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  4. The magnitude of the de-Broglie wavelength (λ) of electron (e), proton (p), neutron (n) and α-particle (a) all having the same energy of 1 MeV, in the increasing order will follow the sequence

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Q. The magnitude of the de-Broglie wavelength $(\lambda)$ of electron $(e)$, proton $(p)$, neutron $(n)$ and $\alpha$-particle $(a)$ all having the same energy of $1\, MeV$, in the increasing order will follow the sequence

JIPMERJIPMER 2019Dual Nature of Radiation and Matter

Solution:

$\lambda=\frac{h}{\sqrt{2mE}} So \lambda \propto\frac{1}{\sqrt{m}}$
since $m_{\alpha}>m_{n}>m_{p}>m_{e}$
so de-Broglie wave length in increasingorder will be $\lambda_{\alpha}<\lambda_{n}<\lambda_{e}$