Electron beam used in an electron microscope, when accelerated by a voltage of 20   kV has a de-Broglie…

The de-Broglie wavelength of an electron accelerated through potential $V$ is 

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2 \mathrm{m}\left(\mathrm{KE}\right)}}=\frac{h}{\sqrt{2 \mathrm{meV}}}$

where, $h$ is Planck's constant, $e$ is charge of electron and $m$ is mass of electron.

From above relation, $\lambda \propto \frac{1}{\sqrt{V}}$

$\therefore \frac{\mathrm{\lambda }}{{\mathrm{\lambda }}_0}=\sqrt{\frac{20}{40}}$

$\Rightarrow \lambda =\frac{{\lambda }_0}{\sqrt{2}}$.