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Q.
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
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Solution:
de-Broglie wavelength
$\lambda=\frac{h}{m v}$
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
$\frac{K_{e}}{K_{p}}=\frac{\frac{1}{2} m_{e} v_{e}^{2}}{\frac{1}{2} m_{p} v_{p}^{2}}$
$\frac{K_{e}}{K_{p}}=\frac{1}{2}$