# Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage.

Question:

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 Å, which is of the order of inter-atomic spacing in the lattice) (me= 9.11 × 10−31 kg).

Solution:

An X-ray probe has a greater energy than an electron probe for the same wavelength.

Wavelength of light emitted from the probe, λ = 1 Å = 10−10 m

Mass of an electron, me = 9.11 × 10−31 kg

Planck’s constant, h = 6.6 × 10−34 Js

Charge on an electron, e = 1.6 × 10−19 C

The kinetic energy of the electron is given as:

$E=\frac{1}{2} m_{e} v^{2}$

$m_{e} v=\sqrt{2 E m_{e}}$

Where,

v = Velocity of the electron

mev = Momentum (p) of the electron

According to the de Broglie principle, the de Broglie wavelength is given as:

$\lambda=\frac{h}{p}=\frac{h}{m_{e} v}=\frac{h}{\sqrt{2 E m_{e}}}$

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

$=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(10^{-10}\right)^{2} \times 9.11 \times 10^{-31}}=2.39 \times 10^{-17} \mathrm{~J}$

$=\frac{2.39 \times 10^{-17}}{1.6 \times 10^{-19}}=149.375 \mathrm{eV}$

Energy of a photon, $E^{\prime}=\frac{h c}{\lambda e} \mathrm{eV}$

$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-10} \times 1.6 \times 10^{-19}}$

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

Hence, a photon has a greater energy than an electron for the same wavelength.