Question

An electron and alpha particle have the same kinetic energy. How are the de-Broglie wavelengths associated with them related?

Answer 1

Dear Student,
$$\begin{gathered} \mathrm{Debroglie}\mathrm{wavelength}: \\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}} \\ =\frac{\mathrm{h}}{\sqrt{2{\mathrm{mE}}_{\mathrm{k}}}} \\ \mathrm{Here}, \\ \mathrm{h}\mathrm{is}\mathrm{Planck}\text{'}\mathrm{s}\mathrm{constant}, \\ \mathrm{m}\mathrm{is}\mathrm{mass}, \\ \mathrm{and} \\ {\mathrm{E}}_{\mathrm{k}}\mathrm{is}\mathrm{kinetic}\mathrm{energy} \\ \mathrm{For}\mathrm{a}\mathrm{given}\mathrm{kinetic}\mathrm{energy}, \\ \mathrm{\lambda }\mathrm{\alpha }\frac{1}{\sqrt{\mathrm{m}}} \\ \frac{{\mathrm{\lambda }}_{\mathrm{electron}}}{{\mathrm{\lambda }}_{\mathrm{alpha}}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{alpha}}}{{\mathrm{m}}_{\mathrm{electron}}}} \\ =\sqrt{\frac{2{\mathrm{m}}_{\mathrm{electron}}}{{\mathrm{m}}_{\mathrm{electron}}}}[\mathrm{since}{\mathrm{m}}_{\mathrm{alpha}=}2\times {\mathrm{m}}_{\mathrm{electron}}] \\ =\sqrt{2} \\ \mathrm{Hence}, \\ {\mathbf{\lambda }}_{\mathbf{electron}}\mathbf{=}\sqrt{\mathbf{2}}{\mathbf{\lambda }}_{\mathbf{alpha}} \end{gathered}$$
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