The de-Broglie wavelength of an electron moving with a velocity $c / 2$ ( $c=$ velocity of light in vacuum)…
The de-Broglie wavelength of an electron moving with a velocity $c / 2$ ( $c=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is
$1: 4$
$1: 2$
$1: 1$
$2: 1$
Solution
de-Broglie wavelength
$\lambda=\frac{h}{m v}$
Here, $\lambda_{e}=\frac{h}{m_{e} \frac{c}{2}}$ and $\lambda_{p}=\frac{h}{m_{p} c}$
Given, $\quad \lambda_{e}=\lambda_{B}$
So, $\quad \frac{h}{m_{e} \frac{c}{2}}=\frac{h}{m_{p} c}$
$\frac{m_{e}}{m_{p}}=2$
Ratio of KE
$\begin{array}{l}
\frac{K_{e}}{K_{p}}=\frac{\frac{1}{2} m_{e} v_{e}^{2}}{\frac{1}{2} m_{p} v_{p}^{2}} \\
\frac{K_{e}}{K_{p}}=\frac{1}{2}
\end{array}$