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:
Show Hint
- \( \frac{\lambda_0}{\sqrt{1 - \frac{e^2 B^2 t^2}{m^2}}} \)
- \( \frac{\lambda_0}{\sqrt{1 + \frac{e^2 B^2 t^2}{m^2}}} \)
- \( \lambda_0 \sqrt{1 + \frac{e^2 B^2 t^2}{m^2}} \)
- \( \lambda_0 \)
The Correct Option is D
Approach Solution - 1
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.
Approach Solution -2
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