Question

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 $1.54\,\mathring{A}$

Answer 1

If accelerated by potential difference of $V$ volt, then
$\frac{1}{2} m v^{2}=e V$
$\Rightarrow \frac{p^{2}}{2 m}=e V$,
here $p=$ momentum $(m v)$
Using de-Broglie equation, $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m e V}}$
$\Rightarrow 1.54 \times 10^{-10}=\frac{6.625 \times 10^{-34}}{\left(2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} V \right)^{1 / 2}}$
Solving for $V$ gives: $V=63.56\, V$.