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Question #42817

Difficulty: 8/10 | 🟢 Free

The Question

An electron of mass $m$ and charge $-e$ is injected at $x=0$ with zero initial velocity into a semiconductor region of length $L$ . A non-uniform electric field $\vec{E} = -E_0 \left(1 - \frac{x}{L}\right) \hat{i}$ is applied across the region ( $0 \le x \le L$ ). Due to collisions with the lattice, the electron experiences a drag force. In the 'hot electron' regime, the mean free path $\lambda$ is constant, and the momentum relaxation time is given by $\tau = \frac{\lambda}{v}$ , where $v$ is the instantaneous speed of the electron. Which of the following expressions gives the square of the electron's velocity, $v^2$ , at $x=L$ ?

Options

A.
$\frac{eE_0\lambda}{m} \left[ \frac{\lambda}{2L} - \left(1 + \frac{\lambda}{2L}\right) e^{-2L/\lambda} \right]$
B.
$\frac{eE_0 L}{m}$
C.
$\frac{eE_0\lambda}{m} \left[ \frac{\lambda}{2L} + \left(1 - \frac{\lambda}{2L}\right) e^{2L/\lambda} \right]$
D.
$\frac{eE_0\lambda}{m} \left[ \frac{\lambda}{L} - \left(1 + \frac{\lambda}{L}\right) e^{-L/\lambda} \right]$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Relaxation Time Dependency|Non-Uniform Acceleration Constraint|Temperature and Kinetic Energy

Why This Question Is Tricky

Common Mistakes

This question targets the misconception of treating electrical resistance (or momentum relaxation) as a static parameter.

Students often use a constant $\tau$ or ignore the drag force entirely when calculating work done by a non-uniform field (leading to Option B via the Work-Energy Theorem).

By linking $\tau$ to the local kinetic energy ( $\tau = \lambda/v$ ), the drag force becomes non-linear ( $F_d \propto v^2$ ), forcing students to recognize resistance as a dynamic state variable.

This requires setting up a spatial first-order ODE rather than relying on simple kinematics.

Option D specifically traps students who forget the $1/2$ factor when converting $v \frac{dv}{dx}$ to $\frac{1}{2} \frac{d(v^2)}{dx}$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Relaxation Time Dependency
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free