An electron of mass $m_{e}$ and a proton of mass $\mathrm{m}_{\mathrm{p}}$ are accelerated through t

Question

An electron of mass $m_{e}$ and a proton of mass $\mathrm{m}_{\mathrm{p}}$ are accelerated through the same potential. Then the ratio of their de Broglie wavelengths is\

Professor PC Thomas Classes · Answer

de Broglie wavelength $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ Since potential ( V ) is same for both the particle $	herefore \lambda \propto-\frac{1}{\sqrt{\mathrm{mq}}}$\ Thus $\frac{\lambda_{e}}{\lambda_{p}}=\sqrt{\frac{m_{p} q_{p}}{m_{e} q_{e}}}=\sqrt{\frac{m_{p}}{m_{e}}}\left(\because q_{p}=q_{e}=eight)$\