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The calculated rotational constant gives ${\theta }_{\mathrm{rot}}=1.18721\mathrm{K}$. At room temperature, taken as $300\mathrm{K}$, the continuous maximum is $J_{\ast }=10.7404$; comparing the adjacent integer populations gives $J_{\max }=11$. Taking $298.15\mathrm{K}$ instead gives $J_{\ast }=10.7057$ and the same integer answer.

At $8.7\mathrm{K}$, $J_{\ast }=1.41417$. In particular,and $N_1>N_0$, so $J_{\max }=1$. The integer comparison avoids treating a noninteger angular momentum quantum number as a physical state.

For a linear rigid rotor with moment of inertia $I$, the energy eigenvalue at angular momentum quantum number $J$ is ${\hslash }^{2}J(J+1)/(2I)$. Each level has statistical weight $2J+1$ before additional symmetry or nuclear-spin restrictions.