Physics - Modern Physics Question with Solution | TestHub

de-Broglie wavelength, $$\begin{array}{l}\lambda =\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2\cdot \mathrm{m}\cdot (\mathrm{K}\cdot \mathrm{E})}}\\ \therefore \lambda \propto \frac{1}{\sqrt{\mathrm{K}\cdot \mathrm{E}}}\end{array}$$ If $\mathrm{K}.\mathrm{E}$ is doubled, then de-Broglie wavelength becomes $\frac{\lambda }{\sqrt{2}}$