Electron beam used in an electron microscope, when accelerated by a...

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

$\lambda = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2m ( KE )}} = \dfrac{h}{\sqrt{2meV}}$

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

From above relation, $\lambda ∝ \dfrac{1}{\sqrt{V}}$

$∴ \dfrac{\lambda}{{\lambda}_{0}} = \sqrt{\dfrac{20}{40}}$

$⇒ \lambda = \dfrac{{\lambda}_{0}}{\sqrt{2}}$.