Q. Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E.$ If the kinetic energy is to be increased to $2E,$ the wavelength must be changed to $\lambda'$ where
A) $\lambda^{'} = \frac{\lambda}{2}$
B) $\lambda^{'} = 2\lambda$
C) $\frac{\lambda}{2} < \lambda^{'} < \lambda$
D) $\lambda^{'} > \lambda$
Explanation
(c) $E = \frac{hc}{\lambda} - W_{0}\ and\ 2E = \frac{hc}{\lambda^{'}} - W_{0}$
$\Rightarrow \frac{\lambda^{'}}{\lambda} = \frac{E + W_{0}}{2E + W_{0}} \Rightarrow \lambda^{'} = \lambda\left( \frac{1 + W_{0}/E}{2 + W_{0}/E} \right)$
Since $\frac{(1 + W_{0}/E)}{(2 + W_{0}/E)} > \frac{1}{2}\ $so$\ \lambda^{'} > \frac{\lambda}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.