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

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 \( t = 0 \): \[ \lambda(t) = \lambda_0 \]