de Broglie wavelength of an electron after beings accelerated by a potential difference of 'V' volt from rest is:

De Broglie found that an electron's wavelength after being accelerated by a V-volt potential difference from rest equals

$\begin{array}{l}\lambda =\frac{\mathrm{h}}{\mathrm{p}}\\ =\frac{\mathrm{h}}{\sqrt{2\mathrm{mev}}}\end{array}$

Hence, 

$\lambda =\frac{12.26}{\sqrt{v}}$

${\lambda }_e=\frac{12.3}{\sqrt{V}}\text{Å}$

Hence, the option B is the correct answer.