Q. In the following diagrams if $V_{2} > V_{1}$ then
A) $\lambda_{1} = \sqrt{\lambda_{2}}$
B) $\lambda_{1} < \lambda_{2}$
C) $\lambda_{1} = \lambda_{2}$
D) $\lambda_{1} > \lambda_{2}$
Explanation
(d) $\because V_{0} = \left( \frac{h}{e} \right)v - \left( \frac{W_{0}}{e} \right).$ From the graph $V_{2} > V_{1}$
$\Rightarrow \frac{hv_{2}}{e} - \frac{W_{0}}{e} > \frac{hv_{1}}{e} - \frac{W_{0}}{e} \Rightarrow v_{2} > v_{1}$
$\Rightarrow \lambda_{1} > \lambda_{2}\ (as\ \lambda \propto \frac{1}{v})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.