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)

$\lambda_{e} > \lambda_{p}$

Answer 2

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