If the kinetic energy of an electron is increased four times, the wavelength of de – Broglie wavelength associated with it is

Expression of kinetic energy:

$$\begin{gathered} \mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{v}=\sqrt{\frac{2\mathrm{KE}}{\mathrm{m}}}...\left(1\right) \end{gathered}$$

de Broglie wavelength is given by:

 $$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}} \\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{m}\times \sqrt{{\displaystyle \frac{2\mathrm{KE}}{\mathrm{m}}}}}=\frac{\mathrm{h}}{\mathrm{m}}\times \sqrt{\frac{\mathrm{m}}{2\mathrm{KE}}} \\ \mathrm{\lambda }\infty \frac{1}{\sqrt{\mathrm{K}.\mathrm{E}}} \end{gathered}$$

When K.E becomes four times the wavelength becomes half.

$$\begin{gathered} \mathrm{\lambda }\infty \frac{1}{\sqrt{4}} \\ \mathrm{\lambda }\infty \frac{1}{2} \end{gathered}$$