Dual Nature Of Matter And Radiation Short Answer Questions

Dual Nature Of Matter And Radiation Short Answer Questions

Question 1. Photoelectric current continues when the anode is even at a slight negative potential concerning the photocathode. Explain.
Answer:

Emitted photoelectrons possess some initial kinetic energy. Hence, a few electrons can reach the anode overcoming the repulsive force of the negative potential. If the value of the negative potential is not too high, the photoelectric current continues to flow

Question 2. When radiation of wavelength 2000A° is incident on a nickel plate, the plate gets positively charged. But when the wavelength of Incident radiation is raised to 3440 A°, the plate remains neutral. Explain
Answer:

The photoelectric effect takes place for incident radiation of wavelength 2000A and due to the emission of photoelectrons, the plate becomes positively charged. No photoelectric effect takes place when radiation of wavelength 3440A is incident on the plate, as the threshold wavelength for photoelectric effect is more than 2000 A but less than 3440A

Question 3. Give examples of the production of

  1. An electron by a photon and
  2. Photon by an electron.

Answer:

  1. When a photon is incident on the metal surface, the electron is emitted in a photoelectric effect.
  2. In the process of production of X-rays, when high-speed electrons in cathode rays are incident on a metal surface, an X-ray photon is emitted.

(Hence photoelectric effect and X-ray production are two opposite site effects)

Question 4. A metal plate that emits photoelectrons under the influence of blue light, may not emit photoelectrons under the influence of red light. Explain.
Answer:

  • For the emission of electrons from a metal surface, the frequency of the incident light should be higher than the threshold frequency ofthe metal.
  • The frequency of blue light is greater than the frequency of red light.
  • If the frequency of blue light is higher than the threshold frequency for the metal, electrons will be emitted.
  • But if the frequency of red light is less than the threshold frequency, red light will not be able to emit electrons from that metal surface

Question 5. Radiation of frequency 1015 Hz is Incident separately on two photosensitive surfaces P and Q. The following observations were made:

  1. Surface P: Photoemission occurs but the photoelectrons have zero kinetic energy
  2. Surface Q: Photoemission occurs and electrons have non-zero kinetic energy.

Question 6. Which of these two has a higher work function? If the frequency of incident light Is reduced, what will happen to photoelectron emission In the two cases?
Answer:

The energy of the incident photon, hf = Ek + W0; here W0 – work function of the emitting surface, and Ek = kinetic energy of the emitted photoelectron

. Given, for the surface P, Ek = 0; but for the surface Q, Ek> 0. For the same incident light, hf= constant. So the work function W0 is higher for the surface P.

If incident light of a lower frequency is taken, he would reduce. Then it would be less than the work function of the surface P; so, this surface would emit no photoelectron. The surface Q may emit photoelectrons, provided hf > W0 for that surface

Question 7. State two important properties of photons which
Answer:

The properties of photons used to write Einstein’s photoelectric equation are

  1. The rest mass of the photon is zero,
  2. The energy ofthe photon is E = hv

Question 8. An electron (charge = e, mass = m ) is accelerated from rest through a potential difference of V volt. What will be the de Broglie wavelength ofthe electron?
Answer:

The kinetic energy gained by the electron

= eV = ½mv²

Or, v = \(\sqrt{\frac{2 e V}{m}}\)

Hence, de Broglie wavelength,

λ = \(\frac{h}{m \nu}=\frac{h}{m} \sqrt{\frac{m}{2 e V}}\)

= \(\frac{h}{\sqrt{2 m e V}}\)

Question 9. When light incident on a metal ha* energy le** than the work function of the metal, then no electron h emitted from the surface of the metal. Mathematically justify this statement
Answer:

Einstein’s photoelectric equation,

½mv²max = hf – W0 = work function

If hf < W0, the maximum kinetic energy of electrons (½mv²max )is negative and so v²max   is negative. However, the magnitude of the square of a physical quantity cannot be negative. Hence photoelectrons are not emitted from the surface ofthe metal

Question 10. Is matter wave an electromagnetic wave?
Answer:

No, matter wave is not an electromagnetic wave, because the matter waves are not associated with periodic vibrations of the electric and magnetic fields

Question 11. A beam of photons of energy 5.0 eV falls on a free metal surface of work function 3.0 eV. As soon as the photo-electrons are emitted, they are removed. But the emission comes to a stop after a while explain the reasons, i
Answer:

When the photons of energy 5.0 eV fall on a free metal surface of work function 3.0 eV, electrons are emitted from the surface instantaneously and the surface becomes positively charged. After a while, the number of free electrons in the metal surface decreases and more energy is required for the emission of electrons from the metal surface, i.e., the work function of the metal increases.

When the work function becomes greater than 5.0 eV, the photons of energy 5.0 eV are not able to eject electrons from the metal surface. So, the emission of electrons ceases after some time. To overcome this difficulty/the metal surface is connected to a negative pole of an electric source to make it negatively charged.

Question 12. Does the photoelectric emission take place due to the incidence of visible light and ultraviolet rays on the faces of different types of metals in our everyday experience?
Answer:

Yes, the photoelectric emission does take place. But in the absence of any positively charged collector, the emitted photoelectrons accumulate on the metal surface and make a layer of negative charges. Within a very short period, the emission of photoelectrons comes to a stop due to the repulsion of the negatively charged layer. Hence the photoelectric emission stops after a while

Question 13. Why the photoelectric emission can not be performed by using X-rays and gamma rays?
Answer:

When a photon of sufficient energy falls on a metal surface, the photon vanishes by imparting all its energy to an electron and the electron comes out of the metal surface. This phenomenon is known as the photoelectric effect. But in the case of X-rays and gamma rays, the photon energy is too high for the electron of the metal to absorb so that the photon, is not annihilated by imparting its whole energy to an electron. Hence the photoelectric emission does not take place with X-ray or gamma-ray photons.

Question 14. What is the relation of the de Broglie wavelength of a moving particle with temperature?
Answer:

If the thermal energy is not a source of kinetic of a proving,partic)e, then the de Broglie wavelength of that particle is independent of the temperature. Hence the de Broglie wavelength emitted particles in photoelectric effect radioactive radiation etc. have no relation with temperature. T is given by v ∝ \(\nu \propto \sqrt{T}\).

Therefore, the de Broglie wavelength

λ = \(\lambda=\frac{h}{m v} \quad \text { or, } \lambda \propto \frac{1}{\sqrt{T}}\)

Question 15. The threshold frequency for a certain, metal is 3.3 × 10 14 Hz. If the light of frequency 8.2 × 10 -14 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:

We know, eV0 = hf-hf0

Or, V0 = \(\frac{h}{e}\left(f-f_0\right)\)

= \(\frac{6.626 \times 10^{-34}\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}}\)

= 2.03V

Question 16. What is the de Broglie wavelength of a bullet of mass 0.040 kg traveling at the speed of 1.0 km s-1?
Answer:

Here, m = 0.04 J kg, v = 1.0 km. s-1  = 103 m. s-1

∴ λ = \(\frac{h}{m v}=\frac{6.626 \times 10^{-34}}{0.040 \times 10^3}\)

= 1.66 ×10-35m

Question 17. Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic?
Answer:

The work function of a metal is the minimum energy required to knock out an electron from the highest Tilled level of the conduction band. The conduction band comprises different energy levels forming a continuous band of levels. Electrons in different energy levels need different energy for emission. Thus, the electron, once emitted has different kinetic energies depending on the energy supplied to the emitter.

Question 18. Write the expression for the de Broglie wavelength associated with a charged particle having charge q and mass m, when it is accelerated by a potential V.
Answer:

Kinetic energy acquired, E = qV

Momentum, p = \(\sqrt{2 m E}=\sqrt{2 m q V}\)

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

Question 19. If light of wavelength 412.5 nm is incident on each of the metals given in the table, which ones will show photoelectric emission and why?
Answer:

Dual Nature Of Matter And Radiation Metal And Work Function

The energy of incident light,

E = 12400/λ(in Å)

= \(\frac{12400}{4125}\)

= 3 eV

Na and K will show photoelectric emission because their work functions are less than the energy of incident light

Question 20. Does the stopping potential depend on

  1. The intensity and
  2. The frequency of the incident light? Explain

Answer:

  1. Stopping potential does not depend on the intensity of the incident light.
  2. Stopping potential Is directly proportional to the frequency of the incident light

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