Q. Photoelectric emission is observed from a metallic surface for frequencies $v_{1}$ and $v_{2}$ of the incident light rays $\left( v_{1} > v_{2} \right).$ If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1\ :k,$ then the threshold frequency of the metallic surface is
A) $\frac{v_{1} - v_{2}}{k - 1}$
B) $\frac{kv_{1} - v_{2}}{k - 1}$
C) $\frac{kv_{2} - v_{1}}{k - 1}$
D) $\frac{v_{2} - v_{1}}{k}$
Explanation
(b) By using $hv - hv_{0} = K_{\max}$
$\Rightarrow h\left( v_{1} - v_{0} \right) = K_{1}$ …(i)
And $h\left( v_{2} - v_{0} \right) = K_{2}$ …(ii)
$\Rightarrow \frac{v_{1} - v_{0}}{v_{2} - v_{0}} = \frac{K_{1}}{K_{2}} = \frac{1}{K},$ Hence $v_{0} = \frac{Kv_{1} - v_{2}}{K - 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.