First slide

Law of equipartition of energy

Question

The average thermal energy for a mono-atomic gas is: ( $K_{B}$ is Boltzman constant and T, absoulte temperature)

Moderate

A

$\frac{3}{2} k_{B} T$
B

$\frac{5}{2} k_{B} T$
C

$\frac{7}{2} k_{B} T$
D

$\frac{1}{2} k_{B} T$

Solution

Average thermal energy $= \frac{3}{2} K_{B} T$ , as the number of degrees of freedom is 3

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