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Question #42589

Difficulty: 8/10 | 🟢 Free

The Question

Consider an ideal gas consisting of identical particles, each of mass $m$ , in thermal equilibrium at absolute temperature $T$ . Let $\langle \lambda \rangle$ be the true average de Broglie wavelength of the gas molecules, calculated by averaging over the Maxwell-Boltzmann speed distribution. Let $\lambda_{rms}$ be the de Broglie wavelength of a molecule moving exactly with the root-mean-square speed of the gas. What is the ratio $\frac{\langle \lambda \rangle}{\lambda_{rms}}$ ?

Options

A.
$\sqrt{\frac{6}{\pi}}$
B.
$\sqrt{\frac{8}{3\pi}}$
C.
$\sqrt{\frac{3}{2}}$
D.
1

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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

Temperature and Kinetic Energy|De Broglie Wavelength Scaling|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental distinction between the function of an average and the average of a function in statistical mechanics.

Students commonly fall into the trap of assuming $\langle \lambda \rangle = h / (m v_{rms})$ or $h / (m \langle v \rangle)$ , directly substituting a characteristic speed into the de Broglie equation.

However, because wavelength is inversely proportional to speed ( $\lambda \propto 1/v$ ), the correct approach requires evaluating the expectation value $\langle 1/v \rangle$ over the Maxwell-Boltzmann distribution.

This highlights the non-linear relationship between momentum and matter-wave properties in a thermal ensemble, a quintessential JEE Advanced conceptual hurdle.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Kinetic Theory - Class 11 Part 2
Concept	Temperature and Kinetic Energy
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free