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

By using $E=W_0+K_{max}\Rightarrow E_A=hf=W_A+K_A$ and $E_B=h\left(2f\right)=W_B+K_B$
So, $E_B=h\left(2f\right)=W_B+K_B\frac{1}{2}=\frac{W_A+K_A}{W_B+K_B}$ .....(i) also it is given that $\frac{W_A}{W_B}=\frac{1}{2}$ .....(ii)
From equation (i) and (ii) we get $\frac{K_A}{K_B}=\frac{1}{2}$.