A diatomic molecule has $2.6 \times 10 ^ { - 5 } \mathrm { eV }$ of rotational energy in the $L = 2$ quantum state. What is its rotational energy in the $L = 1$ quantum state? $\left( 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$

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Quizplus · Accepted Answer

The rotational energy levels of a diatomic molecule are given by $E_L = \frac{L(L+1)h^2}{8\pi^2I}$, where $L$ is the quantum number, $h$ is Planck's constant, and $I$ is the moment of inertia. The energy is proportional to $L(L+1)$. For $L=2$, the energy is proportional to $2(2+1)=6$. For $L=1$, it is proportional to $1(1+1)=2$. Since the energy for $L=2$ is $2.6 \times 10^{-5} \mathrm{eV}$, for $L=1$, it should be $\frac{2}{6}$ of that value, which is $\frac{2}{6} \times 2.6 \times 10^{-5} \mathrm{eV} = 8.7 \times 10^{-6} \mathrm{eV}$.

A Diatomic Molecule Has 2.6×10−5eV2.6 \times 10 ^ { - 5 } \mathrm { eV }2.6×10−5eV

A Diatomic Molecule Has $2.6 \times 10 ^ { - 5 } \mathrm { eV }$