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प्रश्न
Light of intensity ‘I’ and frequency ‘v’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when (ii) the frequency of incident radiation is increased. In each case, all other factors remain the same. Explain, giving justification in each case.
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उत्तर
For photoelectric emission to occur, there is a minimum cut off frequency of the incident radiation called the threshold frequency below which no photoelectric emission occurs. This frequency is independent of the intensity of the incident light. With an increase in the frequency of the incident radiation, the kinetic energy of the photoelectrons ejected increases, whereas it is independent of the number of photoelectrons ejected. Hence, with the increase in the frequency of incident radiation, there will not be any change in the anode current.
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संबंधित प्रश्न
Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies vA > vB.
(i) In which case is the stopping potential more and why?
(ii) Does the slope of the graph depend on the nature of the material used? Explain.
The photoelectric work function for a metal is 4.2 eV. If the stopping potential is 3V, find the threshold wavelength and maximum kinetic energy of emitted electrons.
(Velocity of light in air = 3 x 108m/s,
Planck's constant = 6·63 x10-34 J -s,
Charg.e ori electron = 1·6 x 10 -19 C)
Light of intensity ‘I’ and frequency ‘v’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when (i) the intensity of light is gradually increased. In each case, all other factors remain the same. Explain, giving justification in each case.
Use Einstein's photoelectric equation to explain the observations from this graph ?
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