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Question #43618

Difficulty: 8/10 | 🟢 Free

The Question

A non-linear polyatomic ideal gas, consisting of molecules of mass $m$ , is in thermal equilibrium at absolute temperature $T$ . Each molecule has exactly $6$ active degrees of freedom ( $3$ translational and $3$ rotational, with all vibrational modes frozen). If $E$ represents the average total energy per molecule, which of the following expressions gives the effective thermal de Broglie wavelength ( $\lambda$ ) of the gas molecules based on their root-mean-square momentum? (Here $h$ is Planck's constant)

Options

A.
$\frac{h}{\sqrt{mE}}$
B.
$\frac{h}{\sqrt{2mE}}$
C.
$\frac{h}{\sqrt{3mE}}$
D.
$\frac{h}{\sqrt{6mE}}$

Check Your Answer

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

De Broglie Wavelength Scaling|Molecular Degrees of Freedom|Kinetic Theory

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of the Equipartition Theorem and its application to the de Broglie wavelength.

A common trap is assuming that the total internal energy of a polyatomic molecule contributes to its translational momentum, leading to the incorrect use of $p = \sqrt{2mE}$ (Option B).

Students must recognize that the de Broglie wavelength is determined strictly by the translational degrees of freedom ( $f=3$ ), which partition only a fraction of the total energy ( $K_{\text{trans}} = E/2$ for $f=6$ ).

By expressing the wavelength in terms of total energy $E$ , the problem forces students to explicitly decouple translational kinetic energy from rotational kinetic energy.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Dual Nature of Radiation and Matter - Class 12
Concept	De Broglie Wavelength Scaling
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free