Question

The kinetic energy of electron and photon arc the same. The relation between their De - Broglie wavelengths $ \lambda _e $ and $ \lambda _p $ is

• A. $ \lambda_{e} = \lambda_{p} $
• B. $ \lambda_{e} < \lambda_{p} $
• C. $ \lambda_{e} > \lambda_{p} $
• D. $ \lambda_{e} =2 \lambda_{p} $

Answer 1

Answer: (C)

De-broglie wavelength of electron,
$\lambda_{e}=\frac{h}{\sqrt{2m_{e}k_{e}}}$
De-broglie wavelength of photon, $\lambda_{p}=\frac{hc}{k_{p}}$
$\frac{\lambda_{e}}{\lambda_{p}}=\frac{K_{p}}{c \sqrt{2m_{e}K_{e}}}$
$=\sqrt{\frac{K}{2m_{e} c^{2}}}>\,1$ $(\because K_{e}=K_{p}=K)$
$\therefore \lambda_{e} >\, \lambda_{p}$