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Question #43719

Difficulty: 8/10 | 🟢 Free

The Question

A semiconductor rod of length LL is placed between two heat reservoirs maintained at temperatures T1T_1 (at x=0x = 0) and T2T_2 (at x=Lx = L). The thermal conductivity of the rod is uniform and constant, and steady-state heat flow is established. The electron concentration in the rod is maintained as n(x)=n0e−αxn(x) = n_0 e^{-\alpha x}, where n0n_0 and α\alpha are positive constants. The generalized one-dimensional electron charge current density is given by J=enμE+eddx(Dn)J = e n \mu E + e \frac{d}{dx}(D n), where ee is the elementary charge, μ\mu is the constant electron mobility, EE is the electric field, and DD is the diffusion coefficient. The diffusion coefficient satisfies the local Einstein relation D(x)=kBT(x)eμD(x) = \frac{k_B T(x)}{e} \mu, where kBk_B is the Boltzmann constant. Assuming the net electron current density is zero in the steady state, what is the built-in potential difference ΔV=V(L)−V(0)\Delta V = V(L) - V(0) across the rod?

Options

  • A.

    kBe[(T2−T1)−αL2(T1+T2)]\frac{k_B}{e} \left[ (T_2 - T_1) - \frac{\alpha L}{2} (T_1 + T_2) \right]

  • B.

    −kBαL2e(T1+T2)-\frac{k_B \alpha L}{2e} (T_1 + T_2)

  • C.

    kBe[(T1−T2)+αL2(T1+T2)]\frac{k_B}{e} \left[ (T_1 - T_2) + \frac{\alpha L}{2} (T_1 + T_2) \right]

  • D.

    kBe[(T2−T1)−αL(T1+T2)]\frac{k_B}{e} \left[ (T_2 - T_1) - \alpha L (T_1 + T_2) \right]

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Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Drift vs Diffusion Current|Potential Gradient Direction|Steady State Temperature Gradient

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to synthesize thermal conduction (Fourier's Law) with semiconductor drift-diffusion mechanics under non-isothermal conditions.

The primary trap lies in treating the diffusion coefficient DD as a spatial constant and pulling it out of the derivative ddx(Dn)\frac{d}{dx}(Dn), which is only valid for isothermal systems.

By doing so, students miss the thermoelectric (Seebeck) contribution dTdx\frac{dT}{dx} to the built-in electric field.

Furthermore, it rigorously tests the correct application of the potential gradient E=−dV/dxE = -dV/dx and the integration of a linear temperature profile over the rod's length.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterSemiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
ConceptDrift vs Diffusion Current
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree