question_answer

An electron, \[\alpha \]-particle and a proton have the same kinetic energy. Which of these particles has the shortest de-Broglie wavelength?

Answer:

                The kinetic energy of a particle, \[K=\frac{1}{2}m{{\upsilon }^{2}}=\frac{1}{2}\frac{{{(m\upsilon )}^{2}}}{m}=\frac{{{p}^{2}}}{2m}\] \[\therefore \] Linear momentum, \[p=\sqrt{2mK}\] de-Broglie wavelength, \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mk}}\] For the particles possessing same kinetic energy, \[\lambda \propto \frac{1}{\sqrt{m}}\] AS \[{{m}_{e}}<<{{m}_{p}}<{{m}_{\alpha }}\] \[\therefore \]  \[{{\lambda }_{e}}>>{{\lambda }_{p}}>{{\lambda }_{\alpha }}\] Hence the \[\alpha \]-particle has the shortest de-Broglie wavelength.