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

Answer: (B)

1 : 2

Answer 2

By using $E = W_{0} + K_{\max}$ ⇒ $E_{A} = hf = W_{A} + K_{A}$ and

$E_{B} = h(2f) = W_{B} + K_{B}$

So, $\frac{1}{2} = \frac{W_{A} + K_{A}}{W_{B} + K_{B}}$ ……(i) also it is given that

$\frac{W_{A}}{W_{B}} = \frac{1}{2}$ ……..(ii)

From equation (i) and (ii) we get $\frac{K_{A}}{K_{B}} = \frac{1}{2}.$