Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies $v_{A} > v_{B}$ .

(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.

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Solution

(i) We know that,

$V_{0} = \frac{h}{e} (v - v_{0})$
For the same value of v, stopping potential is more for which threshold frequency ( $v_{0}$ ) is less.

Given, $v_{A} > v_{B}$

Therefore, threshold frequency for metal B is less than the threshold frequency for metal A. Hence, stopping potential is more for metal B.

(ii) No, the slope of the graph tells us the value of $h / e$ which is same for both the materials. So, it does not depend on the nature of the materials.

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