The de Broglie wavelengths for an electron and a photon are \lambda e and \lambda p respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two ?

\lambda _p \propto \sqrt \lambda _e

\lambda _p \propto \sqrt 1 \over \lambda _e

Dual Nature of Radiation PYQ — JEE Main 2022 · 28 Jun · Shift 1 · Q56 with Solution

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Correct answer: A

For a non-relativistic electron, de Broglie wavelength is

\lambda_e = \frac{h}{p}

and its kinetic energy is

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

So,

p = \sqrt{2mK}.

Hence,

\lambda_e = \frac{h}{\sqrt{2mK}}.

Rearranging,

\lambda_e \propto \frac{1}{\sqrt{K}}.

Therefore,

K \propto \frac{1}{\lambda_e^2}.

For a photon,

E = pc = \frac{hc}{\lambda_p}.

A photon has no rest mass, so its entire energy is kinetic in nature. Thus for the same given energy $K$ ,

K = \frac{hc}{\lambda_p}.

So,

\lambda_p = \frac{hc}{K}.

Thus,

\lambda_p \propto \frac{1}{K}.

Since

K \propto \frac{1}{\lambda_e^2},

we get

\lambda_p \propto \frac{1}{K} \propto \lambda_e^2.

The correct relation is

\lambda_p \propto \lambda_e^2.

So the correct option is A.

Stored correct answer: A

This matches the derived answer.