MCQ
If ${\lambda _p}$ and ${\lambda _\alpha }$ be the de-Broglie's wavelengths associated with protons and $\alpha  - $ particles of equal kinetic energies, then
  • A
    ${\lambda _p} = \frac{{{\lambda _\alpha }}}{4}$
  • B
    ${\lambda _p} = \frac{{{\lambda _\alpha }}}{2}$
  • C
    ${\lambda _p} = {\lambda _\alpha }$
  • ${\lambda _p} = 2{\lambda _\alpha }$

Answer

Correct option: D.
${\lambda _p} = 2{\lambda _\alpha }$
d
$\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}_{\mathrm{p}} \mathrm{k}}}$

$\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2\left(4 \mathrm{m}_{\mathrm{p}} \mathrm{k}\right)}}=\frac{\lambda_{\mathrm{p}}}{2}$

$\lambda_{\mathrm{p}}=2 \lambda_{\alpha}$

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