Question

An electron of mass \( m \) with an initial velocity \( \vec{v} = v_0 \hat{i} (v_0>0) \) enters an electric field \( \vec{E} = -E_0 \hat{k} \). If the initial de Broglie wavelength is \( \lambda_0 \), the value after time \( t \) would be:

• A. \( \lambda_0 \sqrt{\frac{1}{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \)
• B. \( \lambda_0 \sqrt{\frac{1}{1 - \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \)
• C. \( \lambda_0 \)
• D. \( \lambda_0 \left( 1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2} \right) \)

Hint:

In this problem, the momentum of the electron changes due to the electric field, which in turn changes the de Broglie wavelength. Use the relation \( \lambda = \frac{h}{p} \) to find the new wavelength at time \( t \).

Answer 1

The Correct Option is A

The de Broglie wavelength \( \lambda \) of a particle is related to its momentum \( p \) by: \[ \lambda = \frac{h}{p} \] In the electric field, the electron’s velocity changes due to the force exerted by the field. The force on the electron is \( F = eE \), and the acceleration is: \[ a = \frac{F}{m} = \frac{eE}{m} \] The velocity of the electron at time \( t \) is: \[ v(t) = v_0 + \frac{eE}{m} t \] The momentum at time \( t \) is: \[ p(t) = m \cdot v(t) = m \left( v_0 + \frac{eE}{m} t \right) \] Using the de Broglie relation: \[ \lambda(t) = \frac{h}{p(t)} = \frac{h}{m \left( v_0 + \frac{eE}{m} t \right)} \] Since \( \lambda_0 = \frac{h}{m v_0} \), we can rewrite this as: \[ \lambda(t) = \lambda_0 \frac{1}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \] Thus, the correct answer is \( \boxed{1} \).