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Question #43658

Difficulty: 8/10 | 🟢 Free

The Question

A cubical container of volume VV holds NN atoms of an ideal monatomic gas, each of mass mm. The gas is initially in thermal equilibrium at temperature TT with pressure P0P_0. At time t=0t=0, every atom simultaneously emits a single photon of energy EE in the +x+x direction. Assume the emission is instantaneous and the atoms undergo elastic collisions with the container walls without colliding with each other. Let ΔPx\Delta P_x be the change in the average pressure exerted by the gas on the two walls perpendicular to the x-axis, and ΔPy\Delta P_y be the change in pressure on the walls perpendicular to the y-axis. Which of the following is correct? (Here cc is the speed of light and n=N/Vn = N/V)

Options

  • A.

    ΔPx=nE2mc2,ΔPy=0\Delta P_x = \frac{n E^2}{m c^2}, \quad \Delta P_y = 0

  • B.

    ΔPx=nE23mc2,ΔPy=nE23mc2\Delta P_x = \frac{n E^2}{3 m c^2}, \quad \Delta P_y = \frac{n E^2}{3 m c^2}

  • C.

    ΔPx=nE22mc2,ΔPy=0\Delta P_x = \frac{n E^2}{2 m c^2}, \quad \Delta P_y = 0

  • D.

    ΔPx=2nE23mc2,ΔPy=0\Delta P_x = \frac{2 n E^2}{3 m c^2}, \quad \Delta P_y = 0

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Recoil of Atom|Pressure-Kinetic Energy Relation|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental microscopic definition of pressure as the momentum flux tensor, rather than the macroscopic scalar relation P=23KVP = \frac{2}{3} \frac{K}{V}.

Students often fall into the trap of calculating the total increase in kinetic energy (ΔK=Np22m\Delta K = N \frac{p^2}{2m}) and blindly applying the isotropic pressure formula to conclude ΔP=23ΔKV=nE23mc2\Delta P = \frac{2}{3} \frac{\Delta K}{V} = \frac{n E^2}{3 m c^2} for all walls.

However, the photon emission is highly directional (anisotropic), meaning the velocity distribution only shifts along the x-axis.

Consequently, the pressure increase is entirely confined to the x-walls, requiring an integration of the momentum transfer over the shifted velocity distribution.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAtoms - Class 12
ConceptRecoil of Atom
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree