Physics - Modern Physics Question with Solution | TestHub

The de-Broglie's wavelength is given by $\lambda =\frac{h}{mv}$, where, $h$ is Planck's constant and $v$ is velocity.

Using relation between momentum and kinetic energy, we have $\lambda =\frac{h}{\sqrt{2 m\left(\mathrm{KE}\right)}}$

Here, kinetic energy of electron is stated as $KE=qV$.

Then, we have $\lambda =\frac{h}{\sqrt{2mqV}}$

Putting the values of $m=9.1\times {10}^{-31} \mathrm{kg} \mathrm{and} q=1.62\times {10}^{-19 }\mathrm{C}$ of electron

We get, $\lambda =\frac{1.22}{\sqrt{\mathrm{V}}} \mathrm{nm}$.

So, $x$ stand for $\sqrt{V}$.