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Physics · JEE | KINETIC THEORY

हिन्दी

According to the law of equipartition of energy, each translational degree of freedom contributes an average energy per molecule equal to:

  1. A
    \( k_B T \)
  2. B
    \( 3 k_B T \)
  3. C
    \( \frac{3}{2} k_B T \)
  4. D
    \( \frac{1}{2} k_B T \)
View Answer & Solution

Correct answer · Option D

\( \frac{1}{2} k_B T \)

Explanation

Each translational degree of freedom contributes \( \frac{1}{2} k_B T \) of average energy per molecule.
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