WBCHSE Class 12 Physics Dual Nature Of Matter And Radiation Multiple Questions

Dual Nature Of Matter And Radiation Multiple Questions And Answers

Question 1. A particle is dropped from a height of H. The de Broglie wavelength of the particle as a function of height is proportional to

1. H
2. H^½
3. H
4. H^-½

Answer: 4. H^-½

Question 2. Consider a beam of electrons (each electron with energy E₀) incident on a metal surface kept in an evacuated chamber. Then

1. No electrons will be emitted as only photons can emit electrons
2. Electrons can be emitted but all with an energy E_p
3. Electrons can be emitted with any energy, with a maximum of E_p– Φ (Φ is the work function)
4. Electrons can be emitted with any energy, with a maximum of E₀

Answer: 4. Electrons can be emitted with any energy, with a maximum of E₀

Question 3. An electron (mass m ) with an \(\vec{V}=v_0 \hat{i}\) is in an electric (\(\vec{E}=E_0 \hat{j}\)  constant E₀> 0 ). Its de Broglie wavelength at time t is given by

1. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m} \cdot \frac{t}{v_0}\right)}\)
2. \(\lambda_0\left(1+\frac{e E_0 t}{m v_0}\right)\)
3. λ₀
4. λ₀t

Answer: 1. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m} \cdot \frac{t}{v_0}\right)}\)

Question 4. An electron (mass m ) with an initial velocity \(\vec{V}=v_0 \hat{i}\)  is in an electric field \(\vec{E}=E_0 \hat{j}\). \(\lambda_0=\frac{h}{m v_0}\)  If,  its de Broglie wavelength at time t is given by

1. λ₀
2. \(\lambda_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m_0^2 v_0^2}}\)
3. \(\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m_0^2 \nu_0^2}}}\)
4. \(\frac{\lambda_0}{\left(1+\frac{e^2 E_0^2 t^2}{m_0^2 v_0^2}\right)}\)

Answer: 3. \(\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m_0^2 \nu_0^2}}}\)

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 5. An electron is moving with an initial velocity \(\vec{V}=v_0 \hat{i}\)  and is in a magnetic field \(\vec{B}=B_0 \hat{i}\),  its de Broglie wavelength

1. Remains constant
2. Increase with time
3. Decreases with time
4. Increases and decreases periodically

Answer: 1. Remains constant

Question 6. Relativistic corrections become necessary when the expression for kinetic energy ½mv² becomes comparable to me2. At what de Broglie wavelength will relativistic corrections become important for an electron?

1. 10 nm
2. 10 nm
3. 10 nm
4. 10-4 nm

Answer: 3 and 4

Question 7. Photons absorbed in matter are converted to heat. A source emitting n photons of frequency f is used to convert 1 kg of ice at 0°C to water at 0°C. Then the time T taken for the conversion

1. Decreases with increasing n, with f fixed
2. Decreases with n fixed, f Increasing
3. Remains constant with n and f changing such that nf= constant
4. Increases when product nf increases

Answer: 1, 2, And 3

Question 8. A particle moves in a closed orbit around the origin, due to a force that is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values, λ₁,λ₂ with  λ₁>λ₂ Which of the following statements is true?

1. The particle could be moving in a circular orbit with the origin as the center.
2. The particle could be moving in an elliptic orbit with the origin as the focus.
3. When the de Broglie wavelength is λ₁, the particle is the origin than when its value is λ₂.
4. When the de Broglie wavelength is λ₂, the particle is nearer the origin than when its value is λ₂

Question 9. A monochromatic source of light is kept at a distance of 0.2 m from a photoelectric cell. Stopping potential V0 and sat¬ uration current I0 are 0.6V and 18.0 mA, respectively. Now the source is kept at a distance of 0.6m from the cell. Then

1. V₀ = 0.2V, I₀= 18.0mA
2. V₀ = 0.2 V, I₀= 2.0mA
3. V₀ = o.6V,I₀ = 18.0mA
4. V₀ = 0.6V, I₀ = 2.0mA

Answer: 4. V₀ = 0.6V, I₀ = 2.0mA

Question 10. For a monochromatic light incident on a metal surface, the stopping potential is V. Then the kinetic energy of the fast¬ est photoelectrons emitted from that surface is

1. eV
2. 2eV
3. \(\frac{2 e V}{m}\)
4. \(\sqrt{\frac{2 e V}{m}}\)

Answer: eV

Question 11. If in a photo-electric experiment, the wavelength of inci¬ dent radiation is reduced from 6000 A to 4000 A then

1. Stopping potential will decrease
2. Stopping potential will increase
3. The kinetic energy of emitted electrons will decrease
4. The value of the work function will decrease

Answer: 2. Stopping potential will increase

Question 12. If the stopping potential for photoelectric emission is 0.75 V,’ the kinetic energy of the fastest photoelectrons is

1. 0.75 V
2. 7.5eV
3. 0. 7.5eV
4. 0.75 × 10^-19 J

Answer: 3. 0. 7.5eV

Question 13. For a monochromatic light incident on a metal surface, the maximum velocity of the emitted photoelectrons is v. Then the stopping potential would be

1. \(\frac{2 m v^2}{e}\)
2. \(\frac{m v^2}{e}\)
3. \(\frac{m v^2}{2e}\)
4. \(\frac{m v^2}{\sqrt{2} e}\)

Answer: 3. \(\frac{m v^2}{2e}\)

Question 14. For two monochromatic radiations incident on the same metal surface, the stopping potentials are 1.0 V and 2.0V. The ratio between the maximum velocities, of the emitted photoelectrons is

1. 2:1
2. √2:1
3. 1:√2
4. 1:2

Answer: 3. 1: √2

Question 15. The energy of photon incident on a metal plate is twice its work function. How many times should be the wavelength of incident light so that the kinetic energy of the fastest elec¬ tron will be doubled?

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

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

Question 16. The work function of zinc is twice that of sodium. If the photo¬ electric threshold wavelength for sodium is 7000A, what will be its value for zinc?

1. 3500A°
2. 14000A°
3. 10500A°
4. 4667A°

Answer: 1. 3500A°

Question 17. The threshold wavelength of a metal for electron emission is 5200A. Which one of the following sources of light will be able to emit electrons from the metal?

1. 50 W infrared
2. 1W infrared
3. 50 W red light
4. 1W ultraviolet

Answer: 4. 1W ultraviolet

Question 18. When photons of energy 6eV are incident on a metal surface, the kinetic energy of the fastest electrons becomes 4eV. The value of stopping potential is (in V)

1. 2
2. 4
3. 6
4. 10

Answer: 2 . 4

Question 19. In the case of the photoelectric effect, the incident photon

1. Vanishes completely
2. Is scattered with lower frequency
3. Is scattered with higher frequency
4. Is scattered with the same frequency

Answer: 1. Vanishes completely

Question 20. Which of the following quantities has the same dimension as that of Planck’s constant?

1. Linear momentum
2. Angular momentum
3. Energy
4. Power

Answer: 2. Angular momentum

Question 21. In the case of the photoelectric effect, the graph of the kinetic energy of the photoelectron concerning the frequency of incident radiation will be a straight line. The slope of this straight line depends on

1. The nature of the metal surface
2. The intensity of the incident radiation
3. The nature of the metal surface as well as the intensity of incident radiation
4. None of the nature of the metal surface or the intensity

Answer: 4. None of the nature of the metal surface or the intensity

Question 22. A body absorbs 5 x 1029 photons of frequency 102° Hz. Which of the following information is correct? [Assume, all the energy of photons is transformed into mass.]

1. The mass of the body remains unchanged
2. The mass of the body increases by 0.00037 kg
3. The mass of the body increases by 0.37 kg
4. The mass of the body increases by 3.7 kg

Answer: 3. Mass of the body increases by 0.37 kg

Question 23. The momentum of a photon (frequency = f, rest mass = 0 )

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

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

Question 24. A monochromatic radiation of wavelength A and intensity f is incident on a plate of area A. Find the number of photons striking the plate per second.

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

Answer: 3. \(\frac{I \lambda A}{h c}\)

Question 25. The threshold frequency for a photosensitive metal is 3.3 × 10¹⁴ Hz. If a light of frequency 8.2 × 10¹⁴Hz is incident on this metal, the cut-off voltage for the photoelectron emission is nearly

1. 1V
2. 2V
3. 3V
4. 5V

Answer: 2. 2V

Question 26. The anode voltage of n photocell I kept fixed, The 24. The wave wavelength λ of the light falling on the cathode Is gradually changed. The plate current I of the photocell varies as follows

Answer: 2

Question 27. A body of mass 60g is moving with a velocity of 10 m. s^-1. The de Broglie wavelength of the body will be approximately (h = 6.63 × 10^-34J.s )

1. 10^-35m
2. 10^-25 m
3. 10^-33m
4. 10^-23m

Answer: 3. 10^-33m

Question 28. If the de Broglie wavelength of a gas molecule at 0°C is λ, what will be its wavelength at 819°C?

1. λ
2. λ/2
3. λ/3
4. λ/4

Answer:  2. λ/2

Question 29. Two moving electrons have a ratio of 1: 2 between their respective kinetic energies. The ratio between their de Broglie wavelength is

1. 2:1
2. √2:1
3. 1: √2
4. 1:2

Answer:  2. √2:1

Question 30. An electron is accelerated by a potential difference V and as a result, its de Broglie wavelength becomes A. If the applied potential difference was 2 V, the de Broglie wave¬ length would have been

1. 2λ
2. √2λ
3. λ/√2
4. λ/2

Answer: 4. λ/2

Question 31. The wavelength associated with an electron of mass m having kinetic energy E is given by

1. \(\frac{2 h}{m E}\)
2. 2mhE
3. \(\frac{2 \sqrt{2 m E}}{h}\)
4. \(\frac{h}{\sqrt{2 m E}}\)

Answer: 4. \(\frac{h}{\sqrt{2 m E}}\)

Question 32. The wavelength of a light is. 0.01 A° .If h is Planck constant, then the momentum of the corresponding photon will be

1. 10^-2 h
2. h
3. 10²h
4. 10¹² h

Answer: 4. 10¹² h

Question 33. Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100kV then the de Broglie wavelength associated with the electrons would

1. Increase by 2 times
2. Decrease by 2 times
3. Decrease by 4 times
4. Increase by 4 times

Answer: 2. Decrease by 2 times

Question 34. The de Broglie wavelength associated with proton changes by 0.25% If its momentum is changed by pQ. The initial momentum was

1. 100 p₀
2. p₀ / 400
3. 401 p₀
4. p₀ /100

Answer: 3. 401 p₀

Question 35. The proton when accelerated through a potential difference of V volt has a wavelength A associated with it. If an a -particle is to have the same wavelength A, it must be accelerated through a potential difference of

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

Answer: 1. \(\)

Question 36. The threshold frequency of a photoelectric effect depends on

1. Nature of the metal surface
2. The intensity of the incident radiation
3. The energy of the incident photon
4. The work function of the metal

Answer: 1 And 4

Question 37. Maximum kinetic energy of photoelectron depends on

1. Nature of the metal surface
2. Intensity of incident radiation
3. The energy of an incident photon
4. The work function of the metal

Answer: 1,3 And 4

Question 38. The work function of a metal surface is 2.0 eV. Light of wavelength 5000 A is incident on it:

1. The energy of each incident photon is 2.48 eV
2. The threshold wavelength for the photoelectric effect is 6200 A°
3. The maximum kinetic energy of the emitted photoelectron is 0.48 eV
4. The stopping potential is 0.48 eV

Answer: 1,2,3 And 4

Question 39. n number of photons of frequency f are emitted per second from a light source of power P (h =Planck’s constant; c = speed of light). Then

1. n = \(\frac{P}{h f}\)
2. Energy of each photon = \(\frac{P}{c n}\)
3. Momentum of each photon = \(\frac{P}{p n}\)
4. The value of n increases if the wavelength of the light increases

Answer: 1,3 And 4

Question 40. The threshold frequency and the threshold wavelength of photoelectric emission from a metal surface are fQ and. The; frequency and the wavelength of incident light are f and λ₀. Then

1. There will be no photoelectric effect If f>f₀
2. There will be no photoelectric effect If  λ> λ₀
3. Stopping potential ∝ ( f – f₀)
4. Maximum kinetic energy of photoelectron ∝ (f>f₀)

Answer: 2 And 3

Question 41. Work functions of two metals .1 and It art’ 3.1 eV and 1.9 eV respectively light of wavelength 3000 A° Is incident on both the surfaces.

1. No photo mission will take place in case of metal A
2. Photoelectrons will be emitted from both metals
3. The maximum kinetic energy of the photoelectron will be higher in metal It
4. Threshold wavelength of photoelectric effect In the case of metal .-1 will lie <1000 A° approximately

Answer: 2,3 And 4

Question 42. The de Broglie wavelength of a moving particle of mass m is A . For a few particles of different masses

1. \(\lambda \propto \frac{1}{m}\) , If their momenta are same
2. \(\lambda \propto \frac{1}{m}\) , If their velocities are same
3. \(\lambda \propto \frac{1}{m}\) , If kinetic energies are same
4. \(\lambda \propto \frac{1}{\sqrt{m}}\), If their kinetic energies are same

Answer: 2 And 4

Question 43. An electron (mass in ) and a proton (mass M) are acceler¬ rated with the same potential difference, then

1. Ratio of their velocities = \(\sqrt{\frac{M}{m}}\)
2. Ratio of their momenta = \(\sqrt{\frac{M}{m}}\)
3. The ratio of their kinetic energies = 1
4. Ratio of their de Broglie wavelengths = \(\sqrt{\frac{M}{m}}\)

Answer: 1, 3 And 4

Question 44. The wavelength of Ka X-ray for lead isotopes Pb208, Ph206, Pb204 are Av A2 and A3 respectively. Then

1. λ₁ = λ₂ =  λ₃
2. λ₁ > λ₂ > λ₃
3. λ₁ < λ₂ <  λ₃
4. \(\sqrt{\lambda_1 \lambda_2}\)

Answer: 1 And 4

Question 45. In which of the following situations the heavier of the two particles have a smaller de Broglie wavelength? The two particles

1. Move with the same speed
2. Move with the same kinetic energy
3. Move with the same linear momentum
4. Have fallen through the same height

Answer: 1,2 And 4

Question 46. The graph represents the variation of maximum kinetic energy with the frequency of an emitted photoelectron. this graph helps us to determine.

1. Work function
2. Planks constant
3. Threshold frequency
4. Charge on an electron

Answer: 1,2 And 3

Question 47. Einstein established the idea of photons or the basis of Planck’s quantum theory. According to his idea? the light of frequency f or wavelength A is a stream of photons The rest mass of each photon is zero velocity is equal to the mass of each photon, and velocity is equal to the velocity of light (c)
= 3 ×10^-8, J.s energy E = hf , where h = planks constant = 6.625 ×10^-34 J.s . Each photon has a momentum p= \(\), although its rest mass is zero. The number of photons increases when the intensity of incident light increases and vice-versa

On the other hand, according to de Broglie, any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is λ = h/p, where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K = eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, a charge of electron e = 1.6 ×10^-19C and mass = 9.1×10^-31kg.

1. The energy (in eV) of each photon associated with the light of wavelength 5893 A

1. 2.1
2. 3.9
3. 4.2
4. 5.89

Answer: 1. 2.1

2. The number of photons emitted per second from a light source of power 40 W and wavelength 5893 A

1. 3.95 ×10¹¹
2. 1.186 ×10²⁰
3. 3.56 ×10²⁰
4. 3.56×10²⁸

Answer: 2. 1.186 ×10²⁰

3. The number of photons emitted per second by a source of light of power 30 W is 10²⁰; the momentum of each photon (in kg – m. s^-1 )

1. 10^-24
2. 10^-25
3. 10^-26
4. 10^-37

Answer: 4. 10^-37

4. Two stationary electrons accelerated with potential differences V₁ and V₂ respectively such that V₁ : V₂ = n . The ratio of their de Broglie Wavelength

1. \(\sqrt{n}\)
2. \(\frac{1}{\sqrt{n}}\)
3. n2
4. \(\frac{1}{n^2}\)

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

5. A proton is 1836 times heavier than an electron and has the same charge as that of the electron. For what velocity of the proton will its de Broglie wavelength’ be 4455 A °?

1. 10⁶ m.s^-1
2. 10⁷ m.s^-1
3. 3 × 10⁶ m.s^-1
4. 3 × 10⁷m.s ^-1

Answer: 3.3 × 10⁶ m.s^-1

Question 48. Einstein’s equation for photoelectric effect is Emax = ty-Wo’ where h = Planck’s constant = 6.625 3 × 10^-34 m.s^-1J.s , f = frequency of light incident on metal surface, WQ = work function of metal and E = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency is less than a minimum value of f₀ or if the wavelength λ is greater than a maximum value λ₀, the value of E_max would be negative, which is impossible. Thus for a particular metal surface, λ₀ is the threshold frequency f₀ is the threshold wavelength for photoelectric emission to take place.

Again the collector plate is kept at a negative potential concerning the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becomes zero is known as the stopping potential, the photoelectric effect stops for a potential lower than this [velocity of light = 3× 10⁸ m.s^-1; the mass of an electron, m = 9.1× 10^-31 kg; charge of an electron, e = 1.6× 10¹⁹ C]

1. The. threshold wavelength of the photoelectric effect for a metal surface is 4600 A. The work function of the metal (in eV) is

1. 2.7
2. 3.45
3. 4.2
4. 6.9

Answer: 1. 2.7

2. Ultraviolet ray of wavelength 1800 A° is incident on the metal surface. The maximum velocity of the emitted photoelectron (in m s_1) is

1. 8.5 × 10⁵
2. 1.2 × 10⁶
3. 1.7 × 10⁶
4. 2.4 × 10⁶

Answer: 2. 1.2 × 10⁶

3. The stopping potential in case of an incident ultraviolet ray of wavelength 1800 A (in V) is

1. 2.7
2. 3.45 V
3. 4.2
4. 6.9

Answer: 3. 4.2

Question 49. The wavelength of matter waves associated with an electron of mass m having kinetic energy E is given by (h is Planck’s constant)

1. \(\frac{2 h}{m E}\)
2. 2mhE
3. \(\frac{2 \sqrt{2 m E}}{h}\)
4. \(\frac{h}{\sqrt{2 m E}}\)

Answer: 4. \(\frac{h}{\sqrt{2 m E}}\)

If p is the momentum, E = \(\frac{p^2}{2 m}\)

Or, p = \(\sqrt{2 m E}\)

Wavelength ofthe associated matter wave = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

Question 50. For a monochromatic light incident on a metal surface, the maximum velocity of the emitted photoelectrons is v . Then the stopping potential would be

1. \(\frac{2 m v^2}{e}\)
2. \(\frac{m v^2}{e}\)
3. \(\frac{m v^2}{2 e}\)
4. \(\frac{m v^2}{\sqrt{2} e}\)

Answer: 3. \(\frac{m v^2}{2 e}\)

If V₀ is the stopping potential, then the maximum energy of emitted photoelectrons = eV₀

∴ eV₀ =½mv²

Or, V₀= \(\frac{m v^2}{2 e}\)

Question 51. When green light is incident on a certain metal surface, electrons are emitted but no electrons are emitted with yellow light. If the red light is incident on the same metal surface

1. More energetic electrons will be emitted
2. Less energetic electrons will be emitted
3. The emission of electrons will depend on the intensity of light
4. No electrons will be emitted

Answer: 4. No electrons will be emitted

The energy of photons of red light is less than that of yellow light

Question 52. The wavelength of de Broglie waves associated with a thermal neutron of mass m at absolute temperature T is given by (k is the Boltzmann constant)

1. \(\frac{h}{\sqrt{m k T}}\)
2. \(\frac{h}{\sqrt{2 m k T}}\)
3. \(\frac{h}{\sqrt{3 m k T}}\)
4. \(\frac{h}{2 \sqrt{m k T}}\)

Answer:  3. \(\frac{h}{\sqrt{3 m k T}}\)

E = \(\frac{3}{2}\) kT

Again  E = ½mv²

Or, mv = \(\sqrt{2 m E}=\sqrt{2 m \cdot \frac{3}{2} k T}\)

= \(\sqrt{3 m k T}\)

de Broglie wavelength

λ = \(\frac{h}{m v}=\frac{h}{\sqrt{3 m k T}}\)

Question 53. Find the correct statements about the photoelectric effect.

1. There is no significant time delay between the absorption of suitable radiation and the emission of electrons.
2. Einstein analysis gives a threshold frequency above which no electron can be emitted.
3. The maximum kinetic energy of the emitted photoelectrons is proportional to the frequency of incident radiation.
4. The maximum kinetic energy of electrons does not depend on die intensity of radiation

Answer: 1 And 4

Question 54. The de Broglie wavelength of an electron is the same as that of a 9.50 KeV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is 03 MeV)

1. 1:50
2. 1:20
3. 20:1
4. 50:1

Answer: 3. 20:1

The energy of photons,

E = \(\frac{h c}{\lambda}\)

= 50 × 10³ eV

The kinetic energy of the electron

K = ½mv²

= \(\frac{h^2}{2 m \lambda^2}\)

Since v = h/mλ

= \(\frac{E}{K}=\frac{h c}{\lambda} \times \frac{2 m \lambda^2}{h^2}\)

= \(\frac{2 m \cdot c^2}{\left(\frac{h c}{\lambda}\right)}=\frac{2 \times 0.5 \times 10^6}{50 \times 10^3}=\frac{20}{1}\)

E:K = 20:1

Question 55. The work function of metals is in the range of 2 eV to 5 eV. Find which of the following wavelengths of light cannot be used for the photoelectric effect (Consider, Planck’constan = 4 × 10^-15 eV s, the velocity of light
= 3 × 10⁸ m/s )

1. 510 nm
2. 600 nm
3. 400 nm
4. 570 nm

Answer: 2. 600 nm

The energy of the incident photons should not be less than 2 eV

The energy of the photon, E = hv = hc/λ

Hence maximum value of the wavelength

λ_m = hc/E = \(\frac{\left(4 \times 10^{-15}\right) \times\left(3 \times 10^4\right)}{2}\)

= 6 × 10^-7m

= 6 × 10^-9m

=  600 nm

Therefore light of wavelength 600 nm cannot be used for a photoelectric effect

Question 56. Consider two particles of different masses. in which of the following situations the beaver of the two particles will have a smaller de Broglie wavelength?

1. Both have a free fall through the same height
2. Both move with the same kinetic energy
3. Both move with the same linear momentum
4. Both move with the same speed

Answer: 1, 2 And 4

de Broglie wavelength

λ = \(\frac{h}{p}=\frac{h}{m v}\)

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

h = Planks constant , m = mass, v= velocity, p= momentum, E = Kinetic energy

I the linear momentum of the two particles are equal their ewavelenght s will also be equal. The velocities of the particles are the same in both cases 1 And 4, hence will be relatively smaller for the heavier particle, Similarly, the E of the two particles i the same for case B, hence will be smaller again for the heavier particle.

Question 57. The potential difference V required for Accelerating an electron to have the de Broglie of 1/λ is wavelength of 1/λ  is

1. 100 V
2. 125 V
3. 150 V
4. 200V

Answer: 3. 150 V

de Broglie wavelength of an electron

λ = \(\frac{h}{\sqrt{2 m q V}}\)

or, V=  \(\frac{h^2}{2 m q \lambda^2}\)

= \(\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(1 \times 10^{-10}\right)^2}\)

= 1.496 × 10² ≈ 150 V

Question 58. The work function of cesium is 2.27 eV. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of 600 nm wavelength is

1. 0. 5 V
2. – 0.2 V
3. -0.5 V
4. 0.2 V

Answer: None of these

The energy of each photon incident on cesium

E = \(\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}\)

= 3.3 × 10^-19 J

= 2.06 eV

Which is less than the work function of cesium (2.27 eV). Hence there will not be any photocurrent

Question 59. The distance between a light source and a photoelectric cell is d. The distance is decreased to \(\) then

1. The emission of electrons per second will be four times
2. The maximum kinetic energy of photoelectrons will be four times
3. Stopping potential will remain the same
4. The emission of electrons per second will be doubled

Answer: 1 And 3

The intensity oflight incident on the photoelectric cell

I ∝ \(\frac{1}{d^2}\)

When the distance between the light source and the photoelectric cell is d.

Again, the number of photoelectrons emitted, n ∝ I

∴ n = \(\frac{1}{d^2}\)

Or = \(\frac{n_2}{n_1}=\left(\frac{d_1}{d_2}\right)^2\)

Or = \(n_2=n_1 \times\left(\frac{d_1}{d_2}\right)^2\)

= \(n_1\left(\frac{d}{\frac{d}{2}}\right)^2\)

= 4n₁

Maximum kinetic energy of photoelectrons,

E_max = eV₀ = hf – W₀

As the frequency (f) of incident light remains the same, the maximum kinetic energy (E_max) and stopping potential (E_max) will remain unchanged.

Question 60. The de Broglie wavelength of an electron is 0.4 ×10^-10m when its kinetic energy is 1.0 keV. Its wavelength will be 1.0  ×10^-10m when its kinetic energy is

1. 0.2 keV
2. 0.8 keV
3. 0.63 keV
4. 0.16 keV

Answer: 4. 0.16 keV

de Broglie wavelength, λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E_k}}\)

[p = momentum, E_k =  kinetic energy]

∴ \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\left(E_k\right)_2}{\left(E_k\right)_1}}\)

∴ \(\left(E_k\right)_2=\left(\frac{\lambda_1}{\lambda_2}\right)^2 \times\left(E_k\right)_1\)

= \(\left(\frac{0.4 \times 10^{-10}}{1.0 \times 10^{-10}}\right)^2 \times 1\)

= 0.16 keV

Question 61. When light of frequency ν₁ is incident on a metal with work function W (where hν₁>W), the photocurrent falls to zero at a stopping potential of V₁. If the frequency of light is increased to ν₂, the stopping potential changes to V₂. Therefore the change of an electron is given by.

1. \(\frac{W\left(\nu_2+\nu_1\right)}{\nu_1 V_2+\nu_2 V_1}\)
2. \(\frac{W\left(\nu_2+\nu_1\right)}{\nu_1 V_1+\nu_2 V_2}\)
3. \(\frac{W\left(\nu_2-\nu_1\right)}{\nu_1 V_2-\nu_2 V_1}\)
4. \(\frac{W\left(\nu_2-\nu_1\right)}{\nu_2 V_2-\nu_1 V_1}\)

Answer: 3.\(\frac{W\left(\nu_2-\nu_1\right)}{\nu_1 V_2-\nu_2 V_1}\)

According to Einstein’s photoelectric equation

h = W+eV₁ …………………………(1)

And = W+eV₂ …………………….(2)

Solving equation (1) and (2) we get

e = \(\frac{W\left(\nu_2-\nu_1\right)}{\nu_1 V_2-\nu_2 V_1}\)

Question 62. An electron accelerated through a potential of 10000 V from rest has a de Broglie wavelength A. What should be the accelerating potential so that the wavelength is 4 doubled? a

1. 20000 V
2. 40000 V
3. 5000 V
4. 2500 V

The stored energy in the electron accelerated through a potential of 10000 V

E = hf

Or, 10000 e = \(\) Where e = charge of electron

= de Broglie wavelength of the electron

Or, 5000 e = hc/2

The accelerating potential should be 5000 V so that the wavelength is doubled

Question 63. Radiation of wavelength A is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to \(\frac{3}{4}\), the speed of the fastest emitted electron will be

1. \(>v\left(\frac{4}{3}\right)^{1 / 2}\)
2. \(<v\left(\frac{4}{3}\right)^{1 / 2}\)
3. \(=v\left(\frac{4}{3}\right)^{1 / 2}\)
4. \(=v\left(\frac{3}{4}\right)^{1 / 2}\)

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

In the first case \(\frac{1}{2} m v^2=\frac{h c}{\lambda}-W_0\) ………. (1)

In the second case,

⇒ \(\frac{1}{2} m v^2=\frac{h c}{\lambda}-W_0\)

[where = speed of fastest emitted electron when wavelength is \(\frac{3}{4}\)

Or, \(\frac{1}{2} m v_1^2=\frac{1}{2} m v^2+\frac{h c}{3 \lambda}\)

Using equation (1)

Or, \(\sqrt{v^2+\frac{2 h c}{3 \lambda m}}=\sqrt{v^2+\frac{2}{3 m}\left(\frac{1}{2} m v^2+W_0\right)}\)

= \(\sqrt{\frac{4 v^2}{3}+\frac{2 W_0}{3 m}}\)

= \(v_1>\sqrt{\frac{4}{3} v^2} \quad \text { or, } v_1>\sqrt{\frac{4}{3}} v\)

Question 64. A particle A of mass m and initial velocity v collides with a particle B of mass y which is at rest. The collision is head-on and elastic. The ratio of the de Broglie wavelengths AA to AB after collision is

1. \(\frac{\lambda_A}{\lambda_B}=\frac{1}{3}\)
2. \(\frac{\lambda_A}{\lambda_B}\) = 2
3. \(\frac{\lambda_A}{\lambda_B}=\frac{2}{3}\)
4. \(\frac{\lambda_A}{\lambda_B}=\frac{1}{2}\)

Answer: 2. \(\frac{\lambda_A}{\lambda_B}\) = 2

Let after collision the velocities of A and B respectively vA and vB.

According to the law of conservation of momentum

mv = \(m v_A+\frac{m}{2} v_B\)

According to the law of conservation of relative velocity

v = v_B – v_A

Solving (1) and (2) we get, vA = \(\) and \(\frac{4 u}{3}\)

= \(\frac{\lambda_A}{\lambda_B}=\frac{\frac{h}{p_A}}{\frac{h}{p_B}}=\frac{p_B}{p_A}\)

= \(\frac{\frac{m}{2} v_B}{m v_A}=\frac{\frac{4 v}{3 \times 2}}{\frac{v}{3}}\)

= 2

Question 65. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increases from 0.5eV to 0.8eV. The work function of the metal is

1. 0. 65 eV
2. 1.0 eV
3. 1.3 eV
4. 1.5 eV

Answer: 2. 1.0 eV

According to Einstein’s equation,

E = hf – W₀ or,hf = E + W₀

For the first case, hf₁ = E₁ + W₀

For the second case, hf₂ = E₂ + W₀

⇒ \(\frac{h f_2}{h f_1}=\frac{E_2+W_0}{E_1+W_0} \text { or, } \frac{120}{100}=\frac{0.8+W_0}{0.5+W_0}\)

Or, 3+ 6W₀ = 4+5 W₀

9W₀ = 9W₀

W₀ =  \(\frac{9}{9}\)

W₀ =  1eV

Question 66. If the kinetic energy of the 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%

Wavelength = \(\)

= \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}} \quad \text { or, } \lambda \propto \frac{1}{\sqrt{E}}\)

= \(\frac{\lambda_2}{\lambda_1}=\sqrt{\frac{E_1}{E_2}}=\frac{1}{4}\)

Or, = 4λ₂

∴ Percentage change in wavelength

= \(\frac{\lambda_1-\lambda_2}{\lambda_1} \times 100=\frac{4 \lambda_2-\lambda_2}{4 \lambda_2} \times 100\)

= 75

Question 67. Radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (c = velocity of light)

1. E/c
2. 2E/c
3. 2E/c²
4. E/c²

Answer: 2. 2E/c

Transfer of momentum for an incident photon of energy hf is hf/c. Hence, the transfer of momentum for the incidence of energy

E is E/c

Again, the transfer of momentum due to the reaction for total reflection of radiation is E/c

∴ The total momentum transferred = 2E/c

Question 68. A certain metallic surface is illuminated with monochromatic light of wavelength, A. The stopping potential for photoelectric current for this light is 3 V₀. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V₀ The threshold wavelength for this surface for the photoelectric effect is

1. 6λ
2. 4λ
3. λ/4
4. λ/6

Answer: 2. 4λ

eV₀ = h(f-f₀)

Or, eV₀ = \(\left(\frac{c}{\lambda}-f_0\right)\)

f₀ = Threshold frequency

In the first case

e. 3V₀ = \(\left(\frac{c}{\lambda}-f_0\right)\)………………….. (1)

In the second case

= \(\left(\frac{c}{2 \lambda}-f_0\right)\)……………………… (2)

Dividing equation (1 ) by equation (2)

3 =  \(=\frac{\frac{c}{\lambda}-f_0}{\frac{c}{2 \lambda}-f_0}\)

Or, \(\frac{c}{\lambda}-f_0=\frac{3}{2} \cdot \frac{c}{\lambda}-3 f_0\)

Or,  2f₀ = \(\frac{1}{2} \frac{c}{\lambda} \quad \text { or, } f_0=\frac{c}{4 \lambda}\)

∴ Threshold wavelength, λ₀ = c/f₀ = 4λ

Question 69. Which of the following figures represents the variation of the particle momentum and the associated de Broglie wavelength?

Answer: 2.

de Broglie wavelength, \(\)

p = Momentum

So, pλ = h = constant

The λ – p graph will be a rectangular hyperbola.

Question 70. When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface Is illuminated with radiation of wavelength 2 λ, the stopping potential is V/4 The threshold wavelength for the metallic surface is

1. 5λ
2. 5/2λ
3. 3λ
4. 4λ

Answer: 3. 3λ

In the first eV = \(\frac{h c}{\lambda}-W_0\)  ……………………………….(1)

In the second case = \(e \frac{V}{4}=\frac{h c}{2 \lambda}-W_0\) ………………………………(2)

Subtracting equation (2) × 4 from equation (1 ), we get,

0 = \(-\frac{h c}{\lambda}+3 W_0\)

Or, \(W_0=\frac{h c}{3 \lambda}\)

Threshold wavelength, λ₀ = 3λ

Question 71. An electron of mass m and a photon have the same energy E. The ratio of de Broglie wavelengths associated with them is

1. \(\left(\frac{E}{2 m}\right)^{1 / 2}\)
2. \(c(2 m E)^{1 / 2}\)
3. \(\frac{1}{c}\left(\frac{2 m}{E}\right)^{1 / 2}\)
4. \(\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}\)

Answer: 4.

de Broglie wavelength of electron = \(\lambda_e=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

de Broglie wavelength of photon = \(\lambda_p=\frac{h c}{E}\)

∴ \(\frac{\lambda_e}{\lambda_p}=\frac{h}{\sqrt{2 m E}} \times\)\(\frac{E}{h c}\)

= \(\frac{1}{c} \sqrt{\frac{E}{2 m}}\)

Question 72. If the mass of a neutron is 1.7 × 10^-27 kg, then the de (*ÿ50 Broglie wavelength of neutron of energy 3 eV is h = 6.6 × 10^-34 J.s)

1. 1.4 × 10^-1111m
2. 1.6 × 10^-10 m
3. 1.65 × 10^-11m
4. 1.4 × 10^-10m

Answer: 4. 1.4 × 10^-10m

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

= \(\frac{6.6 \times 10^{-34}}{\sqrt{\left[2 \times\left(1.7 \times 10^{-27}\right) \times\left(3 \times 1.6 \times 10^{-19}\right)\right]}}\)

1.634 × 10^-10m

Question 73. In an experiment of photoelectric effect, the stopping potential was measured to be V₁  and V₂ with incident light of wavelength λ and λ/2 respectively. The relation V₁ and V₂ is

1. V₂ >2V₁
2. V₂ >V₁
3. V₁ <V₂< 2V₁
4. V₂ = 2V₁

Answer: 1. V₂ >2V₁ 

ev = hf – W₀

Or \(\frac{h c}{\lambda}=e V+W\)

For these two cases \(\frac{h c}{\lambda}=e V_1+W\) eV1+ W

And \(\frac{h c}{\lambda / 2}=e V_2+W\) hc/ = ev₁+W₀

Hence, \(\frac{2 h c}{\lambda}=e V_2+W\)

∴ eV₁ + W = 2eV₁ + 2W

V₂ = 2V₁ + \(\frac{W}{e}\)

So, V₂ > 2V₁

Question 74. An electron of mass m with an initial velocity \(\vec{V}=V_0 \hat{i}\), field (V>0) enters an electric \(\vec{E}=E_0 \hat{i}\) E₀ = constant > 0) at t = 0. If λ₀ is its de Broglie de Broglie wavelength of electron

1. λ₀t
2. \(\lambda_0\left(1+\frac{e E_0}{m V_0} t\right)\)
3. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m V_0} t\right)}\)
4. λ₀ 

Answer: 3. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m V_0} t\right)}\)

Initial de Broglie wavelength of the electron,

⇒ \(\lambda_0=\frac{h}{m V_0}\)

Velocity ofthe electron after time t

V = u+ at = \(V_0+\frac{F}{m} t=V_0+\frac{e E_0}{m} t\)

= \(V_0\left[1+\frac{e E_0}{m V_0} t\right]\)

= \(\frac{h}{m V}=\frac{h}{m V_0\left[1+\frac{e E_0}{m V_0} t\right]}\)

= \(\frac{\lambda_0}{\left[1+\frac{e E_0}{m V_0} t\right]}\)

Question 75. When the light of frequency 2ν₀ (where l/Q is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is v₁. When the frequency of the incident radiation is increased to 5ν₀, the maximum velocity ofelectrons emitted from the same plate is v₂. The ratio of v₁ to v₂ is

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

Answer: 3. 1:2

E = \(W_0+\frac{1}{2} m v^2\)

First case \(h \nu_0+\frac{1}{2} m v_1^2\)

Or, \(h \nu_0=\frac{1}{2} m \nu_1^2\) …………….. (1)

For the second case \(=h \nu_0+\frac{1}{2} m \nu_2^2\)

Or, \(-4 h \nu_0=\frac{1}{2} m v_2^2\)

Now dividing (1) by (2), we have

⇒ \(\frac{\nu_1^2}{\nu_2^2}=\frac{h \nu_0}{4 h \nu_0}=\frac{1}{4}\)

Or, \(\frac{v_1}{v_2}\) = 1:2

Question 41. The wavelength of de Broglie waves associated with a thermal neutron of mass m at absolute temperature T is given by (k is the Boltzmann constant)
Answer:

1. \(\frac{h}{\sqrt{m k T}}\)
2. \(\frac{h}{\sqrt{2 m k T}}\)
3. \(\frac{h}{\sqrt{3 m k T}}\)
4. \(\frac{h}{2 \sqrt{m k T}}\)

Answer: 3. \(\frac{h}{\sqrt{3 m k T}}\)

E \(\frac{3}{2}\)

Again E = ½ mv²

Or, mv = \(\sqrt{2 m E}\)

= \(\sqrt{2 m \cdot \frac{3}{2} k T}=\sqrt{3 m k T}\)

de Broglie wavelength \(\frac{h}{m v}=\frac{h}{\sqrt{3 m k T}}\)

Unit 7 Dual Nature Of Matter And Radiation Synopsis

When the surface of a substance is irradiated by light of a suitable wavelength, electrons are emitted from that surface. This phenomenon is known as the photoelectric effect or photo¬ electric emission.

1. The electrons emitted in the photoelectric effect are called
photoelectrons.

2. With the help of suitable arrangement a stream of unidirectional photoelectrons can be obtained and the electric current thus produced is called photoelectric current.

3. The minimum amount of energy required to remove an electron from the surface of a particular substance, is called the work function of that substance.

4. The minimum negative potential of the anode concerning the photocathode, for which photoelectric current becomes zero is called the stopping potential.

5. The value of stopping potential depends on

• The nature of the metallic surface of the photocathode and
• The wavelength or frequency of the incident light.

6. The minimum frequency of the incident radiation required to emit photoelectrons from the surface of a substance, is called the threshold frequency for that substance.

7. Photoelectric emission is an instantaneous process.

8. The cell made based on the photoelectric effect where light energy is converted into electrical energy, is called a photoelectric cell.

9. According to de Broglie’s hypothesis, matter also behaves as waves.

10. The photoelectric effect cannot be explained in terms of the wave theory oflight. Einstein first introduced the concept of photon particles by using Planck’s quantum theory. Elec¬ tromagnetic radiation consists of a stream of particles. These particles are called photons

Unit of Planck’s constant (h) :

Unit of h in SI = J .s

Unit of h in CGS system = org-s

Unit of h In eV = eV.s

11. Einstein’s photoelectric equation is based on the quantum theory of radiation. This equation correctly explains the following observations in the photoelectric effect.

1. Maximum kinetic energy of photoelectrons,
2. Threshold frequency,
3. Photoelectric emission is Instantaneous,
4. Dependence of photoelectric current on Intensity of incident light.

12. Radiation sometimes behaves like waves and sometimes like a stream of particles. Thus, wave theory and particle theory are not contradictory but complementary to each other.

13. According to de Broglie’s hypothesis, the concept of matter waves is only important in the case of particles of atomic dimensions.

14. In 1927, two American scientists Davisson and Germer first experimentally, demonstrated diffraction of electrons to prove the existence of matter waves.

15. The matter wave cannot be represented by a pure sinusoidal wave. Matter waves can be represented by duly formed wave groups or wave packets.

16. Matter wave is neither a type of elastic wave nor a magnetic wave.

17. At any instant, a moving particle is located at a specific point but at that instant matter wave associated with that particle extends over some space. This is an inherent property of matter-wave

18. The relation between the kinetic energy of photoelectrons and the stopping potential is,

½ mv²_max = eV₀

Or, \(\sqrt{\frac{2 e V_0}{m}}\)

m – mass of electron having charge e, v_max = maximum initial velocity of photoelectron and V₀ =stoppingpotential

4 If the threshold frequency is f₀,

Threshold wavelength,

λ₀ = \(\frac{c}{f_0}\) = c( velocity of light)

Amount of energy carried by a photon, E = hf.

Here, f = frequency of the radiation, h = Planck’s constant

19. According to the theory of relativity, if the rest mass of a particle is m₀ and its momentum is p, the energy of the particle

E = \(\sqrt{p^2 c^2+m_0^2 c^4}\)

In the case of a photon m₀ = 0

Hence E = pc

∴ p = \(\frac{E}{c}=\frac{h f}{c}\)

20. The relation between the wavelength of radiation and photon energy is E = 12400/ λ (inÅ)

21. Einstein’s photoelectric equation can be written as,

E_max  = hf-W₀

Or, ½mv² =  hf-W₀

Or, ev₀ = hf- W₀

22. When a radiation of frequency f is incident on a metal of threshold frequency f₀(f > f₀)’, ten the maximum kinetic energy of emitted photoelectrons

E_max= hf-hf₀= h(f-f₀)

Since (W₀= hf₀)

= hc \(\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\)

Where λ and λ₀ are the wavelength of the incident light and threshold wavelength for the metal surface, respectively.

This equation is another form of Einstein’s photoelectric equation. Important information

h = 6.625 ×10^-12 erg. s , c = 3 ×10¹⁰ cm .s^-1

1 eV = 1.6 × 10^-12 erg

22. de Broglie wavelength, λ = \(\frac{h}{p}\)

Where, p = mv – momentum of the particle

23. If an electron is accelerated by V volt, then the de Broglie wavelength associated with the moving electron

λ =   \(\frac{12.27}{\sqrt{V}}\) Å

24. de Broglie wavelength of any molecule (mass m ) of gas at temperature TK,

λ =   \(\frac{h}{\sqrt{3 m k T}}\)

(k = Boltzmann constant)

25. Number of photons of wavelength X emitted from a lamp of power P in time t is

n = \(\frac{p t \lambda}{h c}\)

26. Photoclectrons are easily produced from the surface of metals that have a low work function. For example, cesium.