Sol. When electron is accelerated through potential difference $V$, then K.E. $=\mathrm{eV}$ $$ \begin{array}{ll} \Rightarrow & \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\ \therefore & \lambda \propto \frac{1}{\sqrt{\mathrm{~V}}} \\ \therefore & \frac{\lambda}{\lambda_0}=\sqrt{\frac{20}{40}} \\ \therefore & \lambda=\frac{\lambda_0}{\sqrt{2}} \end{array} $$