A moving proton and electron have the same

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 :

  1. $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$ and $\mathrm{P}_{\mathrm{p}}=\mathrm{P}_{\mathrm{e}}$

  2. $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}$ and $\mathrm{P}_{\mathrm{p}}=\mathrm{P}_{\mathrm{e}}$

  3. $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$ and $\mathrm{P}_{\mathrm{p}}<\mathrm{P}_{\mathrm{e}}$

  4. $\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{e}}$ and $\mathrm{P}_{\mathrm{p}}=\mathrm{P}_{\mathrm{e}}$

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)

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