Q. In a photoelectric effect measurement, the stopping potential for a given metal is found to be $V_{0}$ volt when radiation of wavelength $\lambda_{0}$ is used. If radiation of wavelength 2$\lambda_{0}\ $is used with the same metal then the stopping potential (in volt ) will be
A) $\frac{V_{0}}{2}$
B) $2V_{0}\ $
C) $V_{0} + \frac{hc}{2e\lambda_{0}}$
D) $V_{0} - \frac{hc}{2e\lambda_{0}}$
Explanation
(d) From Einstein’s photoelectric equation
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ eV_{0} = \frac{hc}{\lambda_{0}} - W_{0}$
$eV^{'} = \frac{hc}{2\lambda_{0}} - W_{0}$
Subtracting $e\left( V_{0} - V^{'} \right) =$ $\frac{hc}{\lambda_{0}}\left\lbrack 1 - \frac{1}{2} \right\rbrack = \frac{hc}{2\lambda_{0}\ }$
or $V^{'} = \ V_{0} - \frac{hc}{2e\lambda_{0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.