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प्रश्न

On the basis of the graphs shown in the figure, answer the following questions :
(a) Which physical parameter is kept constant for the three curves?
(b) Which is the highest frequency among v1, v2, and v3?
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उत्तर

(a) Intensity is kept constant for all three curves. The given graph shows the variation of photocurrent with collector plate potential for the same intensity of light radiation at various frequencies.
(b) The stopping potential is found to be changing linearly with the frequency of incident light. The stopping potential is more negative for higher frequencies of incident radiation. As an increase in the frequency of the incident light increases the kinetic energy of the emitted electrons, so the greater retarding potential is required to stop them completely.
Hence higher the negative potential, the more the frequency of incident electrons.
As V1 is more negative than V2 and V2 is more negative than V3, hence, V1 > V2 > V3
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संबंधित प्रश्न
Define the term 'intensity of radiation' in terms of photon picture of light.
Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.
The equation E = pc is valid
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(a) photoelectric emission will not take place
(b) photoelectric emission may or may not take place
(c) the stopping potential will increase
(d) the stopping potential will decrease
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(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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Do all the electrons that absorb a photon come out as photoelectrons?
Consider a thin target (10–2 cm square, 10–3 m thickness) of sodium, which produces a photocurrent of 100 µA when a light of intensity 100W/m2 (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m3].
How would the stopping potential for a given photosensitive surface change if the intensity of incident radiation was decreased? Justify your answer.
