Question

The de Broglie wavelength of an electron moving with a velocity $\frac{C}{2}$ (C = velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is

• A. 1 : 4
• B. 1 : 2
• C. 1 : 1
• D. 2 : 1

Answer 1

Answer: (B)

de-Broglie wavelength
$\lambda =\frac{h}{mv}$
Here,
${\lambda }_e=\frac{h}{m_e\frac{c}{2}}\text{ and }{\lambda }_p=\frac{h}{m_{p}c}$
Given, ${\lambda }_e={\lambda }_p$
So, $\frac{h}{m_e\frac{c}{2}}=\frac{h}{m_{p}c}$
$\frac{m_e}{m_p}=2$
Ratio of KE
$\frac{K_e}{K_p}=\frac{\frac{1}{2}{m}_e{v}_e^2}{\frac{1}{2}{m}_p{v}_p^2}$
$\frac{K_e}{K_p}=\frac{1}{2}$