Question:
A moving proton and electron have the same deBroglie wavelength. If $\mathrm{K}$ and $P$ denote the $\mathrm{K} . \mathrm{E}$. and momentum respectively. Then choose the correct option :
Correct Option:
Solution:
$\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{P}_{\mathrm{p}}} \quad \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{P}_{\mathrm{e}}}$
$\because \lambda_{\mathrm{P}}=\lambda_{\mathrm{e}}$
$\Rightarrow P_{P}=P_{e}$
$(\mathrm{K})_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{P}}^{2}}{2 \mathrm{~m}_{\mathrm{P}}}$
$(\mathrm{K})_{\mathrm{e}}=\frac{\mathrm{P}_{\mathrm{e}}^{2}}{2 \mathrm{~m}_{\mathrm{e}}}$
$\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$ as $\mathrm{m}_{\mathrm{P}}>\mathrm{m}_{\mathrm{e}}$
Option (1)