An electron with mass 'm' with an initial velocity $(t=0) \vec{v}=v_{0} \hat{i}$ $(v_{0}>0)$ enters a magnetic field $\vec{B}=B_{0} \hat{j}$. If the initial de-Broglie wavelength at $t=0$ is $\lambda_{0}$ then its value after time 't' would be :

(A) $\frac{\lambda_{0}}{\sqrt{1-\frac{e^{2} B_{0}^{2} t^{2}}{m^{2}}}}$

(B) $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2} B_{0}^{2} t^{2}}{m^{2}}}}$

(C) $\lambda_{0} \sqrt{1+\frac{e^{2} B_{0}^{2} t^{2}}{m^{2}}}$

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Modern Physics

An electron with mass 'm' with an initial velocity

(t = 0) v = v_{0} \hat{i}

(v_{0} > 0)

enters a magnetic field

B = B_{0} \hat{j}

. If the initial de-Broglie wavelength at

t = 0

λ_{0}

then its value after time 't' would be :

(A) $\frac{λ _{0}}{1 - \frac{e ^{2} B _{0}^{2} t ^{2}}{m ^{2}}}$
(B) $\frac{λ _{0}}{1 + \frac{e ^{2} B _{0}^{2} t ^{2}}{m ^{2}}}$
(C) $λ_{0} 1 + \frac{e ^{2} B _{0}^{2} t ^{2}}{m ^{2}}$
(D) $λ_{0}$ ✓ Correct

Answer: (D)

Solution

A magnetic field does no work on a charged particle because the magnetic force is always perpendicular to the velocity. Since no work is done, the kinetic energy and hence the speed of the electron remain constant. The de-Broglie wavelength

λ = h / (m v)

depends only on the momentum magnitude. Since the speed is unchanged, the de-Broglie wavelength remains

λ_{0}

at all times.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Modern Physics

Solution

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