$E=kT$ or $\frac32kT$? Basically, which is the correct formula for thermal energy, and is this the same as kinetic energy? My notes are pretty conflicting on this topic, and I'm getting pretty confused.
 A: Prahar is correct that generally we have an energy contribution of ${1 \over 2} kT$ per degree of freedom in a system - so that atoms in a gas of atoms (e.g. Helium) will have an average energy of ${3 \over 2} kT$.
Often people talk about thermal energy being '$kT$' because of the exponential expression in 
$N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}}$
where $N_i$ is the number of atoms in a state $i$ with degeneracy $g_i$ and energy $E_i$ above the ground state which has $N_0$ atoms and a degeneracy of $g_0$.
So often people compare the energy of a state $E_i$ with the 'available thermal energy' $kT$, because the term 
$e^{-{E_i \over kT}}$
is dominant in the expression above and if $kT$ << $E_i$ then the ratio of population in state $i$ to the population in the ground state ($N_i$/$N_0$) will be small or perhaps close to zero. 
A: The thermal energy of a system is
$$
E = f \frac{1}{2} k T
$$
where $f$ is the number of degrees of freedom of the theory - which is roughly speaking the number of dimensions it is allowed to move in. 
For instance, if you are talking about an atom in 3 space dimensions, then the atom can move along the 3 axes and hence $f=3\implies E = \frac{3}{2} kT$. If I have a gas with $N$ atoms each of which can move in 3 dimensions, then $f = 3 N \implies E = \frac{3N}{2} kT$.
A: kT is the energy of collision between two particles, since each particle carries (on average) 1/2kT energy in the direction of the collision. Thus, 1/2kT+1/2kT=kT.
