Question

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0 \hat{i}$ ($E_0 = \text{constant}>0$), with an initial velocity $\vec{v} = v_0 \hat{i}$ ($v_0 = \text{constant}>0$). If $\lambda_0 = \frac{h}{4mv_0}$, its de Broglie wavelength at time $t$ is:

• A. $\frac{4\lambda_0}{\left[ 1 - \frac{E_0 e t}{2m v_0} \right]}$
• B. $\frac{4\lambda_0}{\left[ 1 + \frac{E_0 e t}{2m v_0} \right]}$
• C. $\frac{4\lambda_0}{\left[ 1 + \frac{2E_0 e t}{m v_0} \right]}$
• D. $\frac{4\lambda_0}{\left[ 1 - \frac{2E_0 e t}{m v_0} \right]}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
An electron (charge $-e$) is moving in a uniform electric field. This field causes an acceleration, changing the electron's velocity over time, which in turn changes its de Broglie wavelength.
Step 2: Key Formula or Approach:
1. Force $\vec{F} = q\vec{E} = (-e)(-2E_0 \hat{i}) = 2eE_0 \hat{i}$.
2. Acceleration $a = \frac{F}{m} = \frac{2eE_0}{m}$.
3. Velocity at time $t$: $v(t) = v_0 + at$.
4. de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{m v(t)}$.
Step 3: Detailed Explanation:
Velocity at time $t$:
\[ v(t) = v_0 + \left( \frac{2eE_0}{m} \right) t = v_0 \left( 1 + \frac{2eE_0 t}{m v_0} \right) \]
Now, find the wavelength at time $t$:
\[ \lambda = \frac{h}{m v(t)} = \frac{h}{m v_0 \left( 1 + \frac{2eE_0 t}{m v_0} \right)} \]
Given $\lambda_0 = \frac{h}{4mv_0} \implies \frac{h}{mv_0} = 4\lambda_0$.
Substitute this into the expression for $\lambda$:
\[ \lambda = \frac{4\lambda_0}{1 + \frac{2eE_0 t}{m v_0}} \]
Step 4: Final Answer:
The wavelength is $\frac{4\lambda_0}{1 + \frac{2E_0 e t}{m v_0}}$.