The figure shows the variation of photocurren | vedclass

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Question

(C) In the photoelectric effect,the stopping potential depends only on the frequency of the incident radiation and the work function of the metal surf

Vedclass · Accepted Answer

$f_a = f_b = f_c = f_d$ and $I_a < I_b < I_c < I_d$

Vedclass · Answer

(C) In the photoelectric effect,the stopping potential depends only on the frequency of the incident radiation and the work function of the metal surface. Since all four curves intersect the potential axis at the same point (the same stopping potential),the frequency of all four radiations must be the same. Therefore,$f_a = f_b = f_c = f_d$.The saturation photocurrent is directly proportional to the intensity of the incident radiation. From the graph,the saturation current values are $I_a < I_b < I_c < I_d$. Therefore,the intensities follow the same order: $I_a < I_b < I_c < I_d$.

The figure shows the variation of photocurrent with anode potential for four different radiations. Let $I_a, I_b, I_c$ and $I_d$ be the intensities for the curves $a, b, c$ and $d$ respectively $[f_a, f_b, f_c$ and $f_d$ are frequencies respectively].

Similar Questions

Photoelectric effect occurs when ultraviolet rays are incident on a metal plate. It does not occur when which of the following are incident?

Assertion : Photoelectric saturation current increases with the increase in frequency of incident light.
Reason : Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.

If the intensity of incident light on a metal surface is decreased,then . . . . . .

Which one of the following is true in photoelectric emission?

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