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प्रश्न
Answer the following question.
Why is the wave theory of electromagnetic radiation not able to explain the photoelectric effect? How does a photon picture resolve this problem?
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उत्तर
There are three main drawbacks:
- Intensity: If we consider light as a wave then as the intensity of the light is increased the amplitude of the oscillation of the electron will increase. Thus, as the intensity of the incident light is increased the maximum kinetic energy of the emitted electron will also increase. But it was observed that the kinetic energy of the emitted electrons does not depend on the intensity whereas the magnitude of the photoelectric current increases with the frequency.
- Frequency: If we consider the light as a wave then the photoelectric emission should happen on any frequency, but it was observed that the electrons are emitted after a particular frequency. If the frequency of the incident light is lesser than this frequency there is no photoelectric emission observed.
- Time Delay: According to the wave theory the energy is uniformly distributed over the wavefront. As the light falls on the metallic surface, it will take some time for the electron to gain sufficient energy to get emitted. But experimentally it was observed that the electrons are emitted instantaneously as the light falls on the metallic surface.
How the photon theory can explain the photoelectric effect:
- According to photon theory increasing the intensity means increasing the number of photons that do not change the maximum kinetic energy but changes the number of ejected electrons.
- The energy of a photon is given as E = hf that explains the dependence of the energy on the frequency, after a particular frequency of a photon that is threshold frequency there is photoelectric emission.
- As soon as a photon falls on the metallic surface it is absorbed, hence the electron is ejected instantaneously. Hence, all these are in accordance with the experimental observations.
संबंधित प्रश्न
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the a radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1?
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?
| Metal | Work Function (eV) |
| Na | 1.92 |
| K | 2.15 |
| Ca | 3.20 |
| Mo | 4.17 |
Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron?
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
In the arrangement shown in the figure, y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Work function of aluminium is 4.2 eV. If two photons each of energy 2.5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.
Cathode rays can be deflected by
In photoelectric effect, the photoelectric current
