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

(B) An $X-$ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of the probe,$\lambda = 1\,\mathring{A} = 10^{-10}\,m$
Mass of an electron,$m_{e} = 9.11 \times 10^{-31}\,kg$
Planck's constant,$h = 6.63 \times 10^{-34}\,Js$
Charge on an electron,$e = 1.6 \times 10^{-19}\,C$
Speed of light,$c = 3 \times 10^{8}\,m/s$
$1$. Energy of an electron $(E_{e})$:
Using the de Broglie relation,$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_{e}E_{e}}}$,we get $E_{e} = \frac{h^{2}}{2m_{e}\lambda^{2}}$.
$E_{e} = \frac{(6.63 \times 10^{-34})^{2}}{2 \times 9.11 \times 10^{-31} \times (10^{-10})^{2}} \approx 2.41 \times 10^{-17}\,J$.
Converting to $eV$: $E_{e} = \frac{2.41 \times 10^{-17}}{1.6 \times 10^{-19}} \approx 150.6\,eV$.
$2$. Energy of an $X-$ray photon $(E_{ph})$:
$E_{ph} = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{10^{-10}} = 1.989 \times 10^{-15}\,J$.
Converting to $eV$: $E_{ph} = \frac{1.989 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 12431\,eV \approx 12.43\,keV$.
Comparing the two,$E_{ph} > E_{e}$. Thus,the $X-$ray probe has significantly greater energy.