Please see the attached file.
We can solve this problem by computing the partition function of the system. The partition function is:
Z(N) = Z1^N (1)
where Z(N) is the partition function of the system and Z1 is the partition function of a single atom. We must now not divide by N! because the atoms are assumed to be distinguishable. So, let's compute Z1. The atom can be in two states, one with energy 0 and one with energy e. We thus have:
Z1 = sum over all states of exp(-beta energy of the state) = exp(0) + exp(-beta e) = 1 + exp(-beta e) (2)
If we plug this in Eq. (1) we get:
Z(N) = [1 + exp(-beta e)]^N (3)
The energy per atom of the system at some temperature is not a constant value but it has a probability distribution that becomes sharper and sharper as we make N larger and larger (the spread is of order 1/sqrt(N) ). What we can compute is the average value (a.k.a. expectation value) of the energy and this will then also be "the energy" because of the sharpness of the probability distribution. The general formula for the average of the total energy of the system is:
E(N) = -d Log[Z(N)]/dbeta (4)
To see this, let's work out the derivative (I'm suppressing the argument N of Z for the moment):
d Log(Z)/dbeta = 1/Z dZ/dbeta
Insert in here Z = Sum over r of exp(-beta E_r) (here by r we denote the states of the system and E_r are the value of the energy of state r)
d Log(Z)/dbeta = 1/Z d/dbeta [Sum over r of exp(-beta E_r)] = (bring derivative inside summation) =
1/Z Sum over r of exp(-beta E_r) (-E_r) ...
A detailed solution is given.