91Ó°ÊÓ

The energy of a molecule in a quantum state is given by the sum of its rotational and vibrational energies. The rotational energy \(E_{r}\) of a molecule can be found with the formula \[E_{r} = \frac{1}{2} l(l+1) \hbar^{2}/I\]where \(l\) is the quantum number of the state, \(I\) is the rotational inertia, and \(\hbar\) is the reduced Planck's constant. The vibrational energy \(E_{v}\) is given by\[E_{v} = (n+\frac{1}{2})h\nu\]where \(n\) is the vibrational quantum number, \(h\) is Planck's constant, and \(\nu\) is the vibrational frequency.For the \(n=0, l=3\) state, we find \(E_{r, i} = \frac{1}{2} \cdot 3 \cdot (3+1) \cdot \hbar^{2}/I\) and \(E_{v, i} = (0+\frac{1}{2})h\nu\).For the \(n=1, l=2\) state, we find \(E_{r, f} = \frac{1}{2} \cdot 2 \cdot (2+1) \cdot \hbar^{2}/I\) and \(E_{v, f} = (1+\frac{1}{2})h\nu\).Thus the energy of the initial state \(E_{i}\) is \(E_{r, i} + E_{v, i}\) and the energy of the final state \(E_{f}\) is \(E_{r, f} + E_{v, f}\).

Next, calculate the energy difference \(\Delta E\) between the final and initial states, which is just the energy of the final state minus that of the initial state. If the energy of the final state is higher than that of the initial state, \(\Delta E\) will be positive and vice versa. Therefore, \(\Delta E = E_{f} - E_{i}\).

The energy difference \(\Delta E\) pertains to the energy of the photon required to effect the transition. This translates to the wavelength of infrared radiation mentioned in the question. The relationship between energy and wavelength is given by \(E = h c /\lambda\), where \(c\) is the speed of light, \(h\) is Planck's constant, and \(\lambda\) is the wavelength. Rearranging this equation for the desired wavelength gives \(\lambda = h c / \Delta E\). Plug in the calculated value for \(\Delta E\) to find the sought wavelength.