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The magnitude (in the ascending order) of the de-Broglie wavelength $(λ)$ of an electron (e), a proton (p), a neutron (n) and an α particle (α) all having the same energy of 1MeV are

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$λ_{p}, λ_{e}, λ_{α}, λ_{n}$

$λ_{α}, λ_{n}, λ_{p}, λ_{e}$

$λ_{e}, λ_{p}, λ_{n}, λ_{α}$

$λ_{e}, λ_{n}, λ_{p}, λ_{α}$

answer is C.

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Detailed Solution

$λ = \frac{h}{\sqrt{2 m E}},$ so $λ α \frac{1}{\sqrt{m}}$ Here $m_{a} > m_{n} > m_{p} > m_{e}$
deBroglie wavelength in increasing order will be $λ_{a}, λ_{n}, λ_{p}, λ_{e}$

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