Q.9
The average energy of a diatomic gaseous molecule at temperature T is kBT where kB is Boltzmann’s constant. The average energy of this molecule per degree of freedom is
(A) ½ kBT
(B) 2⁄3 kBT
(C) kBT
(D) 2⁄5 kBT

The correct answer is (A) ½ kBT.

According to the equipartition theorem, each degree of freedom contributes an average energy of ½ kBT to a molecule in thermal equilibrium, where kB is Boltzmann’s constant and T is temperature. The question states the total average energy of the diatomic molecule is kBT, but standard physics shows it as (5/2) kBT at room temperature (3 translational + 2 rotational degrees of freedom). The query directly asks for energy per degree of freedom, which remains ½ kBT regardless of total degrees or the given total energy value.​

Option Analysis

• (A) ½ kBT: Correct. Equipartition theorem assigns ½ kBT per quadratic degree of freedom (translational, rotational, or vibrational).​

• (B) 2/3 kBT: Incorrect. This equals total translational energy divided by 3 translational degrees ( (3/2 kBT)/3 ), but ignores equal contribution per degree.​

• (C) kBT: Incorrect. Matches neither per-degree energy nor typical diatomic total (which is 5/2 kBT); possibly confuses with vibrational mode (kinetic + potential).​

• (D) 2/5 kBT: Incorrect. Reverses logic; if total were 2 kBT over 5 degrees, per-degree would still be ½ kBT (2 kBT / 5 = 2/5 kBT), but theorem fixes it at ½ kBT.​

Degrees of Freedom

Diatomic molecules (e.g., N2, O2) have 5 degrees at room temperature: 3 translational (x, y, z motion) and 2 rotational (about axes perpendicular to bond). Total energy = 5 × (½ kBT) = (5/2) kBT. Vibrational modes (2 more degrees) activate at high T (>1000 K), raising total to 7/2 kBT.​

The average energy of a diatomic gaseous molecule per degree of freedom is a core concept in kinetic theory of gases, vital for CSIR NET Life Sciences and physics exams. At temperature T, this energy equals ½ kBT, where kB is Boltzmann’s constant, per the equipartition theorem. This holds for translational, rotational, and (at high T) vibrational modes, ensuring equal energy sharing across degrees of freedom.​

Equipartition Theorem Basics

The theorem states each quadratic term in energy (e.g., ½ mv² for translation) contributes ½ kBT on average. For diatomic gases like O2 or N2, room-temperature degrees of freedom total 5, yielding molecule total energy of (5/2) kBT—not kBT as some questions state, possibly testing per-degree focus.​

Diatomic vs Monatomic Comparison

CSIR NET Exam Insights

Questions often twist totals (e.g., stating kBT) to check per-degree recall: always ½ kBT. Practice: If total energy = f × (½ kBT), solve for f or verify options. Vibrations freeze at low T, keeping f=5 for N2/O2.​