question_answer

An electron and a proton 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 momentum? Justify your answer.

Answer:

                (i) de ? Broglie wavelength, \[\lambda =\frac{h}{\sqrt{2mqV}}\]                 For same \[V\], \[\lambda \propto \frac{1}{\sqrt{mq}}\] \[\therefore \]  \[\frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\sqrt{\frac{{{m}_{p}}\times e}{{{m}_{e}}\times e}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{e}}}}\] As \[{{m}_{p}}>{{m}_{e}}\], so \[{{\lambda }_{e}}>{{\lambda }_{p}}\]i.e., electron has a greater de- Broglie wavelength. (ii) Momentum , \[p=\sqrt{2meV}\] or            \[p\propto \sqrt{m}\] As           \[{{m}_{e}}<{{m}_{p}},\]              \[{{p}_{e}}<{{p}_{p}}\] i.e., electron has less momentum.