StemCET Logo

Physics Question 33 – JEE-MAIN 2025

An electron with mass 'm' with an initial velocity (t=0) v=v0i^ (v0>0) enters a magnetic field B=B0j^. If the initial de-Broglie wavelength at t=0 is λ0 then its value after time 't' would be :

Recall that the magnetic force on a charged particle is always perpendicular to its velocity.

🥷
Ninja StrategyConservation of Speed

Recognize that a magnetic force does no work on a charged particle, thus its speed and kinetic energy remain constant, leading to a constant de-Broglie wavelength.

Step 1: Analyze the effect of magnetic force on speed✦ Active

The magnetic force F=q(v×B) is always perpendicular to the velocity vector v. Therefore, the work done by the magnetic force, W=Fdr=F(vdt), is always zero because Fv=0. Since no work is done, the kinetic energy of the electron, K=12mv2, remains constant.

K(t)=K(0)12mv(t)2=12mv02

This implies that the speed of the electron, v(t), remains constant and equal to its initial speed v0.

Step 2: Relate speed to de-Broglie wavelength○ Expand

The de-Broglie wavelength is given by λ=hp, where h is Planck's constant and p is the momentum of the particle. The momentum is p=mv. Since the speed v remains constant (v(t)=v0), the momentum p(t)=mv(t)=mv0 also remains constant.

Step 3: Determine the de-Broglie wavelength at time t○ Expand

As the momentum remains constant, the de-Broglie wavelength at any time t will be the same as the initial de-Broglie wavelength λ0.

λ(t)=hmv(t)=hmv0=λ0
✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.