An electron (mass m) with an initial velocity v = v₀î (v₀ 0) is in an electric field E = -E₀î (E₀ = constant

Question

An electron (mass m) with an initial velocity v = v₀î (v₀ > 0) is in an electric field E = -E₀î (E₀ = constant > 0). Its de Broglie wavelength at the time t is given by:

Accepted Answer

1. **Identify Forces:** The electron (charge $q = -e$) is in an electric field $\vec{E} = -E_0 \hat{i}$. The force on the electron is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.
2. **Determine Acceleration:** According to Newton's second law, acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{eE_0}{m} \hat{i}$.
3. **Calculate Velocity at time t:** Since the force is in the direction of the initial velocity ($v_0 \hat{i}$), the electron accelerates. The velocity at time $t$ is $\vec{v}(t) = v_0 \hat{i} + a t \hat{i} = (v_0 + \frac{eE_0 t}{m}) \hat{i}$.
4. **Calculate Momentum:** The magnitude of momentum is $p(t) = m v(t) = m(v_0 + \frac{eE_0 t}{m}) = mv_0 + eE_0 t = mv_0(1 + \frac{eE_0 t}{mv_0})$.
5. **Apply de Broglie Relation:** The de Broglie wavelength is $\lambda = \frac{h}{p}$. The initial wavelength is $\lambda_0 = \frac{h}{mv_0}$.
6. **Substitute and Simplify:** $\lambda(t) = \frac{h}{mv_0(1 + \frac{eE_0 t}{mv_0})} = \frac{\lambda_0}{1 + \frac{eE_0 t}{mv_0}}$.

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