Question

An electron of mass \( m \) with initial velocity \( \vec{v} = v_0 \hat{i} \) (\( v_0>0 \)) enters in an electric field \( \vec{E} = -E_0 \hat{i} \) (\( E_0 \) is constant \(>0 \)) at \( t = 0 \). If \( \lambda \) is its de-Broglie wavelength initially, then the de-Broglie wavelength after time \( t \) is:

• A. \( \frac{\lambda}{1 + \frac{eE_0 t}{m v_0}} \)
• B. \( \frac{\lambda}{\left(1 - \frac{eE_0 t}{m v_0}\right)^2} \)
• C. \( \left(1 + \frac{eE_0 t}{m v_0}\right) \lambda \)
• D. \( \left(1 + \frac{eE_0 t}{m v_0}\right)^2 \lambda \)

Hint:

For charged particles moving in an electric field, use Newton’s second law to determine velocity change over time. The de-Broglie wavelength is inversely proportional to momentum, allowing derivation of the new wavelength using \( \lambda = \frac{h}{mv} \).

Answer 1

The Correct Option is A

Step 1: Equation of Motion for the Electron The force on the electron due to the electric field is given by: \[ F = eE_0 \] Using Newton’s second law: \[ m \frac{dv}{dt} = -eE_0 \] Integrating both sides from \( v_0 \) to \( v \) over time \( 0 \) to \( t \): \[ v = v_0 - \frac{eE_0 t}{m} \] Step 2: De-Broglie Wavelength Relation The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{mv} \] After time \( t \), the new wavelength \( \lambda' \) is: \[ \lambda' = \frac{h}{m v} \] Substituting \( v = v_0 - \frac{eE_0 t}{m} \): \[ \lambda' = \frac{h}{m \left( v_0 - \frac{eE_0 t}{m} \right)} \] Dividing the numerator and denominator by \( h / m v_0 \), we get: \[ \lambda' = \frac{\lambda}{1 + \frac{eE_0 t}{m v_0}} \] Step 3: Conclusion Thus, the de-Broglie wavelength after time \( t \) is \( \frac{\lambda}{1 + \frac{eE_0 t}{m v_0}} \), which matches option (1).