A) \[\frac{3}{2}{{k}_{B}}T\]
B) \[\frac{5}{2}{{k}_{B}}T\]
C) \[\frac{7}{2}{{k}_{B}}T\]
D) \[\frac{1}{2}{{k}_{B}}T\]
Correct Answer: A
Solution :
[a] For monoatomic gases, degree of freedom is 3. |
Hence average thermal energy per molecule is |
\[K{{E}_{avg}}=\frac{3}{2}{{k}_{B}}T\] |
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