JEE Physics 2025 Question with Solution

Given: An electron of mass $m$ and charge $-e$ enters a uniform magnetic field $\vec{B}=B\hat{j}$ with initial velocity $\vec{v}=\vec{v_0}$. The initial de-Broglie wavelength is $\lambda_0$.

Find: The de-Broglie wavelength after time $t$.

Concept used: The magnetic force is always perpendicular to the instantaneous velocity, so it changes only the direction of motion, not the speed.

Using Lorentz force,

$$\vec{F}=-e(\vec{v}\times\vec{B})$$

Since this force is perpendicular to $\vec{v}$, the magnitude of velocity remains constant.

The de-Broglie wavelength is

$$\lambda=\frac{h}{p}=\frac{h}{mv}$$

Because the speed $v$ remains unchanged, the momentum magnitude $p=mv$ also remains unchanged.

Therefore,

$$\lambda(t)=\frac{h}{mv_0}=\lambda_0$$

So the de-Broglie wavelength remains constant with time.

The solution states that the correct option is D. However, in the listed options, the visible value $\lambda_0$ appears more than once, so the answer is taken from the solution authority.

Therefore, the correct option is D, and the wavelength after time $t$ is $\lambda_0$.

Given: The electron moves only under a magnetic field.

Find: Whether its de-Broglie wavelength changes with time.

A magnetic field does no work on a charged particle, so it cannot change the particle's kinetic energy or speed. Since

$$\lambda=\frac{h}{mv}$$

and $m$ is constant, an unchanged speed means an unchanged de-Broglie wavelength.

Hence,

$$\lambda(t)=\lambda_0$$

Therefore, the correct option is D.