A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

\lambda_\alpha\lt \lambda_p\lt \lambda_e

\lambda_p\gt \lambda_e\gt \lambda_\alpha

\lambda_e\gt \lambda_\alpha\gt \lambda_p

\lambda_p\lt \lambda_e\lt \lambda_\alpha

Dual Nature of Radiation PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q80 with Solution

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Correct answer: B

For a particle,

\lambda = \frac{h}{p}

where $h$ is Planck’s constant and $p$ is momentum.

Since the proton, electron and alpha particle have the same kinetic energy $K$ ,

K = \frac{p^2}{2m}

so,

p = \sqrt{2mK}

Substitute into de-Broglie wavelength:

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

For fixed $K$ ,

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

So, smaller mass means larger wavelength.

Thus,

m_e < m_p < m_\alpha

Hence,

\lambda_e > \lambda_p > \lambda_\alpha

Option B says:

\lambda_\alpha < \lambda_p < \lambda_e

This is exactly the same ordering.

Therefore, the correct answer is:

\boxed{\text{B}}