Question: Q. 3. Explain giving reasons for the following :
(i) Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation.
(ii) The stopping potential $\left(V_{0}\right)$ varies linearly with the frequency $(v)$ of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces.
(iii) Maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
U [OD II 2017]
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Solution:
Ans. (i) The collision of a photon can cause emission of a photoelectron (if the frequency is above the threshold frequency). As intensity if the frequency increases, number of photons increases. Hence, the current increases.
(ii) We have,
Graph of $2 s, v$ is a straight line and slope $\left(=\frac{h}{e}\right)$
is a constant.
$1 / 2$
(iii) Maximum K.E. for different surfaces $=h\left(v-v_{0}\right) \quad 1 / 2$ Hence, it depends on the frequency and not on the intensity of the incident radiation.
[CBSE Marking Scheme 2017]
