Q. Two identical metal plates shown photoelectric effect by a light of wavelength $\lambda\mathring{\mathrm{A}}$ falls on plate A and $\lambda_{B}$ on plate $B\left( \lambda_{A} = 2\lambda_{B} \right).$ The maximum kinetic energy is
A) $2\ K_{A} = K_{B}$
B) $K_{A} < K_{B}/2$
C) $K_{A} = 2K_{B}$
D) $K_{A} = K_{B}/2$
Explanation
(b) $K_{A} = \frac{hc}{\lambda_{A}} - \phi_{0}$ and $K_{B} = \frac{hc}{\lambda_{B}} - \phi_{0}$
$\frac{K_{A}}{K_{B}} = \frac{\frac{hc}{2\lambda_{B}}}{\frac{hc}{\lambda_{B}}} < \frac{1}{2}\ or\ K_{A} < K_{B}/2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.