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)

• A. 1 : 1
• B. 1 : 2
• C. 1 : 3
• D. 1 : 4

Answer 1

The correct option is B

1 : 2

$E=W_0+K_{max}\cdots \left(i\right)$
$\Rightarrow hf=W_A+K_A\cdots \left(ii\right)$
$and2hf=W_B+K_B=2W_A+K_B(\because \frac{W_A}{W_B}=\frac{1}{2})$
Dividing equation (i) by (ii)
$\frac{1}{2}=\frac{W_A+K_A}{2W_A+K_B}\Rightarrow \frac{K_A}{K_B}=\frac{1}{2}$