1. First we'll figure out the spacing for the vibrational levels:

    hν((v+1)+1/2)- hν(v+1/2) = hν

    We're looking for J such that the energy of the rotational motion, BJ(J + 1) is greater than or equal to hν. Since the rotational energy increases as J increases, we just need to find the value of J for which the energy is exact, and round up to the next integer. So we solve the quadratic equation:

    BJ(J + 1) = hν
    BJ2 + BJ - hν = 0
    J = 1/2 (-1 + (1 + 4hν/B)1/2)

    where the positive root is chosen because J must be positive. Plugging in numbers for N2, we obtain J = 33.9, so the lowest physically meaningful J for which this is true is J = 34.

  2. As above, the vibrational energy spacing is hν. Now we need to find the rotational energy spacing; that is, the energy difference between the rotational state J + 1 and the rotational state J.

    EJ+1-EJ = B(J + 1)(J + 2) - BJ(J + 1) = B(J + 1)((J + 2) - J) = 2B(J + 1)

    Setting this equal to hν and rearranging a bit, we obtain J = (hν)/(2B) - 1. Plugging in the numbers for N2, we find that J = 589.

When we compare the energy spacings to the thermal energy, we note that the average thermal energy at room temperature is about a tenth of the vibrational energy spacing, but is a hundred times the rotational energy constant. The thermal energy would correspond to a rotational quantum number of J = 9 or 10, substantially less than the J = 34 of part 1.

As mentioned in the problem, the ratios of different states at a given temperature can be calculated using techniques which you will learn in Chem 120B. For the vibrational distribution, you can see the calculation done on page 742 of McQuarrie and Simon, which gives 1.31×10-5 as the fraction in the first excited vibrational state, with nearly everything else in the ground state.

The rotational states have a more interesting distribution. The details of the calculation can be found on pages 744-748 of McQuarrie and Simon (but be careful, because Prof. Miller defines B in a different way than McQuarrie and Simon). The following graph plots the fraction of molecules in the first few energy levels. As you can see, the peak is around J = 6 or 7. A rotational quantum number of J = 34 would be extremely unusual at room temperature. In fact, it would contribute a little less to the total population than the first excited vibrational state does, since what actually determines the contribution is the energy (and the energy for J=34 is a little higher than the energy for v=1).

N_2 rotational distribution