Undergraduate quantum chemistry bonus problems

The point of this exercise is to take small sections of the Morse potential and see how they compare to the potentials we know. For low values of $n$ (energies less than around -0.8 in the diagram) the potential looks like what you see to the right.

As you can see, this looks a like a slightly lopsided harmonic oscillator potential. Since the energy level spacing depends on the shape, we would expect the energy levels to be spaced about evenly (as with the harmonic oscillator). Since the anharmonicity (fancy word meaning “not shaped like a harmonic oscillator”) makes it look a little like a hydrogen atom potential, we’d expect levels to get a little more closely spaced.

At higher energies, the Morse potential looks more like a hydrogen atom potential. For the hydrogen atom, we know that the energy levels get more closely packed together, so we would expect similar behavior in the high-energy Morse potential. A plot focusing on the Morse potential at high energies is shown to the right.

1. First we’ll figure out the spacing for the vibrational levels: \[\hbar \omega ((v+1) + 1/2) - \hbar\omega (v + 1/2) = \hbar\omega\]

   We’re looking for $J$ such that the energy of the rotational motion, $B J(J + 1)$ is greater than or equal to $\hbar\omega$. 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: \[\begin{align*} BJ(J + 1) &= \hbar\omega \\ BJ^2 + J - \hbar\omega &= 0 \\ J &= (-1 + \sqrt{1 + 4B\omega\hbar}) / 2 \end{align*}\]

   where the positive root is chosen because $J$ must be positive. Plugging in the numbers for N$_2$, 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 $\hbar\omega$. Now we need to find the rotational energy spacing; that is, the energy difference between the rotational state $J + 1$ and the rotational state $J$. \[\begin{align*} E_{J+1} - E_J &= B(J + 1)(J + 2) - BJ(J+1) \\ &= B(J+1)((J+2) - J) = 2B(J+1) \end{align*}\]

   Setting this equal to the vibrational spacing $\hbar\omega$ and rearranging a bit, we obtain $J = \hbar\omega / 2B - 1$. Plugging in the numbers for N$_2$, 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 your introductory course on statistical mechanics and thermodynamics. For the vibrational distribution, you can see the calculation done on page 742 of McQuarrie and Simon, which gives $1.31\times 10^{-5}$ 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 the $B$ described in class is defined a little differently than in a different way than McQuarrie and Simon). The graph to the right 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 $v=1$).