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.
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