Question

an electron, a proton and an alpha particle have kinetic energies of 16E, 4E and E respectively. what is the quantitative order of their de broglie wavelengths?

Answer 1

$$\begin{gathered} \text{Dear Student,} \\ \mathrm{De}\mathrm{Broglie}\mathrm{wavelength}\mathrm{is}\mathrm{\lambda }=\frac{\mathrm{h}}{\left(\mathrm{mv}\right)} \\ \mathrm{Kinetic}\mathrm{energy}\mathrm{is}\mathrm{K}=\frac{1}{2}\mathrm{mv}² \\ \mathrm{so}\mathrm{v}=\surd (2\mathrm{K}/\mathrm{m})\mathrm{\lambda } \\ \mathrm{v}=\frac{\mathrm{h}}{\left(\mathrm{m}\sqrt{{\displaystyle \frac{2\mathrm{K}}{\mathrm{m}}}}\right)} \\ \mathrm{v}=\frac{\mathrm{h}}{\sqrt{{\displaystyle \frac{2{\mathrm{Km}}^2}{\mathrm{m}}}}} \\ \mathrm{v}=\frac{\mathrm{h}}{\sqrt{2\mathrm{Km}}} \\ \mathrm{v}=\frac{\mathrm{h}}{\sqrt{2}}\frac{1}{\sqrt{\mathrm{Km}}} \\ \mathrm{So}\mathrm{\lambda }\mathrm{is}\mathrm{proportional}\mathrm{to}\frac{1}{\sqrt{\mathrm{Km}}} \\ \mathrm{For}\mathrm{the}\mathrm{electron}\mathrm{m}=1/2000\mathrm{amu}\mathrm{approx}\mathrm{and}\mathrm{K}=16\mathrm{E}.\frac{1}{\sqrt{\mathrm{Km}}}=\frac{11.2}{\sqrt{\mathrm{E}}} \\ \mathrm{For}\mathrm{the}\mathrm{proton}\mathrm{m}=1\mathrm{amu}\mathrm{approx}\mathrm{and}\mathrm{K}=16\mathrm{E}\frac{1}{\sqrt{\mathrm{Km}}}=\frac{0.25}{\sqrt{\mathrm{E}}} \\ \mathrm{For}\mathrm{the}\mathrm{alpha}\mathrm{particle}\mathrm{m}=4\mathrm{amu}\mathrm{approx}\mathrm{and}\mathrm{K}=4\mathrm{E}\frac{1}{\sqrt{\mathrm{Km}}}=\frac{0.25}{\sqrt{\mathrm{E}}} \\ {\mathrm{\lambda }}_{\mathrm{e}}>{\mathrm{\lambda }}_{\mathrm{p}}={\mathrm{\lambda }}_{\mathrm{a}} \end{gathered}$$