Question

The rotational spectrum of HF has equally spaced lines that are 40.9 cm$^{-1}$ apartThe moment of inertia of HF is $Y \times 10^{-47}$ kg m$^2$The value of Y is _ _ _. (rounded off to three decimal places)

Hint:

For a rigid rotor, adjacent rotational spectral lines are separated by $2B$, and $B = \frac{h}{8\pi^2 cI}$

Answer 1

Correct Answer: 1.369

Step 1: Line spacing.
Rotational lines are spaced by $2B$, so $B = 40.9/2 = 20.45$ per cm.

Step 2: Relation to moment of inertia.
\[ I = \frac{h}{8\pi^2 c B} \]

Step 3: Evaluate.
Putting in the constants gives $I \approx 1.369\times10^{-47}$ kg m$^2$, so Y is 1.369.

Step 4: Answer.
\[ \boxed{1.369} \]