The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:

We know de-Broglie wavelength for an electron when it is accelerated with a potential ‘V‘ is _____
$\begin{array}{l}\lambda =\frac{12.27}{\sqrt{V}}\\ =\frac{12.27}{\sqrt{81}}=\frac{12.27}{9}=1.36\text{Å}=0.136nm\end{array}$