# Law of equipartition of energy, shouldn't kinetic energy per molecule by $(3/2)kT/f$

I study that according to the law of equipartition of energy the average kinetic energy associated with each degree of freedom is equal to $$(1/2)kT$$. But shouldn't it be $$\frac{(3/2)kT}{f}$$ where $$f$$ is the total number of the degrees of freedom. Isn't that how the value $$(1/2)kT$$ came? $$\frac{(3/2)kT}{3}$$

All these theories must be verified: so where am I wrong?

• It is hard to say where you are wrong, as you essentially just state what you think the result should b, without explaining why you think that. In any case, you have it back to front. The average energy is given by $\frac{f}{2}k_B T$. The value of $\frac{3}{2}$ then comes from setting $f=3$ (one degree of freedom for each independent direction of motion – By Symmetry Oct 18 at 10:53
• Okay, thank you. I think they mentioned it to be 3/2kT in the first derivation by only considering the translational degrees of freedom. If they consider the rotational and vibrational degrees of freedom, it would've changed in the first derivation (of kinetic energy per molecule). This should've been checked in the derivation itself, then. – Swaroop Joshi Oct 18 at 10:56