Question

The moment of inertia for an axis through the center of mass of a diatomic molecule calculated from the wavelength emitted in an $I=19\to I=18$ transition is different from the moment of inertia calculated from the wavelength of the photon emitted in an $I=1\to I=0$ transition. Explain this difference. Which transition corresponds to the larger moment of inertia?

Short answer

At the transition, $I=1\to I=0$ the moment of inertia is larger.

Step-by-step solution

Definition of moment of inertia

A measurement of a body's resistance to angular acceleration about a specific axis that is equal to the product of each mass component's mass and its squared distance from the axis.

Determine which transition has greater energy (of photon)

Use the formula of rotational energy levels of a diatomic molecule.

$E_I=I(I+1)\frac{h^2}{2l}\text{with}I=0,1,2,\dots \dots$

Now,

$\Delta E_{big}~19(19+1)-18(18+1)=38$

And

$$\begin{gathered} \Delta E_{\text{smaII}}~1(1+1)-0(0+1)=2 \\ \Delta E_{big}>\Delta E_{\text{smaII}} \end{gathered}$$

Therefore, this can also be verified by measuring the wavelength of the emitted photon at each transition. In the first case, the wavelength is shorter than in the second case.

Determine which transition has greater moment of inertia

Now, since $\Delta E~\frac{h}{2l}\Rightarrow I~\frac{h}{2\Delta E}\mid$

Thus, the photon which is the least energetic has the longest wavelength which corresponds to the greatest moment of inertia. Whereas the photon, which is more energetic has a shorter wavelength, thus corresponding to smaller moment of inertia.

Therefore, at transition $I=1-I=0$ the moment of inertia is larger.