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## Q. 10.1

Estimate the value of $E_{ rot }$ for the lowest rotational energy state of $N _{2}$, which has a bond length 0.110 nm.

Strategy If we consider the nitrogen atoms to be point masses (each with mass m) separated by a distance R (see Figure 10.4), then the rotational inertia about an axis passing through the center of the molecule and perpendicular to the line joining the atoms is

$I=m\left(\frac{R}{2}\right)^{2}+m\left(\frac{R}{2}\right)^{2}=\frac{m R^{2}}{2}$

With this numerical estimate of rotational inertia I, the energy levels follow from Equation (10.2).

$E_{ rot }=\frac{\hbar^{2} \ell(\ell+1)}{2 I}$ (10.2) ## Verified Solution

For nitrogen, $m=2.33 \times 10^{-26} kg \text {, and } R=1.10 \times 10^{-10} m$.

Thus,

\begin{aligned}I &=\frac{\left(2.33 \times 10^{-26} kg \right)\left(1.10 \times 10^{-10} m \right)^{2}}{2} \\&=1.41 \times 10^{-46} kg \cdot m ^{2}\end{aligned}

For $\ell=1$,

\begin{aligned}E_{ rot } &=\frac{2 \hbar^{2}}{2 I}=\frac{\hbar^{2}}{I}=\frac{\left(1.055 \times 10^{-34} J \cdot s \right)^{2}}{1.41 \times 10^{-46} kg \cdot m ^{2}} \\&=7.89 \times 10^{-23} J =4.93 \times 10^{-4} eV\end{aligned}

Remember, however, that this is just the energy of the lowest state. The energy of the $\ell=4 \text { state }[\text { with } \ell(\ell+1)=20]$ is 10 times higher, or about 0.005 eV. This energy is still at least two orders of magnitude less than the energy associated with an electronic transition in hydrogen (the kind that produces visible photons). Therefore we can expect photons generated by transitions between adjacent rotational states to be in the infrared or microwave portion of the spectrum.