Answer
See answers.
Work Step by Step
a. Calculate the moment of inertia of the oxygen molecule about its center of mass.
See Figure 29-15. Each oxygen atom is a distance r/2 from the center of mass.
$$I=2m_O\left(\frac{1}{2}r \right)^2=\frac{1}{2}m_Or^2$$
Express the characteristic rotational energy of $O_2$ about its center of mass.
$$E_{rot}=\frac{\hbar ^2}{2I}=\frac{\hbar ^2}{ m_Or^2}$$
Putting in the given numbers, we find that it is $1.79\times10^{-4}$ eV.
b. Refer to Fig. 29–16. The energy involved in the stated transition is $3\frac{\hbar ^2}{I}$.
$$3\frac{\hbar ^2}{I}=6\frac{\hbar ^2}{2I}=6(1.79\times10^{-4}eV)= 1.07\times10^{-3}eV$$
Find the wavelength of the emitted photon.
$$\lambda=\frac{hc}{E}=\frac{1240 eV \cdot nm}{1.07\times10^{-3}eV }=1.15\times10^{-3}m $$
