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.

(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.