Q. 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 of $B$)
A) 1 :1
B) 1 :2
C) 1 :3
D) 1 :4
Explanation
(b) $E = W_{0} + K_{\max}$
$\Rightarrow hf = W_{A} + K_{A}$ …(i)
and$\ 2hf = W_{B} + K_{B} = 2W_{A} + K_{B}\ \left\lbrack \because\frac{W_{A}}{W_{B}} = \frac{1}{2} \right\rbrack\ \ $…(ii)
Dividing equation (i) by (ii)
$\frac{1}{2} = \frac{W_{A} + K_{A}}{2W_{A} + K_{B}} \Rightarrow \frac{K_{A}}{K_{B}} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.