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 determine how an electron's de Broglie wavelength changes over time under an electric field, we follow these steps:

1. The initial de Broglie wavelength, \(\lambda_0\), is given by \(\lambda_0 = \frac{h}{m v_0}\), where \(h\) is Planck's constant, \(m\) is the electron's mass, and \(v_0\) is its initial velocity.
2. Upon entering the electric field \(\vec{E} = -E_0 \hat{k}\), the electron experiences a force \(\vec{F} = e \vec{E} = -e E_0 \hat{k}\), with \(e\) being the electron's charge.
3. This force causes an acceleration \(\vec{a} = \frac{\vec{F}}{m} = -\frac{e E_0}{m} \hat{k}\).
4. The electron's velocity at time \(t\) is then \(\vec{v}(t) = v_0 \hat{i} + \vec{a}t = v_0 \hat{i} - \frac{eE_0 t}{m} \hat{k}\).
5. The electron's speed at 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 calculated as \(\lambda = \frac{h}{m v(t)} = \frac{h}{m \sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2}}\).
7. Expressing \(\lambda\) in terms of \(\lambda_0\), we get \(\lambda = \frac{\lambda_0 m v_0}{m \sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2}} = \lambda_0 \cdot \frac{v_0}{\sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2}}\).
8. Further simplification yields \(\lambda = \lambda_0 \sqrt{\frac{1}{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}\).

This result, \(\lambda_0 \sqrt{\frac{1}{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}\), indicates that the de Broglie wavelength decreases as the electron's speed increases due to the electric field component along the \(\hat{k}\) direction, thus confirming the correct option.