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
$\frac{1}{2}m{v}^2=eV$
$\Rightarrow \frac{p^2}{2m}=eV$,
here $p=$ momentum $(mv)$
Using de-Broglie equation, $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2meV}}$
$\Rightarrow 1.54\times 10^{-10}=\frac{6.625\times 10^{-34}}{{\left(2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}V\right)}^{1/2}}$
Solving for $V$ gives: $V=63.56V$.