Question

The stopping potential as a function of frequency is plotted for two photoelectric surfaces A and B. The graph show that the work function of A is :

• A. greater than that of B
• B. smaller than that of B
• C. same as that of B
• D. such that no comparison can be done from given graphs

Answer 1

The correct option is B smaller than that of B

The relationship between the stopping potential $V_0$ and the work function $h{\nu }_0$ is
$e{V}_0=h\nu -h{\nu }_0$.
Rearrange this equation
$h{\nu }_0=h\nu -e{V}_0$
Thus, higher is the value of the stopping potential $V_0$, lower will be the value of the work function $h{\nu }_0$ and vice versa.
From the graph, it can be seen that for given value of frequency $\nu$, the stopping potential of A is higher than the stopping potential of B.
Hence, the work function of A is lower than the work function of B.
Hence, option B is correct.