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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:

A

\lambda ₀ / (1 + eE₀t / mv₀)

B

\lambda ₀ (1 + eE₀t / mv₀)

C

\lambda ₀

D

\lambda ₀t

Step-by-Step Solution

  1. Identify Forces: The electron (charge q=−eq = -e) is in an electric field E⃗=−E0i^\vec{E} = -E_0 \hat{i}. The force on the electron is F⃗=qE⃗=(−e)(−E0i^)=eE0i^\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}.
  2. Determine Acceleration: According to Newton's second law, acceleration a⃗=F⃗m=eE0mi^\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 (v0i^v_0 \hat{i}), the electron accelerates. The velocity at time tt is v⃗(t)=v0i^+ati^=(v0+eE0tm)i^\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)=mv(t)=m(v0+eE0tm)=mv0+eE0t=mv0(1+eE0tmv0)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 λ=hp\lambda = \frac{h}{p}. The initial wavelength is λ0=hmv0\lambda_0 = \frac{h}{mv_0}.
  6. Substitute and Simplify: λ(t)=hmv0(1+eE0tmv0)=λ01+eE0tmv0\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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