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: Use spacing relation.
For rotational spectrum:
\[ \Delta \bar{\nu} = 2B \]
\[ 2B = 40.9 \text{ cm}^{-1} \]

Step 2: Find rotational constant.
\[ B = \frac{40.9}{2} = 20.45 \text{ cm}^{-1} \]
\[ B = 2045 \text{ m}^{-1} \]

Step 3: Use relation between B and moment of inertia.
\[ B = \frac{h}{8\pi^2 c I} \]
\[ I = \frac{h}{8\pi^2 cB} \]

Step 4: Substitute values.
\[ I = \frac{6.63 \times 10^{-34}}{8\pi^2 \times 3 \times 10^8 \times 2045} \]

Step 5: Calculate I.
\[ I = 1.369 \times 10^{-47} \text{ kg m}^2 \]

Step 6: Compare with given form.
\[ I = Y \times 10^{-47} \]
\[ Y = 1.369 \]

Step 7: Conclusion.
\[ \boxed{1.369} \]