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

de broglie wavelenght is given by $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mV}}$   ........ (i)

$\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2\mathrm{v}=\sqrt{\frac{2\mathrm{KE}}{\mathrm{m}}}$ ...... (ii)

Substitute (2) in (1) $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{m}}\sqrt{\frac{\mathrm{m}}{2\mathrm{KE}}}\mathrm{\lambda }=\mathrm{h}\sqrt{\frac{1}{2\mathrm{mkE}}}=\mathrm{\lambda \alpha }\frac{1}{\sqrt{\mathrm{K}\cdot \mathrm{E}}}$
When K.E becomes four times the wave length becomes half.