Physics - Modern Physics Question with Solution | TestHub

de-Broglie wavelength is given by $\begin{array}{l}\lambda =\frac{\mathrm{h}}{\mathrm{p}},\text{ where }\mathrm{h}=\text{ Planck’s constant and }\\ \mathrm{p}=\text{ momentum }\end{array}$ Also, energy (E) and momentum are related as $\begin{array}{l}\mathrm{E}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}\\ \therefore \mathrm{p}=\sqrt{2\mathrm{mE}}\end{array}$ $\therefore \lambda =\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}\times \frac{1}{\sqrt{\mathrm{E}}}$ as $\mathrm{h}$ and $\mathrm{m}$ are constants Hence, $\frac{{\lambda }_0}{{\lambda }^{\prime }}=\sqrt{\frac{{\mathrm{E}}^{\prime }}{\mathrm{E}}}=\sqrt{\frac{2\mathrm{E}}{\mathrm{E}}}=\sqrt{2}$ $\therefore {\lambda }^{\prime }=\frac{{\lambda }_0}{\sqrt{2}}$