1. 1/2k_BT 2. 3/2k_BT 3. 5/2k_BT 4. 7/2k_BT

The average thermal energy for a mono-atomic gas is : (k_B is Boltzmann constant and T, absolute temperature)

1. $\frac{1}{2}k_BT$

2. $\frac{3}{2}k_BT$

3. $\frac{5}{2}k_BT$

4. $\frac{7}{2}k_BT$

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

Best answer

Answer: 2. $\frac{3}{2}k_BT$

A monoatomic gas molecule has only 3 translational degrees of freedom. By the equipartition law each gives $\tfrac12k_BT$.

$$E=3\times\tfrac12k_BT=\tfrac32k_BT$$

Answer: option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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