An electron is accelerated to potential V. If mass of electron is m = 9.1 × 10^–31 kg its charge is e = 1.6 × 10^–19 C, then de-Broglie wavelength of electron is

\( λ = \frac{1.227}{\sqrt{v}} nm\)

We know,

$\lambda=\frac{h}{p}$ and $p=\sqrt{2 m q V}$

∴ \(\lambda=\frac{h}{\sqrt{2 m q V}} \)

For an electron, $\frac{h}{\sqrt{2 m e }} = 12.27 A^o \text { or } 1.227 nm$

\( \Rightarrow \lambda = \frac{12.27}{\sqrt{V}} \) Å

or $\frac{1.227}{\sqrt{V}} \mathrm{~nm}$