A proton, a neutron, an electron and an \alpha particle have same energy. If \lambda p, \lambda n, \lambda e and \lambda a are the de Broglie's wavelengths of proton, neutron, electron and \alpha particle respectively, then choose the correct relation from the following :

\lambda e p = \lambda n > \lambda a

\lambda p = \lambda n > \lambda e > \lambda a

Dual Nature of Radiation PYQ — JEE Main 2022 · 25 Jun · Shift 2 · Q61 with Solution

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Correct answer: C

For a particle,

\lambda = \frac{h}{p}

If all particles have the same kinetic energy $K$ , then for non-relativistic motion,

K = \frac{p^2}{2m} \quad \Rightarrow \quad p = \sqrt{2mK}

So,

\lambda = \frac{h}{\sqrt{2mK}}

Hence, for same $K$ ,

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

The masses satisfy approximately:

m_e \ll m_p \approx m_n < m_\alpha

where:

Since

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

smaller mass means larger wavelength.

Thus,

\lambda_e > \lambda_p \approx \lambda_n > \lambda_\alpha

And since proton and neutron masses are taken equal for such comparisons,

\lambda_p = \lambda_n

So the final relation is

\boxed{\lambda_e > \lambda_p = \lambda_n > \lambda_\alpha}

This corresponds to Option C.

Stored correct answer is B, but the physically correct relation is C.

So the stored answer appears to be incorrect.