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Question #42555

Difficulty: 8/10 | 🟢 Free

The Question

A quantum engine uses NN non-interacting particle pairs as its working substance. Each pair consists of two particles of masses mm and 3m3m, confined in a 1D infinite potential well. The system is restricted to its lowest two energy states. The engine operates in a cycle consisting of:

  1. Isothermal expansion at THT_H from well width LL to 2L\sqrt{2}L.
  2. Isochoric (constant width) cooling to TLT_L.
  3. Isothermal compression at TLT_L back to width LL.
  4. Isochoric heating back to THT_H. Given that the temperatures are high enough such that kT≫ΔEkT \gg \Delta E (where ΔE\Delta E is the energy gap between the two states), what is the efficiency η\eta of this engine?

Options

  • A.

    3(TH−TL)8TH−5TL\frac{3(T_H - T_L)}{8T_H - 5T_L}

  • B.

    1−TLTH1 - \frac{T_L}{T_H}

  • C.

    3(TH−TL)5TH−2TL\frac{3(T_H - T_L)}{5T_H - 2T_L}

  • D.

    TH−TL2TH−TL\frac{T_H - T_L}{2T_H - T_L}

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Correct Answer

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

Carnot Engine Efficiency|Reduced Mass Effect|Binomial Approximation

Why This Question Is Tricky

Common Mistakes

This question traps students who blindly assume any thermodynamic cycle operating between two temperatures has the Carnot efficiency η=1−TL/TH\eta = 1 - T_L/T_H.

By defining a Stirling-like cycle for a quantum 2-level system, students must derive the macroscopic thermodynamic state variables (UU, SS, FF) from the microscopic partition function ZZ.

The trap lies in the high-temperature Taylor expansion: failing to expand to the second order in βΔE\beta \Delta E results in a constant entropy, leading to zero isothermal heat transfer.

Furthermore, the reduced mass modifies the energy gap, linking quantum mechanics to statistical thermodynamics, ensuring only those with a deep understanding of the fundamental relations F=−kTln⁡ZF = -kT \ln Z and S=−∂F/∂TS = -\partial F/\partial T can extract the correct efficiency.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermodynamics - Class 11 Part 2
ConceptCarnot Engine Efficiency
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree