An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

Dual Nature of Radiation PYQ — JEE Main 2021 · 22 Jul · Shift 2 · Q60 with Solution

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Correct answer: D

For a particle, the de-Broglie wavelength is

\lambda = \frac{h}{p}

where $p$ is the momentum.

When a charged particle of charge magnitude $e$ is accelerated through potential difference $V$ , it gains kinetic energy

K = eV

For non-relativistic motion,

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

So,

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

p = \sqrt{2meV}

Substitute into de-Broglie formula:

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

Thus,

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

for the same accelerating potential $V$ .

Therefore,

\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2m_e eV}}{h/\sqrt{2m_p eV}} = \sqrt{\frac{m_p}{m_e}}

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}

This corresponds to Option D.

Stored correct answer: D

My derived answer also gives D, so they agree.