Crystal diffraction experiments can be performed using $X$-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to $1 \AA$, which is of the order of interatomic spacing in the lattice) $\left(m_{e}=9.11 \times 10^{-31} \mathrm{~kg}\right)$

For Electrons, we have the relation for kinetic energy as,
$\mathrm{K} . \mathrm{E}=(1 / 2) \mathrm{m}_{\mathrm{e}} \mathrm{V}^{2}$

$=\left(\mathrm{m}_{\mathrm{e}} \mathrm{V}\right)^{2} / 2 \mathrm{~m}$

$\mathrm{K} . \mathrm{E}=\mathrm{p}^{2} / 2 \mathrm{~m}_{\mathrm{e}}$

$\Rightarrow p=\sqrt{2 m_{e} K \cdot E}$

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m_{c} K . E}}$

$\lambda^{2}=\frac{h^{2}}{2 m_{e} K \cdot E}$

$K \cdot E=\frac{h^{2}}{2 m_{e} \lambda^{2}}$

$K . E=\frac{\left(6.64 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times\left(10^{-10}\right)^{2}}$

$\mathrm{K} . \mathrm{E}=2.4 \times 10^{-17} \mathrm{~J}$

$\mathrm{K} . \mathrm{E}=\frac{2.4 \times 10^{-17}}{1.6 \times 10^{-19}}$

$=149.375 \mathrm{eV}$

For photon of $X$-rays, the expression for energy is,
$E=h c / \lambda e$

$=\left(6.63 \times 10^{-34} \times 3 \times 10^{8}\right) /\left(10^{-10} \times 1.6 \times 10^{-19}\right)$

$=12.375 \times 10^{3} \mathrm{eV}$

$=12.375 \mathrm{keV}$

The energy of X-ray photons is greater than that of the electron.