First slide

Photo Electric Effect

Question

The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies f and, 2f are incident on the surfaces of A and.B respectively, the ratio of the maximum kinetic energies of photoelectrons emitted is (f is greater than threshold frequency of A, 2f is greater than threshold frequency of B)

Moderate

A

1 : 1
B

1 : 2
C

1 : 3
D

1 : 4

Solution

$\begin{array}{l} E = W_{o} + K_{\max} . . . . . (i) \\ \Rightarrow hf = W_{A} + K_{A} . . . . . (ii) \\ and 2 hf = W_{B} + K_{B} = 2 W_{A} + K_{B} (∵ \frac{W_{A}}{W_{B}} = \frac{1}{2}) \\ Dividing equation (i) by (ii) \\ \frac{1}{2} = \frac{W_{A} + K_{A}}{2 W_{A} + K_{B}} \Rightarrow \frac{K_{A}}{K_{B}} = \frac{1}{2} \end{array}$

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