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(C) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by $eV_0 = h
u - W_0$,where $W_0$ is the work function.This

Vedclass · Accepted Answer

$(iv) > (iii) > (ii) > (i)$

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(C) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by $eV_0 = h
u - W_0$,where $W_0$ is the work function.This can be rewritten as $V_0 = \frac{h}{e}
u - \frac{W_0}{e}$.The graph between $V_0$ and $
u$ is a straight line with a slope of $\frac{h}{e}$.The $x$-intercept of this graph is the threshold frequency $
u_0$,where $V_0 = 0$,so $h
u_0 = W_0$.From the graph,the threshold frequencies are in the order: $(
u_0)_{(iv)} > (
u_0)_{(iii)} > (
u_0)_{(ii)} > (
u_0)_{(i)}$.Since the work function $W_0 = h
u_0$,the work functions follow the same order: $(W_0)_{(iv)} > (W_0)_{(iii)} > (W_0)_{(ii)} > (W_0)_{(i)}$.Therefore,the correct ranking is $(iv) > (iii) > (ii) > (i)$.

The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

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