Question

An electron of specific charge \( \frac{e}{m} \) enters an electric field, \( \vec{E} = -E_0 \hat{i} \) at a time \( t = 0 \) with an initial velocity \( \vec{v}_1 \). If \( \lambda_0 \) is its initial de Broglie wavelength, then its de Broglie wavelength at a time ‘t’ is

• A. \( \frac{\lambda_0}{\left(1 + \frac{eE_0 t}{mv} \right)} \)
• B. \( \lambda_0 \left[ 1 + \frac{eE_0 t}{mv} \right] \)
• C. \( \frac{\lambda_0}{\left(1 - \frac{eE_0 t}{mv} \right)} \)
• D. \( \frac{\lambda_0 e E_0}{mv} \)

Hint:

The de Broglie wavelength varies inversely with momentum: \( \lambda = \frac{h}{p} \). Use Newton's second law to relate momentum and time in fields.

Answer 1

The Correct Option is A

The momentum of the electron changes due to the electric field: \[ F = ma = eE_0 \Rightarrow a = \frac{eE_0}{m} \] Since \( v = v_1 + at = v + \frac{eE_0 t}{m} \), the momentum becomes: \[ p = m \left( v + \frac{eE_0 t}{m} \right) = mv + eE_0 t \] Using de Broglie’s relation: \[ \lambda = \frac{h}{p} = \frac{h}{mv + eE_0 t} = \frac{h/mv}{1 + \frac{eE_0 t}{mv}} = \frac{\lambda_0}{1 + \frac{eE_0 t}{mv}} \]