question_answer

A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de-Broglie wavelength associated with it, and (ii) less kinetic energy? Justify your answer.                           

Answer:

                (i) de Broglie wavelength, \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mqV}}\]                 For same V, \[\lambda \propto \frac{1}{\sqrt{mq}}\]                 \[\therefore \]                  \[\frac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\sqrt{\frac{{{m}_{\alpha }}{{q}_{\alpha }}}{{{m}_{p}}{{q}_{p}}}}\] \[=\sqrt{\frac{4{{m}_{p}}}{{{m}_{p}}}.\frac{2e}{e}}=\sqrt{8}=2\sqrt{2}\] Hence  \[{{\lambda }_{p}}>{{\lambda }_{\alpha }}\]i.e., proton has a greater value of de-Broglie wavelength. (ii) Kinetic energy             \[K=qV\] For same \[V\],                 \[K\propto q\] \[\ \therefore \]               \[\frac{{{K}_{p}}}{{{K}_{\alpha }}}=\frac{{{q}_{p}}}{{{q}_{\alpha }}}=\frac{e}{2e}=\frac{1}{2}\] Hence, \[{{K}_{p}}<{{K}_{\alpha }},\]i.e., proton has less kinetic energy.