Physics - Modern Physics Question with Solution | TestHub

Given,$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mKE}}$

$\Rightarrow KE=\frac{1}{2m{\lambda }^2}$

As$\lambda$is same for both electron and photon

So,$KE\propto \frac{1}{m}$

Hence, kinetic energy will be maximum for particle with lesser mass,$ie,$electron.