Physics - Capacitor Question with Solution | TestHub

de-Broglie wavelength $(\lambda )$ of a particle of mass $\mathrm{m}$ and moving with a velocity $\mathrm{v}$ is given by, $\lambda =\frac{\mathrm{h}}{\mathrm{mv}}$, where $\mathrm{h}$ is Planck's constant. When a particle having charge $\mathrm{q}$ is accelerated through a potential $\mathrm{V}$ then $\mathrm{qV}=\frac{1}{2}{\mathrm{mv}}^2$ or, $\mathrm{qV}=\frac{{\mathrm{m}}^2{\mathrm{v}}^2}{2\mathrm{m}}\Rightarrow \mathrm{mv}=\sqrt{2\mathrm{mqV}}$ $\therefore \lambda =\frac{\mathrm{h}}{\sqrt{2\mathrm{mqV}}}$ Hence, de-Broglie wavelength of electron, ${\lambda }_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\mathrm{e}}\mathrm{eV}}}$ and de-Broglie wavelength of proton, ${\lambda }_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\mathrm{p}}\mathrm{eV}}}$, where e represents the charge of electron (or proton) $\therefore \frac{{\lambda }_{\mathrm{e}}}{{\lambda }_{\mathrm{p}}}=\sqrt{\left(\frac{{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{m}}_{\mathrm{e}}}\right)}$