Q: An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 A° .

Solution: For an electron

$\large \frac{1}{2} m u^2 = e V$

(Where V is accelerating potential)

$\large \lambda = \frac{h}{m u} $

$\large \frac{1}{2} m (\frac{h}{m \lambda})^2 = e V $

$\large V = \frac{h^2}{2 m e \lambda^2}$

$\large V = \frac{(6.625 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times (1.54 \times 10^{-10})^2}$

= 63.3 volt