Question

The stopping potential as function of frequency of incident radiation is plotted for two different photoelectric surfaces A and B. The graphs show that the work function of A is

• A. greater than that of B
• B. lesser than that of B
• C. equal than that of B
• D. no comparison

Answer 1

The correct option is B lesser than that of B

From Einstein’s equation ,

$e{V}_0=h\nu –h{\nu }_0$

$V_0=\frac{h\nu }{e}-\frac{h{\nu }_0}{e}$

From graph, threshold frequency of A is less than that of B.

So, as work function = $h{\nu }_0$

Where $h=Planc{k}^{\prime }sConstant$

So, work function of A is less than B.