Solving for HF Bond Length from Microwave Spectrum Data

This problem requires calculating the bond length of Hydrogen Fluoride (HF) using the provided microwave spectrum data. The separation between spectral lines in microwave spectroscopy is directly related to the rotational constant of the molecule.

1. Determine the Rotational Constant

The microwave spectrum of gaseous HF shows lines separated by $41.11$ cm$^{-1}$. This separation corresponds to twice the rotational constant ($\tilde{B}$) when expressed in wavenumbers (cm$^{-1}$).

• Separation $= 2\tilde{B} = 41.11$ cm$^{-1}$
• Rotational Constant, $\tilde{B} = \frac{41.11 \text{ cm}^{-1}}{2} = 20.555$ cm$^{-1}$

2. Calculate the Reduced Mass ($\mu$)

The reduced mass of a diatomic molecule is calculated using the atomic masses of its constituent atoms.

• Atomic mass of H ($m_H$) = $1.008$ amu
• Atomic mass of F ($m_F$) = $18.998$ amu
• Conversion factor: $1$ amu = $1.661 \times 10^{-27}$ kg
• Reduced mass, $\mu = \frac{m_H \times m_F}{m_H + m_F}$
• $\mu = \frac{1.008 \text{ amu} \times 18.998 \text{ amu}}{1.008 \text{ amu} + 18.998 \text{ amu}} = \frac{19.149984}{20.006}$ amu $\approx 0.95722$ amu
• Convert $\mu$ to kg: $\mu = 0.95722 \text{ amu} \times 1.661 \times 10^{-27} \text{ kg/amu} \approx 1.5897 \times 10^{-27}$ kg

3. Calculate the Moment of Inertia ($I$)

The moment of inertia is related to the rotational constant ($\tilde{B}$) by the formula $\tilde{B} = \frac{h}{8\pi^2 c I}$, where $h$ is Planck's constant, $c$ is the speed of light, and $I$ is the moment of inertia. We rearrange this to solve for $I$. Ensure consistent units (SI units are used here).

• $h = 6.626 \times 10^{-34}$ J s
• $c = 2.998 \times 10^8$ m s$^{-1}$
• $\tilde{B}$ must be in m$^{-1}$: $20.555 \text{ cm}^{-1} = 20.555 \times 100 \text{ m}^{-1} = 2055.5$ m$^{-1}$
• $I = \frac{h}{8\pi^2 c \tilde{B}_{m^{-1}}}$
• $I = \frac{6.626 \times 10^{-34} \text{ J s}}{8 \times (\pi)^2 \times (2.998 \times 10^8 \text{ m s}^{-1}) \times (2055.5 \text{ m}^{-1})}$
• $I \approx \frac{6.626 \times 10^{-34}}{4.8525 \times 10^{12}} \text{ kg m}^2 \approx 1.3655 \times 10^{-46}$ kg m$^2$

4. Determine the Bond Length ($r$)

The moment of inertia for a diatomic molecule is given by $I = \mu r^2$. We can now solve for the bond length $r$.

• $r^2 = \frac{I}{\mu}$
• $r^2 = \frac{1.3655 \times 10^{-46} \text{ kg m}^2}{1.5897 \times 10^{-27} \text{ kg}} \approx 0.85895 \times 10^{-19} \text{ m}^2$
• To get an even exponent for easier square root calculation, rewrite as: $r^2 = 85.895 \times 10^{-20}$ m$^2$
• $r = \sqrt{85.895 \times 10^{-20} \text{ m}^2} = \sqrt{85.895} \times 10^{-10}$ m
• $r \approx 9.268 \times 10^{-10}$ m
• Convert to Ångströms (Å), where $1$ Å $= 10^{-10}$ m: $r \approx 0.9268$ Å

Rounding the bond length to two decimal places gives $0.93$ Å. This value lies within the provided range of $0.92$ to $0.94$ Å.