Question

An electron with mass $ m $ with an initial velocity $ (t = 0) \, \vec{v} = \vec{v_0} \, (v_0>0) $ enters a magnetic field $ \vec{B} = B \hat{j} $. If the initial de-Broglie wavelength at $ t = 0 $ is $ \lambda_0 $, then its value after time $ t $ would be:

• A. \( \frac{\lambda_0}{\sqrt{1 - \frac{e^2 B^2 t^2}{m^2}}} \)
• B. \( \frac{\lambda_0}{\sqrt{1 + \frac{e^2 B^2 t^2}{m^2}}} \)
• C. \( \lambda_0 \sqrt{1 + \frac{e^2 B^2 t^2}{m^2}} \)
• D. \( \lambda_0 \)

Hint:

When dealing with an electron in a magnetic field, remember that the magnetic force only changes the direction of motion, not the speed. Thus, the de-Broglie wavelength remains unaffected.

Answer 1

The Correct Option is D

Given: an electron (mass \(m\), charge \(-e\)) enters a uniform magnetic field \(\vec{B}=B\hat{j}\) with initial velocity \(\vec{v}=\vec{v_0}\) (where \(v_0>0\)). The de-Broglie wavelength initially is \(\lambda_0\). We need to find its wavelength after time \(t\).

Concept Used:

The magnetic field exerts a force on the moving electron given by the Lorentz force:

\[ \vec{F} = -e(\vec{v}\times\vec{B}). \]

This force changes the direction of velocity but not its magnitude (since the magnetic force is perpendicular to \(\vec{v}\)). Hence, the speed remains constant.

The de-Broglie wavelength is given by:

\[ \lambda = \frac{h}{p} = \frac{h}{mv}. \] Since \(v\) (the magnitude of velocity) is unchanged, the de-Broglie wavelength also remains constant in time.

 

Step-by-Step Solution:

Step 1: Write the equation of motion:

\[ m\frac{d\vec{v}}{dt} = -e(\vec{v}\times\vec{B}). \]

The magnetic field causes circular motion with angular frequency (cyclotron frequency):

\[ \omega = \frac{eB}{m}. \]

Step 2: The magnitude of velocity remains \(v_0\), only its direction changes with time. Therefore, momentum magnitude \(p=mv_0\) remains constant.

Step 3: Hence, the de-Broglie wavelength at time \(t\) is:

\[ \lambda = \frac{h}{mv_0} = \lambda_0. \]

Final Computation & Result:

\[ \boxed{\lambda(t)=\lambda_0.} \]

The de-Broglie wavelength of the electron remains unchanged with time.

Answer 2

Magnetic field does not work on the speed of the electron because magnetic forces only act perpendicular to the velocity. Thus, the speed of the electron will not change, and consequently, its de-Broglie wavelength will remain the same. The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{mv} \] Since the speed \( v \) does not change due to the magnetic field, the de-Broglie wavelength remains constant at \( \lambda_0 \). Thus, the de-Broglie wavelength at time \( t \) is the same as at \( t = 0 \): \[ \lambda(t) = \lambda_0 \]