produced at a constant rate α per second. Its decay constant is λ. If N₀ is the number of nuclei at a time t = 0, the maximum possible number of nuclei is

1. \(N_0+\frac{\alpha}{\lambda}\)
2. \(N_0\)
3. \(\frac{\lambda}{\alpha}+N_0\)
4. \(\frac{\alpha}{\lambda}\)

Answer: 4. \(\frac{\alpha}{\lambda}\)

The number of nuclei present will be maximum when the decay rate is equal to the rate at which isotopes are being produced.

Now, rate of decay = \(=\left|\frac{d N}{d t}\right|=N_0 \lambda e^{-\lambda t}=\lambda N\) and rate of formation = a.

∴ λN = α

Hence, the maximum number of nuclei is N = α/λ.

Question 24. If the half-life of a radioactive sample is 69.3 h, how much of the initial number will decay between the tenth and eleventh hours?

1. 1%
2. 2%
3. 3%
4. 4%

Answer: 1. 1%

The number of active nuclei at the time t₁ = 10 h is N₁ = N₀e^-10λ, and that at the time t₂ =11 h is N₂ = N₀e^-11λ.

∴ percentage decay = \(\frac{N_1-N_2}{N_1} \times 100 \%\)

⇒ \(\frac{N_0 \mathrm{e}^{-10 \lambda}-N_0 \mathrm{e}^{-11 \lambda}}{N_0 \mathrm{e}^{-10 \lambda}} \times 100 \%=\left(1-\frac{1}{e^\lambda}\right) \times 100 \%\)

Given that half-life = \(T_{1 / 2}=69.3 \mathrm{~h}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda}\)

∴ decay constant = λ = 0.01 h-1 .

∴ \(\% \text { decay }=\left(1-\frac{1}{\mathrm{e}^{0.01}}\right) \times 100 \%=(1-0.99) \times 100 \%=1 \%\)

Question 25. Two radioactive materials A and B have their decay constants 10λ and λ respectively. If initially they have the same number of nuclei, the ratio of the number of nuclei of A and B will be 1/e after a time period of

1. \(\frac{1}{9 \lambda}\)
2. \(\frac{1}{11 \lambda}\)
3. \(\frac{1}{10 \lambda}\)
4. \(\frac{11}{10 \lambda}\)

Answer: 1. \(\frac{1}{9 \lambda}\)

For the element A, \(N_{\mathrm{A}}=N_0 \mathrm{e}^{-10 \lambda t}\)

For the element B, \(N_{\mathrm{B}}=N_0 \mathrm{e}^{-\lambda t}\)

∴ \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{1}{\mathrm{e}}=\frac{\mathrm{e}^{-10 \lambda t}}{\mathrm{e}^{-\lambda t}}=\frac{1}{\mathrm{e}^{9 \lambda t}}\)

∴ \(9 \lambda t=1 \Rightarrow t=\frac{1}{9 \lambda}\)

Question 26. A sample of a radioactive material A that has an activity of 10 mCi (given that 1 Ci = 3.7 x 10¹⁰ decays per second) has twice the number of nuclei as another sample of a different radioactive material B, which has an activity of 20 mCi. The correct choices for the half-lives of A and B would then be respectively

1. 20 d and 10 d
2. 5 d and 10 d
3. 10 d and 40 d
4. 20 d and 5 d

Answer: 4. 20 d and 5 d

The activity (A) of a radioactive sample is the rate of disintegration. So,

⇒ \(A=\left|\frac{d N}{d t}\right|=\lambda N=\left(\frac{\ln 2}{T}\right) N\),

where T is the half-life of the material.

∴ \(\frac{A_{\mathrm{A}}}{A_{\mathrm{B}}}=\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}} \cdot \frac{T_{\mathrm{B}}}{T_{\mathrm{A}}} \Rightarrow \frac{10 \mathrm{mCi}}{20 \mathrm{mCi}}=\frac{2 N_{\mathrm{B}}}{N_{\mathrm{B}}} \cdot \frac{T_{\mathrm{B}}}{T_{\mathrm{A}}}\)

∴ \(\frac{T_{\mathrm{A}}}{T_{\mathrm{B}}}=\frac{4}{1}\)

This is true for the option (4).

Question 27. Using a nuclear counter, the count rate of particles emitted from a radioactive source is measured. At t = 0, it was 1600 per second, and at t = 8 s, it was 100 per second. The count rate observed at t = 6 s is close to

1. 400 s^-1
2. 200 s^-1
3. 150 s^-1
4. 300 s^-1

Answer: 2. 200 s^-1

The count rate is the same as the activity, i.e., A = \(A=\left|\frac{d N}{d t}\right|\)

But \(N=N_0 \mathrm{e}^{-\lambda t}\)

So, \(A=\left|\frac{d N}{d t}\right|=\lambda N_0 \mathrm{e}^{-\lambda t}\)

A(t) = A(0)e^-λt.

Given that A(0) = 1600 s^-1 and A(8) = 100s^-1.

⇒ \(\frac{A(0)}{A(8)}=\frac{1600}{100}=\mathrm{e}^{\lambda .8} \Rightarrow 2^4=\left(\mathrm{e}^{2 \lambda}\right)^4\)

⇒ \(\mathrm{e}^{2 \lambda}=2 \Rightarrow \lambda=\frac{1}{2} \ln 2=\ln \sqrt{2}\)

At t = 6 s,

∴ \(A(t)=A(6 \mathrm{~s})=N_0 \mathrm{e}^{-6 \lambda}=\left(1600 \mathrm{~s}^{-1}\right) \mathrm{e}^{-6 \ln \sqrt{2}}=\frac{1600 \mathrm{~s}^{-1}}{8}=200 \mathrm{~s}^{-1}\).

Question 28. Two radioactive samples A and B have their decay constants 5λ and λ respectively. At a time t = 0, the samples have equal numbers of active nuclei. The time taken for the ratio of the numbers of nuclei to become 1/e² will be

1. \(\frac{2}{\lambda}\)
2. \(\frac{1}{\lambda}\)
3. \(\frac{1}{4 \lambda}\)
4. \(\frac{1}{2 \lambda}\)

Answer: 4. \(\frac{1}{2 \lambda}\)

⇒ \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{N_0 \mathrm{e}^{-5 i t}}{N_0 \mathrm{e}^{-\lambda, t}}=\mathrm{e}^{-4 i t}\) → (1)

Bur \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{1}{\mathrm{e}^2}=\mathrm{e}^{-2}\) → (2)

From (1) and (2)

∴ \(4 \lambda t=2 \Rightarrow t=\frac{1}{2 \lambda}\)

Question 29. The half-lives of two radioactive samples A and B are 10 min and 20 min respectively. If initially a sample has equal numbers of the nuclei of A and B then after 60 min the ratio of the decayed numbers of the nuclei A and B will be

1. 1: 8
2. 9: 8
3. 8: 1
4. 3: 8

Answer: 2. 9: 8

The number of active nuclei at a time t is N = \(N_0 \mathrm{e}^{-\lambda t}\)

⇒ \(\Delta N=N_0-N=N_0\left(1-\mathrm{e}^{-\lambda t}\right)\)

∴ \(\frac{\Delta N_{\mathrm{A}}}{\Delta N_{\mathrm{B}}}=\frac{1-\mathrm{e}^{-\lambda_{\mathrm{A}} t}}{1-\mathrm{e}^{-\lambda_{\mathrm{B}} t}}=\frac{1-\mathrm{e}^{-\left[(\ln 2) / T_{\mathrm{A}}\right] t}}{1-\mathrm{e}^{-\left[(\ln 2) / T_{\mathrm{B}}\right] t}}\)

⇒ \(\frac{1-\mathrm{e}^{-[(\ln 2) / 10] \times 60}}{1-\mathrm{e}^{-[(\ln 2) / 20] \times 60}}=\frac{1-\frac{1}{2^6}}{1-\frac{1}{2^3}}=\frac{63}{64} \times \frac{8}{7}=\frac{9}{8}\)

∴ \(\Delta N_{\mathrm{A}}: \Delta N_{\mathrm{B}}=9: 8\).

Question 30. At a time t = 0, the activity of two radioactive elements A and B are equal. After a time t = T, the ratio of their activities decreases according to e^-3t. If the half-life of A is In 2, the half-life of B will be

1. 4 In 2
2. \(\frac{1}{4} \ln 2\)
3. \(\frac{1}{2} \ln 2\)
4. 2 In 2

Answer: 2. \(\frac{1}{4} \ln 2\)

For the given samples of A and B,

⇒ \(A_{\mathrm{A}}=A_0 \mathrm{e}^{-\lambda_{\mathrm{A}} t} \text { and } A_{\mathrm{B}}=A_0 \mathrm{e}^{-\lambda_{\mathrm{B}} t}\)

∴ \(\frac{A_{\mathrm{B}}}{A_{\mathrm{A}}}=\mathrm{e}^{\left(\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}\right) t}\)

Given that \(\frac{A_{\mathrm{B}}}{A_{\mathrm{A}}}=\mathrm{e}^{-3 t}\)

Hence, \(\left(\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}\right) t=-3 t\)

But \(\lambda_{\mathrm{A}}=\frac{\ln 2}{T_{\mathrm{A}}}=\frac{\ln 2}{\ln 2}=1\)

So, \(\lambda_{\mathrm{B}}=4\)

∴ \(T_{\mathrm{B}}=\frac{\ln 2}{\lambda_{\mathrm{B}}}=\frac{1}{4} \ln 2\).

Question 31. The (9/16)th part of a radioactive sample remains active after a time t. What fraction of the same remains undecayed after a time t/2?

1. \(\frac{9}{16}\)
2. \(\frac{5}{8}\)
3. \(\frac{3}{4}\)
4. \(\frac{4}{7}\)

Answer: 3. \(\frac{3}{4}\)

In the first case,

⇒ \(\frac{9}{16} N_0=N_0 \mathrm{e}^{-\lambda t} \Rightarrow \frac{9}{16}=\mathrm{e}^{-\lambda t}\)

In the second case,

⇒ \(x N_0=N_0 \mathrm{e}^{-\lambda t / 2} \Rightarrow x^2=\mathrm{e}^{-\lambda t}\)

∴ \(x^2=\frac{9}{16} \text { and } x=\frac{3}{4}\)

Question 32. A radioactive nucleus A decays into B (having a half-life of 10 s) and into C (having a half-life of 100 s). The equivalent half-life for both emissions is

1. 6 s
2. 9 s
3. 2 s
4. 3 s

Answer: 2. 9 s

The rate of disintegration is given by

⇒ \(-\frac{d N}{d t}=\lambda_1 N+\lambda_2 N=\left(\lambda_1+\lambda_2\right) N\)

∴ \(\lambda_{\text {eq }} N=\left(\lambda_1+\lambda_2\right) N\)

⇒ \(\lambda_{\text {eq }}=\lambda_1+\lambda_2 \Rightarrow \frac{\ln 2}{T}=\frac{\ln 2}{T_1}+\frac{\ln 2}{T_2}\)

∴ \(T=\frac{T_1 T_2}{T_1+T_2}=\frac{(10 \mathrm{~s})(100 \mathrm{~s})}{110 \mathrm{~s}}=\frac{100}{11} \mathrm{~s} \approx 9 \mathrm{~s}\).

33. The activities of three radioactive samples A, B, and C are represented by the lines A, B, and C in the figure. Their half-lives \(T_{1 / 2}(\mathrm{~A}), T_{1 / 2}(\mathrm{~B}), T_{1 / 2}(\mathrm{C}) \) are in the ratio

1. 2: 1 : 3
2. 4 : 3: 1
3. 3 : 2: 1
4. 2: 1: 1

Answer: 1. 2: 1 : 3

Activity = \(|R|=R_0 \mathrm{e}^{-\lambda t}\)

∴ \(\ln R=\ln R_0-\lambda t=\ln R_0-\left(\frac{\ln 2}{T}\right) t\)

The slope of the given graph is -X.

∴ \(\lambda_{\mathrm{A}}=\frac{6}{10}, \lambda_{\mathrm{B}}=\frac{6}{5} \text { and } \lambda_{\mathrm{C}}=\frac{2}{5}\)

∴ \(T_{\mathrm{A}}: T_{\mathrm{B}}: T_{\mathrm{C}}=\frac{10}{6}: \frac{5}{6}: \frac{5}{2}=2: 1: 3\)

Semiconductor Devices And Communication Systems Synopsis

Conductivity (σ) of a conductor: This is given by

∴ \(\sigma=\frac{\text { current density }(j)}{\text { electric field }(E)}=\frac{n e v}{E}=n e \mu\)

where μ = mobility of electrons and v = drift speed of electrons.

The conductivity of a semiconductor: This is given by

∴ [laytex]\sigma=n_{\mathrm{e}} e \mu_{\mathrm{e}}+n_{\mathrm{h}} e \mu_{\mathrm{h}}[/latex]

where the subscripts e and h respectively represent electrons and holes.

Resistance of a p-n junction:

1. Static (or DC) resistance = \(R=\frac{V}{I}\), where V = operating voltage.
2. Dynamic (or AC) resistance = \(r=\frac{\Delta V}{\Delta I}\)

This is the reciprocal of the slope of the I-V characteristic.

The emitter, collector, and base currents in a transistor are related by the equation I_e = I_c + I_b.

α- and β-parameters of a transistor: These two parameters are given by

∴ \(\alpha=\frac{I_{\mathrm{c}}}{I_{\mathrm{e}}} \text { and } \beta=\frac{I_{\mathrm{c}}}{I_{\mathrm{b}}}=\frac{\alpha}{1-\alpha}\)

Voltage gain = \(\frac{V_{\text {out }}}{V_{\text {in }}}=\frac{I_{\mathrm{c}} R_{\mathrm{L}}}{I_{\mathrm{b}} R_{\mathrm{BE}}}=\beta\left(\frac{R_{\mathrm{L}}}{\dot{R}_{\mathrm{BE}}}\right)\)

Power gain = (voltage gain)(current gain)

⇒ \(\beta\left(\frac{R_{\mathrm{L}}}{R_{\mathrm{BE}}}\right)\left(\frac{I_{\mathrm{c}}}{I_{\mathrm{b}}}\right)=\beta^2\left(\frac{R_{\mathrm{L}}}{R_{\mathrm{BE}}}\right)\)

A logic gate is an electric device that performs a logical operation on one or more binary inputs and produces a single binary output.

Symbols and Boolean expressions for basic logic gates and their truth tables:

(1) OR gate:

(2) AND gate:

(3) NOT gate:

(4) NOR gate:

(5) NAND gate:

Range: The maximum distance from the transmitter (which sends the signals) to the receiving station (where the signals are received) with sufficient strength is called the range. It is given by

⇒ \(d=\sqrt{2 R_{\mathrm{E}} h}\)

where R_E = the earth’s radius and h = height of the antenna.

Maximum line-of-sight (LoS) distance:

⇒ \(d=\sqrt{2 R h_1}+\sqrt{2 R h_2}\)

where h₁ and h₂ are the heights of the transmitting and receiving antennas.

Modulated and carrier waves:

(1) Carrier wave: e_c = E_c cos ω_ct.

(2) Modulating signal: e_m = E_m cos ω_mt.

(3) Amplitude of the modulated wave = E_c + E_m cos ω_mt.

(4) Modulated wave:

e = (E_c + E_m cos ω_mt) cos ω_ct

⇒ \(E_{\mathrm{c}}\left(1+\frac{E_{\mathrm{m}}}{E_{\mathrm{c}}} \cos \omega_{\mathrm{m}} t\right) \cos \omega_{\mathrm{c}} t\)

⇒ E_c (1 + m cos ω_mt) cos ω_ct,

where \(m=\frac{E_m}{E_c}\) is the modulation index

∴ \(E_{\mathrm{c}} \cos \omega_{\mathrm{c}} t+\frac{m E_{\mathrm{c}}}{2} \cos \left(\omega_{\mathrm{c}}-\omega_{\mathrm{m}}\right) t+\frac{m E_{\mathrm{c}}}{2} \cos \left(\omega_{\mathrm{c}}+\omega_{\mathrm{m}}\right) t\)

Transistor

Question 1. For a CE transistor amplifier, the audio-signal voltage across the collector resistor of 2 kΩ resistance is 4 V. The current-amplification factor of the transistor is 100 and the base resistance is1 kΩ then the input-signal voltage is

1. 10 mV
2. 20 mV
3. 30 mV
4. 15 mV

Answer: 2. 20 mV

The voltage gain (or voltage amplification) is given by

⇒ \(A_v=\frac{\text { output voltage }\left(V_{\mathrm{o}}\right)}{\text { input voltage }\left(V_{\mathrm{i}}\right)}\)

⇒ \(A_v=\frac{I_C R_C}{I_B R_B}\) → (1)

Given that collector resistance = R_c = 2 kΩ,

current amplification factor = \(\beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=100\),

base resistance = R_B = 1 kΩ

and output voltage = V_o = 4V.

From (1),

⇒ \(A_v=\beta\left(\frac{R_{\mathrm{C}}}{R_{\mathrm{B}}}\right)=100 \times \frac{2 \mathrm{k} \Omega}{1 \mathrm{k} \Omega}=200\)

∵ \(A_v=\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=200\)

∴ input voltage = \(V_{\mathrm{i}}=\frac{V_{\mathrm{o}}}{200}=\frac{4 \mathrm{~V}}{200}=20 \mathrm{mV}\).

Question 2. In a CE transistor amplifier, the audio-signal voltage across the collector is 3 V. The resistance of the collector is 3 kΩ. If the current gain is 100 and the base resistance is 2 kΩ, the voltage and power gains of the amplifier are respectively

1. 200 and 1000
2. 15 and 200
3. 150 and 15000
4. 20 and 2000

Answer: 3. 150 and 15000

Given that voltage across the collector = output voltage = V_o = 3 V,

collector resistance = output load = R_o = 3 kΩ,

current gain = β =I_o/I_i =100 .

and input resistance = base resistance = R_i = 2 kΩ.

∴ voltage gain = \(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=\frac{I_{\mathrm{o}} R_{\mathrm{o}}}{I_{\mathrm{i}} R_{\mathrm{i}}}=\beta\left(\frac{3 \mathrm{k} \Omega}{2 \mathrm{k} \Omega}\right)=100 \times 1.5=150\)

Now, power gain = (current gain)(voltage gain)

∴ \(\beta\left(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}\right)=100 \times 150=1500\)

Question 3. A transistor is operated in the common-emitter configuration at V_c = 2 V such that a change in the base current from 100 μA to 200 μA produces a change in the collector current from 5 mA to 10 mA. The current gain is

1. 150
2. 75
3. 50
4. 100

Answer: 3. 50

The change in the base current is ΔI_B = 200 μA-100 μA =100 μA.

The corresponding change in the collector current is

ΔI_C=10 mA- 5 mA = 5 mA.

By definition,

∴ current gain = \(\beta=\frac{\Delta I_{\mathrm{C}}}{\Delta I_{\mathrm{B}}}=\frac{5 \mathrm{~mA}}{100 \mu \mathrm{A}}=\frac{5 \times 10^{-3}}{100 \times 10^{-6}}=\frac{5000}{100}=50\)

Question 4. A transistor is operated in the CE configuration at a constant collector voltage of V_c =15 V such that a change in the base current from 100 μA to 150 μA produces a change in the collector current from 5 mA to 10 mA. The current gain β is

1. 75
2. 50
3. 100
4. 67

Answer: 3. 100

Given that ΔI_B =150 μA – 100 μA = 50 μA.

The corresponding collector current is ΔI_C = 10 mA – 5 mA = 5 mA

∴ current gain = \(\beta=\frac{\Delta I_C}{\Delta I_B}=\frac{5 m A}{50 \mu \mathrm{A}}=100\)

Question 5. An n-p-n transistor is connected in the common-emitter (CE) configuration in a given amplifier. A load resistor of resistance 800 Ω, is connected in the collector circuit, and the voltage drop across it is 0.8 V. If the current-amplification factor is 0.96 and the input resistance of the circuit is 192 Ω, the voltage gain and the power gain of the amplifier will respectively be

1. 3.69 and 3.84
2. 4 and 3.84
3. 4 and 4
4. 4 and 3.69

Answer: 2. 4 and 3.84

Given that load resistance = R_o = 800 Ω,

voltage drop across the load = V_o = 0.8 V,

current amplification factor = \(\beta=\frac{I_C}{I_B}=0.96\)

and input resistance = R_B = 192 Ω.

∴ voltage gain = \(\frac{V_0}{V_{\mathrm{i}}}=\frac{I_{\mathrm{C}} R_{\mathrm{C}}}{I_{\mathrm{B}} R_{\mathrm{B}}}=\left(\frac{800 \Omega}{192 \Omega}\right)(0.96)=4\)

Hence, power gain = (current gain)(voltage gain)

∴ \(\left(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}\right)\left(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}\right)=\beta\left(\frac{V_{\mathrm{o}}}{V_{\mathrm{I}}}\right)=0.96 \times 4=3.84\)

Question 6. For a transistor with I_c/I_E = 0.96, the current gain in the common emitter configuration is

1. 12
2. 48
3. 24
4. 6

Answer: 3. 24

In a transistor,

⇒ \(I_{\mathrm{E}}=I_{\mathrm{C}}+I_{\mathrm{B}}\)

⇒ \(\frac{I_{\mathrm{E}}}{I_{\mathrm{C}}}=1+\frac{I_{\mathrm{B}}}{I_{\mathrm{C}}}\)

⇒ \(\frac{1}{0.96}=1+\frac{I_{\mathrm{B}}}{I_{\mathrm{C}}} \Rightarrow \frac{I_{\mathrm{B}}}{I_{\mathrm{C}}}=\frac{4}{96}=\frac{1}{24}\)

∴ current gain = \(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=24\)

Question 7. The correct relationship of the two current gains a and p in a transistor is

1. \(\beta=\frac{1+\alpha}{\beta}\)
2. \(\alpha=\frac{\beta}{1+\beta}\)
3. \(\alpha=\frac{\beta}{1-\beta}\)
4. \(\beta=\frac{\alpha}{1+\alpha}\)

Answer: 2. \(\alpha=\frac{\beta}{1+\beta}\)

In a transistor,

⇒ \(\alpha=\frac{I_{\mathrm{C}}}{I_{\mathrm{E}}} \text { and } \beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}\)

Now, emitter current = collector current + base current

⇒ \(I_{\mathrm{E}}=I_{\mathrm{C}}+I_{\mathrm{B}}\)

⇒ \(\frac{I_{\mathrm{E}}}{I_{\mathrm{C}}}=1+\frac{I_{\mathrm{B}}}{I_{\mathrm{C}}} \Rightarrow \frac{1}{\alpha}=1+\frac{1}{\beta}=\frac{\beta+1}{\beta}\)

∴ \(\alpha=\frac{\beta}{1+\beta}\).

Question 8. In the common-base configuration of a transistor, ΔI_c/ΔI_E = 0.98. Then, the current gain in the common-emitter configuration of the transistor will be

1. 98
2. 24.5
3. 4.9
4. 49

Answer: 4. 49

In the common-base configuration, the current gain is given by

⇒ \(\alpha=\frac{\Delta I_{\mathrm{C}}}{\Delta I_{\mathrm{E}}}=0.98\)

In the common-emitter configuration, the current gain is given by

⇒ \(\beta=\frac{\Delta I_{\mathrm{C}}}{\Delta I_{\mathrm{B}}}=\frac{\Delta I_{\mathrm{C}}}{\Delta I_{\mathrm{E}}-\Delta I_{\mathrm{C}}}\)    [∵I_E = I_C + I_B]

∴ \(\frac{1}{\frac{\Delta I_{\mathrm{E}}}{\Delta I_{\mathrm{C}}}-1}=\frac{1}{\frac{1}{\alpha}-1}=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}=49\)

Question 9. In the circuit shown in the adjoining figure, when the input voltage of the base resistance is 10 V, V_BE is zero, and V_CE is also zero. The value of β is

1. 196
2. 154
3. 105
4. 133

Answer: 4. 133

Given that input voltage = V_i =10 V and base resistance = R_B = 400 kΩ = 400 x l0³ Ω; the load resistance in the collector circuit is R_c = 3 kΩ = 3 x 10³ Ω. Also, V_BE = 0 V, V_CE = 0 V, and V_CC =10 V.

∴ V_i – V_BE = R_BI_B ⇒ 10 V – 0 V = (400 x 103 Ω.)I_B.

∴ base current = \(I_{\mathrm{B}}=\frac{10 \mathrm{~V}}{4 \times 10^5 \Omega}=25 \mu \mathrm{A}\)

Again, in the output circuit,

⇒ \(V_{\mathrm{CC}}-V_{\mathrm{CE}}=I_{\mathrm{C}} R_{\mathrm{C}}\)

⇒ \(I_{\mathrm{C}}=\frac{V_{\mathrm{CC}}-V_{\mathrm{CE}}}{R_{\mathrm{C}}}=\frac{10 \mathrm{~V}-0 \mathrm{~V}}{3 \times 10^3 \Omega}=\frac{10 \mathrm{~V}}{3 \times 10^3 \Omega}=3.33 \mathrm{~mA}\)

∴ \(\beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=\frac{3.33 \mathrm{~mA}}{25 \mu \mathrm{A}}=\frac{3.33}{25} \times 10^3=133\)

Question 10. An amplifier has a voltage gain of A_v = 1000. The voltage gain in dB is

1. 20 dB
2. 60 dB
3. 3 dB
4. 30 dB

Answer: 2. 60 dB

Voltage gain = (20 log A_v) dB = (20 log 1000) dB

= (20 log 10³) dB = 60 dB.

Question 11. Which of the following statements about a transistor is not true?

1. The emitter is heavily doped.
2. The base is thin.
3. The base is lightly doped.
4. The collector region is smaller compared to the emitter region in size.

Answer: 4. The collector region is smaller compared to the emitter region in size.

In a transistor, the middle layer called the base, is very thin (width ≈1 μm) and very lightly doped. The emitter is heavily doped and the collector is moderately doped. As regards the contact area, the collector has a comparatively larger area than the base and the emitter.

Question 12. In the given figure, an n-p-n transistor in the common-emitter configuration has a current gain of β = 100. The output voltage (V_o) of the amplifier will be

1. 10 V
2. 10 mV
3. 1.0 V
4. 0.1 V

Answer: 3. 1.0 V

Given that current gain = \(\beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=100\)

input resistance = R_B = 1 kΩ,

and output load resistance = R_C =10 kΩ.

Now, base current = \(\frac{\text { input voltage }}{\text { input resistance }}\)

⇒ \(I_B=\frac{1.0 \mathrm{mV}}{1 \mathrm{k} \Omega}=\frac{1.0 \times 10^{-3} \mathrm{~V}}{1 \times 10^3 \Omega}=1.0 \times 10^{-6} \mathrm{~A}\)

∴ collector current = I_C = βI_B = 100 x 10^-6 A = 10^-4 A.

Hence, output voltage = V_c = I_CR_C = (10^-4 A)(10 kΩ)

= (10^-4 A)(10⁴ Ω) = 1.0 V.

Question 13. A transistor connected in the common-emitter mode contains a load resistance of 5 kΩ and an input resistance of 1 kΩ. If the input peak voltage is 5 mV and the current gain is 50, find the voltage gain.

1. 500
2. 50
3. 250
4. 125

Answer: 3. 250

Given that load resistance = R_C = 5 kΩ,

input resistance = R_i = 1 kΩ, input voltage = V_i = 5 mV,

and current gain = β = 50.

∴ voltage gain = \(A_v=\frac{\text { output voltage across } \mathrm{R}_{\mathrm{C}}}{\text { input voltage }}\)

∴ \(\frac{R_{\mathrm{C}} I_{\mathrm{C}}}{R_{\mathrm{i}} I_{\mathrm{B}}}=\beta\left(\frac{R_{\mathrm{C}}}{R_{\mathrm{i}}}\right)=50\left(\frac{5 \mathrm{k} \Omega}{1 \mathrm{k} \Omega}\right)=250\).

Question 14. The current gain for a transistor working as a common-base amplifier is 0.96. If the emitter current is 7.2 ma, the base current is

1. 0.39 mA
2. 0.43 mA
3. 0.29 mA
4. 0.35 mA

Answer: 3. 0.29 mA

In the common-base configuration,

current gain = \(\alpha=\frac{I_{\mathrm{C}}}{I_{\mathrm{E}}}=0.96\)

Given that I_E = 7.2 mA. Hence, collector current = I_C = 0.96 x 7.2 mA.

∴ base current = I_B = I_E– I_C

= 7.2 mA -0.96(7.2 mA)

= (7.2 mA)(l-0.96) = (7.2 mA)(0.04)

= 0.288 mA ~ 0.29 mA.

Question 15. For a common-emitter transistor, the input current is 5 pA, the voltage gain is β = 100, and the circuit is operated at a load resistance of 10 kΩ. The voltage between the collector and the emitter will be

1. 12.5 V
2. 5 V
3. 7.5 V
4. 10 V

Answer: 2. 5 V

Given that input current = I_B = 5 μA, current gain = β =100, and load resistance = R_C = 10 kΩ.

∵ \(\beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=100\),

∴collector current = I_C = 100I_B = 100(5 μA) = 500 μA.

∴ output voltage = I_CR_C = (500 μA)(10 kΩ)

= 500 x 10^-6 x 10 x 10³ V = 5 V.

Question 16. Consider an n-p-n transistor amplifier in the common-emitter mode. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in the emitter current?

1. 1.1mA
2. 1.01mA
3. 10 mA
4. 0.01mA

Answer: 2. 1.01mA

For an n-p-n transistor in the CE mode,

current gain = \(\beta=\frac{\Delta I_{\mathrm{C}}}{\Delta I_{\mathrm{B}}} \Rightarrow 100=\frac{1 \mathrm{~mA}}{\Delta I_{\mathrm{B}}}\)

∴ the change in the base current will be

⇒ \(\Delta I_{\mathrm{B}}=\frac{1 \mathrm{~mA}}{100}=0.01 \mathrm{~mA}\)

Hence, the change in the emitter current will be

∴ ΔI_E = ΔI_C + ΔI_B = 1 mA + 0.01 mA = 1.01 mA.

Question 17. If the collector current is 120 mA, the base current is 2 mA, and the resistance gain is 3, what is the power gain?

1. 180
2. 10800
3. 1.8
4. 18

Answer: 2. 10800

Current gain = \(\beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=\frac{120 \mathrm{~mA}}{2 \mathrm{~mA}}=60\)

∴ power gain = β² x resistance gain = 60² x 3 = 10800.

Question 18. In a common-base amplifier, the phase difference between the input-signal voltage and the output voltage is

1. zero
2. \(\frac{\pi}{2}\)
3. \(\frac{\pi}{4}\)
4. π

Answer: 1. zero

In a transistor amplifier in the CB configuration, the input and output voltages are in the same phase. So, the phase difference is zero.

Question 19. To use a transistor as an amplifier,

1. The emitter-base junction is forward-biased and the base-collector junction is reverse-biased
2. No bias voltages are required
3. Both junctions are forward-biased
4. Both junctions are reverse-biased

Answer: 1. The emitter-base junction is forward-biased and the base-collector junction is reverse-biased

The adjoining figure shows an n-p-n transistor in the CB configuration as an amplifier. This is used when a low input impedance is required. Due to the forward bias in the input, the base-emitter junction acts as a forward-biased p-n junction diode and has a low input impedance. The output across the base-collector junction acts as a reversed-biased junction diode. Hence, a common base amplifier has a high output impedance. An amplifier with a low input impedance and a high output impedance provides a high voltage gain.

Question 20. When an n-p-n transistor is used as an amplifier,

1. Holes move from the emitter to the base
2. Holes move from the base to the emitter
3. Electrons move from the collector to the base
4. Electrons move from the base to the collector

Answer: 4. Electrons move from the base to the collector

As shown in the diagram of the preceding question, collector current(I_c) has been shown to flow from the collector to the base. Hence, electrons move from the base to the collector.

Question 21. The voltage gain (A_v) of the given amplifier is

1. 1000
2. 100
3. 10
4. 9.9

Answer: 2. 100

The voltage gain in an operational amplifier is given by

⇒ \(A_v=\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=\frac{R_{\mathrm{f}}}{R_{\mathrm{i}}}\)

∴ \(A_v=\frac{100 \mathrm{k} \Omega}{1 \mathrm{k} \Omega}=100\).

Question 22. The transfer ratio p of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 kΩ. The peak value of the collector AC current for an AC input voltage of 0.01 V peak is

1. 100 μA
2. 0.01 mA
3. 0.25 mA
4. 500 μA

Answer: 4. 500 μA

Given that input resistance = \(R_{\mathrm{B}}=1 \mathrm{k} \Omega\)

and base current = \(I_{\mathrm{B}}=\frac{V_{\mathrm{B}}}{R_{\mathrm{B}}}=\frac{0.01 \mathrm{~V}}{1000 \Omega}=10^{-5} \mathrm{~A}\)

Now, \(\beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}} \Rightarrow 50=\frac{I_{\mathrm{C}}}{10^{-5} \mathrm{~A}}\)

∴ collector current = I_c = 50 x 10^-5 A = 500 x 10^-6 A = 500 μA.

Question 23. An oscillator is nothing but an amplifier with

1. A positive feedback
2. A negative feedback
3. A large gain
4. No feedback

Answer: 1. A positive feedback

In an oscillator, part of the output signal is fed back to the amplifier input in such a way that the feedback signal is regenerated, reamplified, and fed back again to maintain a constant output signal. Hence, an oscillator is an amplifier with a positive feedback

Question 24. The input signal given to a CE amplifier Physics having a voltage gain of 150 is

\(V_{\mathrm{i}}=(2 \mathrm{~V}) \cos \left(15 t+\frac{\pi}{3}\right)\)

The corresponding output signal will be

1. \((300 \mathrm{~V}) \cos \left(15 t+\frac{\pi}{3}\right)\)
2. \((75 \mathrm{~V}) \cos \left(15 t+\frac{2 \pi}{3}\right)\)
3. \((2 \mathrm{~V}) \cos \left(15 t+\frac{5 \pi}{3}\right)\)
4. \((300 \mathrm{~V}) \sin \left(15 t+\frac{4 \pi}{3}\right)\)

Answer: 4. \((300 \mathrm{~V}) \sin \left(15 t+\frac{4 \pi}{3}\right)\)

Voltage gain = \(\frac{V_o}{V_i}=150\)

∴ V_o = 150V_i = 150 x 2 V = 300 V.

The phase difference between the input and the output is n. Hence, the output signal voltage is

∴ \(V_{\mathrm{o}}=(300 \mathrm{~V}) \sin \left(15 t+\frac{\pi}{3}+\pi\right)=(300 \mathrm{~V}) \sin \left(15 t+\frac{4 \pi}{3}\right)\)

Question 25. In a common-emitter(CE) amplifier having volta gain G, the transistor used has a transconductance of 0.03 mho and a current gain of 25. If the above transistor is replaced with another one having a transconductance of 0.02 mho and a current gain of 20, the voltage gain will be

1. \(\frac{2}{3} G\)
2. 1.5G
3. \(\frac{1}{3} G\)
4. \(\frac{5}{4} G\)

Answer: 1. \(\frac{2}{3} G\)

Transconductance = \(g_{\mathrm{m}}=\frac{\Delta I_{\mathrm{C}}}{\Delta V_{\mathrm{B}}}=\frac{\Delta I_{\mathrm{C}}}{\Delta I_{\mathrm{B}} R_{\mathrm{i}}}=\frac{\beta}{R_{\mathrm{i}}}\)

and voltage gain = \(A_v=\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=\frac{I_{\mathrm{C}} R_{\mathrm{L}}}{I_{\mathrm{B}} R_{\mathrm{i}}}=\beta\left(\frac{R_{\mathrm{L}}}{R_{\mathrm{i}}}\right)=\left(\frac{\beta}{R_{\mathrm{i}}}\right) R_{\mathrm{L}}\)

∴ A_v = g_m R_L

Now, since A_v = G (given),

⇒ \(G=g_{\mathrm{m}} R_{\mathrm{L}} \Rightarrow G \propto g_{\mathrm{m}}\)

∴ \(\frac{G^{\prime}}{G}=\frac{g_{\mathrm{m}}^{\prime}}{g_{\mathrm{m}}} \Rightarrow G^{\prime}=G\left(\frac{g_{\mathrm{m}}^{\prime}}{g_{\mathrm{m}}}\right)=\frac{0.02}{0.03} G=\frac{2}{3} G\)

Question 26. A common-emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω, and an output impedance of 200 Ω. The power gain of the amplifier is

1. 50
2. 1000
3. 500
4. 1250

Answer: 4. 1250

Power gain = (voltage gain)(current gain)

⇒ \(\left(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}\right)\left(\frac{I_{\mathrm{o}}}{I_{\mathrm{i}}}\right)=\left(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}\right)\left(\frac{V_{\mathrm{o}} / R_{\mathrm{o}}}{V_{\mathrm{i}} / R_{\mathrm{i}}}\right)=\left(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}\right)^2\left(\frac{R_{\mathrm{i}}}{R_{\mathrm{o}}}\right)\)

Given that voltage gain = \(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=50 \text { and } \frac{R_{\mathrm{i}}}{R_{\mathrm{o}}}=\frac{100 \Omega}{200 \Omega}=\frac{1}{2}\)

∴ power gain = \((50)^2 \times \frac{1}{2}=1250\)

Question 27. The voltage gain of an amplifier with a 9% negative feedback is 10. The voltage gain without any feedback will be

1. 10
2. 100
3. 1.25
4. 90

Answer: 2. 100

The voltage gain of an amplifier is given by = \(A_v=\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}\)

∴ A_oV_i= V_o

If p is the feedback fraction then

A_v (v_i – βv_o) = v_o

⇒ \(\frac{V_0}{V_{\mathrm{i}}}=\frac{A_v}{1+\beta A_v}\)

Given that \(\frac{V_o}{V_i}=10 \text { and } \beta=9 \%=0.09\)

∴ \(10=\frac{A_v}{1+0.09 A_v} \Rightarrow A_v=10+0.9 A_v \Rightarrow A_v=100\)

Hence, the voltage gain without any feedback is 100.

Question 28. An n-p-n transistor conducts when

1. The collector is positive and at the same potential as the base
2. Both the collector and the emitter are negative with respect to the base
3. Both the collector and the emitter are positive with respect to the base
4. The collector is positive and the emitter is negative with respect to the base

Answer: 4. The collector is positive and the emitter is negative with respect to the base

In a transistor, the input side (base-emitter junction) is forward-biased and the output side (collector-base junction) must be reverse-biased. Hence, for an. n-p-n transistor, the collector is positive and the emitter is negative, as shown in the adjoining diagram.

Question 29. One way in which the operation of an n-p-n transistor differs from that of a p-n-p transistor is

1. The emitter junction injects minority carriers into the base region of the p-n-p transistor
2. The emitter injects holes into the base of the p-n-p transistor and electrons into the base region of the n-p-n transistor
3. The emitter injects holes into the base of the n-p-n transistor
4. The emitter junction is reverse-biased in the n-p-n transistor

Answer: 2. The emitter injects holes into the base of the p-n-p transistor and electrons into the base region of the n-p-n transistor

The emitter of a p-n-p transistor injects holes (majority carriers) into the base, whereas in an n-p-n transistor, the emitter injects electrons (majority carriers) into the base region.

Question 30. The input resistance of a silicon transistor is 100 Ω. The base current is changed by 40 μA, which results in a change in the collector current by 2 mA. This transistor is used as a common-emitter amplifier with a load resistance of 4 kΩ. The voltage gain of the amplifier is

1. 4000
2. 1000
3. 2000
4. 3000

Answer: 3. 2000

Given that input resistance =R_i=100 Ω, output load resistance = R_o = 4000Ω, change in the base current = ΔI_B = 40 μA, and the corresponding change in the collector current = ΔI_c — 2 mA.

voltage gain = \(A_{\mathrm{v}}=\frac{\Delta V_{\mathrm{o}}}{\Delta V_{\mathrm{i}}}=\left(\frac{R_{\mathrm{o}}}{R_{\mathrm{i}}}\right)\left(\frac{\Delta I_{\mathrm{C}}}{\Delta I_{\mathrm{B}}}\right)\)

∴ \(\left(\frac{4000 \Omega}{100 \Omega}\right)\left(\frac{2 \times 10^{-3} \mathrm{~A}}{40 \times 10^{-6} \mathrm{~A}}\right)=2000\)

Question 31. For a transistor to be active,

• The base, emitter, and collector regions should have similar sizes and doping concentrations
• The base region must be very thin and lightly doped
• The emitter-base junction is forward-biased and the base-collector junction is reverse-biased
• Both the emitter-base junction and the base-collector junction are forward-biased

Which one of the following pairs of statements is correct?

1. (2) and (3)
2. (3) and (4)
3. (4) and (1)
4. (1) and (2)

Answer: 1. (2) and (3)

The middle layer (base) in a transistor is very thin and very lightly doped. The emitter-base junction is always forward-biased, whereas the collector-base junction is reverse-biased.

Question 32. The given transistor operates in the saturation region. Find the value of V_BB. (Given that R_o = 200 Q, R_i = 100 kΩ, V_CC=3V, V_BE = 0.7 V_CC = 0 and β = 200.)

1. 7.5 V
2. 8.2 V
3. 4.2 V
4. 6.6 V

Answer: 2. 8.2 V

The voltage between the collector and the emitter is

⇒ \(V_{\mathrm{CE}}=V_{\mathrm{CC}}-I_{\mathrm{C}} R_{\mathrm{o}}=0 \text { (given) }\)

∴ \(3 \mathrm{~V}=I_{\mathrm{C}}(200 \Omega) \Rightarrow I_{\mathrm{C}}=\frac{3}{200} \mathrm{~A}=15 \mathrm{~mA}\)

∵ \(\beta=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=200 \text { (given) }\),

∴ base current = \(I_{\mathrm{B}}=\frac{I_{\mathrm{C}}}{200}=\frac{15 \mathrm{~mA}}{200}=0.075 \mathrm{~mA}=75 \mu \mathrm{A}\)

For the closed loop containing the base and the emitter,

⇒ \(V_{\mathrm{BE}}=V_{\mathrm{BB}}-I_{\mathrm{B}} R_{\mathrm{i}}\)

⇒ \(V_{\mathrm{BB}}=0.7 \mathrm{~V}+(75 \mu \mathrm{A})(100 \mathrm{k} \Omega)\)

∴ \(0.7 \mathrm{~V}+\left(75 \times 10^{-6} \times 10^5 \mathrm{~V}\right)=8.2 \mathrm{~V}\)

Question 33. The transfer characteristic curve of a transistor having input and output resistances of 100 Ω and 100 kΩ respectively is given in the adjoining 10 mA figure. The voltage gain and the power gain are respectively

1. 5 x l0⁴ and 2.5 x l0⁶
2. 2.5 x 10⁴ and 2.5 x 10⁶
3. 3.5 x l0⁴ and 5 x l0⁵
4. 4 x 10⁴ and 5 x 10⁶

Answer: 1. 5 x l0⁴ and 2.5 x l0⁶

Voltage gain = \(\frac{\text { output voltage }}{\text { input voltage }}\)

∴ \(\left(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}\right)\left(\frac{R_{\mathrm{o}}}{R_{\mathrm{i}}}\right)=\left(\frac{5 \mathrm{~mA}}{100 \mu \mathrm{A}}\right)\left(\frac{100 \mathrm{k} \Omega}{100 \Omega}\right)\)

⇒ \(\frac{5 \times 10^{-3} \mathrm{~A}}{100 \times 10^{-6} \mathrm{~A}} \times \frac{100 \times 10^3 \Omega}{100 \Omega}=5 \times 10^4\)

∴ power gain = (voltage gain)(current gain)

⇒ \(\left(5 \times 10^4\right)\left(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}\right)=\left(5 \times 10^4\right)\left(\frac{5 \mathrm{~mA}}{100 \mu \mathrm{A}}\right)\)

⇒ (5 x 10⁴)(50) = 2.5 x 10⁶.

Question 34. An n-p-n transistor operates as a common-emitter amplifier with a power gain of 60 dB. The input resistance is 100 Ω and the output load resistance is 10 kΩ. The current gain β is

1. 3
2. 60
3. 10²
4. 6 x l0²

Answer: 3. 10²

Power gain = (current gain)(voltage gain)

⇒ \(\beta\left(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}\right)=\beta\left(\frac{I_{\mathrm{o}} R_{\mathrm{o}}}{I_{\mathrm{i}} R_{\mathrm{i}}}\right)=\beta^2\left(\frac{R_{\mathrm{o}}}{R_{\mathrm{i}}}\right)\)

Given that power gain = 60 dB = 10 log \(\left(\frac{B^2 R_{\mathrm{o}}}{R_{\mathrm{i}}}\right)\)

∴ \(\frac{\beta^2 R_{\mathrm{o}}}{R_{\mathrm{i}}}=10^6 \Rightarrow \beta^2\left(\frac{10 \mathrm{k} \Omega}{100 \Omega}\right)=10^6\)

⇒ β² =10⁴.

∴ current gain = β =100

Question 35. An n-p-n transistor is used in the common-emitter configuration as an amplifier with a load resistance of 1 kΩ. A signal voltage of 10 mV is applied across the base-emitter junction. This produces a 3 mA change in the collector current and a 15 μA change in the base current of the amplifier. The input resistance and the voltage gain are respectively

1. 0.33 kΩ and 300
2. 0.67 kΩ and 300
3. 0.33 kΩ and 1.5
4. 0.67 kΩ and 200

Answer: 2. 0.67 kΩ and 300

Given that R_C=1 kΩ =1000 Ω, input voltage = V_i =10 mV, change in the collector current = ΔI_c = 3 mA, and ΔI_B =15 μA.

∴ input resistance = \(R_{\mathrm{i}}=R_{\mathrm{BE}}=\frac{\Delta V_{\mathrm{i}}}{\Delta I_{\mathrm{B}}}=\frac{10 \mathrm{mV}}{15 \mu \mathrm{A}}=\frac{2}{3} \times 10^3 \Omega\)

∴ voltage gain = \(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=\frac{\Delta I_{\mathrm{C}} R_{\mathrm{C}}}{\Delta I_{\mathrm{B}} R_{\mathrm{i}}}=\frac{\left(3 \times 10^{-3} \mathrm{~A}\right)(1000 \Omega)}{\left(15 \times 10^{-6} \mathrm{~A}\right)\left(\frac{2}{3} \times 10^3 \Omega\right)}=300\)

Question 36. In the given diagram, the supply voltage (V_BB) can vary from zero to 5.0 V, V_CC = 5 V, β_DC = 200, and V_BE =1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation are respectively

1. 25 μA and 2.8 V
2. 2.0 μA and 2.8 V
3. 25 μA and 3.5 V
4. 20 μA and 3.5 V

Answer: 3. 25 μA and 3.5 V

In a transistor, saturation occurs when V_CE = 0.

Applying the loop rule in the given loop, we have

V_CE – V_CC + I_CR_C = 0

⇒ V_CE = 0 = V_CC – I_CR_C

⇒ \(I_{\mathrm{C}}=\frac{V_{\mathrm{CC}}}{R_{\mathrm{C}}}=\frac{5 \mathrm{~V}}{1 \mathrm{k} \Omega}=5 \times 10^{-3} \mathrm{~A}\)

∴ \(\beta_{\mathrm{DC}}=\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=200 \text { (given) }\)

∴ base current = \(I_{\mathrm{B}}=\frac{I_{\mathrm{C}}}{200}=\frac{5 \times 10^{-3} \mathrm{~A}}{200}=25 \mu \mathrm{A}\)

Again applying the loop rule for the base-emitter closed loop, we obtain

V_BE – V_BB + I_BR_B= 0

⇒ V_BB = (25 x 10^-6 A)(100 x10³ Ω) +1 V

= 3.5 V.

Question 37. A common-emitter amplifier circuit built using an n-p-n transistor is shown in the figure. Its DC current gain is 250, R_c =1, and V_CC = 10V. What is the minimum base current for V_CC to reach saturation?

1. 10μA
2. 100 μA
3. 40 μA
4. 7 μA

Answer: 3. 40 μA

Given that DC current gain = \(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=250, R_{\mathrm{C}}=1 \mathrm{k} \Omega \text { and } V_{\mathrm{CC}}=10 \mathrm{~V}\)

For saturation, V_CE = 0.

Thus, V_CC = I_CR_C

⇒ \(I_{\mathrm{C}}=\frac{V_{\mathrm{CC}}}{R_{\mathrm{C}}}=\frac{10 \mathrm{~V}}{1 \times 10^3 \Omega}=1 \times 10^{-2} \mathrm{~A}\)

Now, \(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=250\)

∴ base current = \(I_{\mathrm{B}}=\frac{I_{\mathrm{C}}}{250}=\frac{10^{-2} \mathrm{~A}}{250}=40 \mu \mathrm{A}\)