The de Broglie wavelength of an electron accelerated between two plates having a potential difference of 900 V is nearly

Electron is accelerated to a potential of 400 V.

So the kinetic energy if the electron is given by 400 eV.
$\text{ de-Broglie wavelength }\lambda =\frac{\mathrm{h}}{\sqrt{2\mathrm{mK}}}$

$\begin{array}{l}\therefore \lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\left(9.1\times {10}^{-31}\right)\times 900\times 1.6\times {10}^{-19}}}\\ \text{ ⇒}\lambda =40.7\times {10}^{-12}\mathrm{m}\Rightarrow \lambda =0.04 \mathrm{nm}\end{array}$