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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.
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Define the term "cut off frequency" in photoelectric emission. The threshod frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photo-electrons is v1. When the frequency of the incident radiation is increased to 5f, the maximum velocity of phto-electrons is v2. Find the ratio v1 : v2.
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 silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which light energy of 1.0 × 10−7 J falls on the surface. Assuming that on average, one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball, assuming zero potential at infinity. What is the potential at the centre of the ball?
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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According to the wave picture, what is light?
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According to wave theory, how should greater radiation intensity affect the energy absorbed by each electron?
How should the maximum kinetic energy of photoelectrons change with increasing intensity, according to wave theory?
Why does the energy absorbed by each electron per unit time become very small in the wave picture?
