Question

Find the moments of inertia \(I_{A}, I_{B}\) and \(I_{C}\) of \(^{79} \mathrm{Br}^{19} \mathrm{F}\) (bond length \(=176 \mathrm{pm} ;\) accurate masses: \\[ \left.^{79} \mathrm{Br}=78.92 ;^{19} \mathrm{F}=19.00\right) \\]

Short answer

The moments of inertia are \(I_A = I_B = I_C \approx 4.75 \times 10^{-39} \, \text{kg} \, \text{m}^2\).

Step-by-step solution

Convert Bond Length to Meters

The bond length is given in picometers (pm). Convert this bond length to meters by knowing that 1 pm = \(10^{-12}\) m. For a bond length of 176 pm, the conversion to meters is:\[176 \, \text{pm} = 176 \times 10^{-12} \, \text{m} = 1.76 \times 10^{-10} \, \text{m}\]

Calculate Reduced Mass

The reduced mass \(\mu\) is used in the calculation of moments of inertia and is given by:\[ \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \]where \(m_1\) is the accurate mass of \(^{79} \mathrm{Br}\) and \(m_2\) is the accurate mass of \(^{19} \mathrm{F}\). Plug in the values:\[ \mu = \frac{78.92 \cdot 19.00}{78.92 + 19.00} = \frac{1498.48}{97.92} \approx 15.30 \]

Calculate the Moment of Inertia \(I_A\)

The moment of inertia \(I_A\) is simply the product of the reduced mass and the square of the bond length:\[ I_A = \mu \cdot r^2 \]Substitute \(\mu = 15.30\) and \(r = 1.76 \times 10^{-10}\) m:\[ I_A = 15.30 \times (1.76 \times 10^{-10})^2 \approx 4.75 \times 10^{-39} \, \text{kg} \, \text{m}^2 \]

Assume Moments of Inertia for System (\(I_B\) and \(I_C\))

For a diatomic molecule like \(^{79}Br^{19}F\), the moments of inertia \(I_B\) and \(I_C\) relating to rotation perpendicular to the molecular axis are equal to \(I_A\). Thus:\[ I_B = I_A \approx 4.75 \times 10^{-39} \, \text{kg} \, \text{m}^2 \]\[ I_C = I_A \approx 4.75 \times 10^{-39} \, \text{kg} \, \text{m}^2 \]

Key concepts

Reduced Mass

When dealing with molecular physics, especially in calculating moments of inertia in systems like diatomic molecules, the concept of reduced mass is vital. Reduced mass

• Helps simplify the system by reducing two-body problems into a single equivalent mass.
• Is used in the formula: \[ \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \] where \(m_1\) and \(m_2\) are the masses of the two atoms in question.

For the diatomic molecule \(^{79}\text{Br}^{19}\text{F}\), \(m_1\) is the mass of the bromine atom, and \(m_2\) is the mass of the fluorine atom. By substituting these values into the formula, the reduced mass \(\mu\) becomes approximately 15.30."

Diatomic Molecule

A diatomic molecule, as the name implies, consists of two atoms bonded together. They are fundamental in understanding molecular dynamics and rotational motion concepts.

• Diatomic molecules like \(^{79}\text{Br}^{19}\text{F}\) have a rotational inertia that allows us to study rotational motion straightforwardly.
• Such molecules rotate around an axis perpendicular to the bond, allowing an easier calculation of moment of inertia.

For a molecule consisting of two atoms, the rotation primarily revolves around a center of mass axis, which simplifies calculations of physical properties such as moments of inertia. In cases of homonuclear diatomic molecules, both atoms are the same, which further simplifies their study.

Bond Length Conversion

When working with molecular bonds at the atomic level, it is essential to accurately convert bond length into standard units. Given data is often in picometers (pm), thus converting it into meters is crucial because of the universality and ease of use in calculations.

• The conversion factor is: 1 pm = \(10^{-12}\) meters.
• This makes it easier to use these measurements in physics equations, like calculating moments of inertia.

For instance, converting 176 pm to meters entails multiplying by \(10^{-12}\), resulting in \(1.76 \times 10^{-10}\) meters. This conversion standardizes measurements and ensures calculations align correctly with other standard international units.