Molecular rotation: A diatomic molecule has 2.6 × 10^-5 eV of rotational energy in the l = 2 quantum state. What is its rotational energy in the l = 1 quantum state? A) 3.4 µeV B) 4.1 µeV C) 5.3 µeV D) 7.8 µeV E) 8.7 µeV

Question

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The rotational energy levels of a diatomic molecule are given by $E_l = \frac{h^2}{8\pi^2I}l(l+1)$, where $E_l$ is the energy of the level with quantum number $l$, $h$ is Planck's constant, and $I$ is the moment of inertia of the molecule. The energy difference between levels depends on $l$, so the energy for $l = 1$ can be calculated relative to $l = 2$ given the energy for $l = 2$.Given $E_{l=2} = 2.6 \times 10^{-5}$ eV, to find $E_{l=1}$, we use the ratio of the $l(l+1)$ terms because the rest of the formula (constants and $I$) remains the same for both states. For $l = 2$, $l(l+1) = 2 \times 3 = 6$, and for $l = 1$, $l(l+1) = 1 \times 2 = 2$.The energy for $l = 1$ is thus $\frac{2}{6}$ of the energy for $l = 2$, or $E_{l=1} = \frac{2}{6} \times 2.6 \times 10^{-5}$ eV. This calculation simplifies to $E_{l=1} = \frac{1}{3} \times 2.6 \times 10^{-5}$ eV = $0.866 \times 10^{-5}$ eV, or $8.7 \times 10^{-6}$ eV, which is 8.7 µeV.

Molecular Rotation: a Diatomic Molecule Has 2