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

To solve this problem, we need to determine how the de Broglie wavelength of an electron changes over time when it is subjected to an electric field. Let's break this down step-by-step: 

1. The initial de Broglie wavelength of the electron, \(\lambda_0\), can be expressed as: \(\lambda_0 = \frac{h}{m v_0}\), where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(v_0\) is the initial velocity.
2. When the electron enters the electric field \(\vec{E} = -E_0 \hat{k}\), it experiences a force: \(\vec{F} = e \vec{E} = -e E_0 \hat{k}\), where \(e\) is the charge of the electron.
3. This force affects the electron's motion and results in an acceleration: \(\vec{a} = \frac{\vec{F}}{m} = -\frac{e E_0}{m} \hat{k}\).
4. The velocity of the electron after time \(t\) can thus be given by: \(\vec{v}(t) = v_0 \hat{i} + \vec{a}t = v_0 \hat{i} - \frac{eE_0 t}{m} \hat{k}\).
5. The speed of the electron after time \(t\) is: \(v(t) = \sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2}\).
6. The new de Broglie wavelength \(\lambda\) is then: \(\lambda = \frac{h}{m v(t)} = \frac{h}{m \sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2}}\).
7. We can express \(\lambda\) in terms of \(\lambda_0\): \(\lambda = \frac{\lambda_0 m v_0}{m \sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2}} = \lambda_0 \cdot \frac{v_0}{\sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2}}\).
8. Further simplification gives: \(\lambda = \lambda_0 \sqrt{\frac{1}{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}\).

This corresponds to the correct option: \(\lambda_0 \sqrt{\frac{1}{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}\). This shows that the de Broglie wavelength decreases as the speed of the electron increases due to the electric field component in the \(\hat{k}\) direction, thereby validating the selected option.

Answer 2

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 \( \boxed{1} \).