MCQ
A proton, a neutron, an electron and an $\alpha\,-$particle have same energy. If $\lambda_{ p }, \lambda_{ n }, \lambda_{ e }$ and $\lambda_{\alpha}$ are the de Broglie's wavelengths of proton, neutron, electron and $\alpha$ particle respectively, then choose the correct relation from the following
  • A
    $\lambda_{ p }=\lambda_{ n }>\lambda_{ e }>\lambda_{\alpha}$
  • $\lambda_{\alpha}<\lambda_{ n }<\lambda_{ p }<\lambda_{ e }$
  • C
    $\lambda_{ e }<\lambda_{ p }=\lambda_{ n }>\lambda_{\alpha}$
  • D
    $\lambda_{ e }=\lambda_{ p }=\lambda_{ n }=\lambda_{\alpha}$

Answer

Correct option: B.
$\lambda_{\alpha}<\lambda_{ n }<\lambda_{ p }<\lambda_{ e }$
b
$\lambda=\frac{ h }{\sqrt{2 Em }}$

$\lambda \propto \frac{1}{\sqrt{ m }}$

$\therefore\lambda_{ e }>\lambda_{ p }>\lambda_{ n }>\lambda_{\alpha}$

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