Question

Calculate the reduced mass, the moment of inertia, the angular momentum, and the energy in the \(J=1\) rotational level for \(\mathrm{H}_{2}\), which has a bond length of \(74.14 \mathrm{pm}\).

Short answer

The reduced mass (\(\mu\)) of the hydrogen molecule is approximately \(8.34 \times 10^{-28} \, \text{kg}\), the moment of inertia (\(I\)) is about \(3.64 \times 10^{-47} \, \text{kg.m}^{2}\), the angular momentum (\(L\)) in the \(J=1\) rotational level is roughly \(1.47 \times 10^{-34} \, \text{kg.m}^{2}\text{s}^{-1}\), and the energy (\(E_1\)) in the \(J=1\) rotational level is approximately \(8.6 \times 10^{-24} \, \text{J}\).

Step-by-step solution

Calculate the reduced mass

To find the reduced mass (\(\mu\)) of the hydrogen molecule (\(\mathrm{H}_{2}\)), we can use the formula for reduced mass: \[\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}\] where \(m_1\) and \(m_2\) are the masses of the individual hydrogen atoms. Since both atoms in a hydrogen molecule \(\mathrm{H}_{2}\) are the same, both \(m_1\) and \(m_2\) will be the mass of a single hydrogen atom, which is approximately \(1 \, \text{amu} = 1.67 \times 10^{-27} \, \text{kg}\). Substitute the values into the formula: \[\mu = \frac{(1.67 \times 10^{-27})(1.67 \times 10^{-27})}{(1.67 \times 10^{-27})+(1.67 \times 10^{-27})}\] After calculating, we get: \[\mu \approx 8.34 \times 10^{-28}\, \text{kg}\]

Calculate the moment of inertia

Next, we need to calculate the moment of inertia (\(I\)) of the molecule. The formula for the moment of inertia of a diatomic molecule is: \[I = \mu \cdot R^2\] where \(R\) is the bond length, which is given as \(74.14 \, \mathrm{pm}\). Convert bond length to meters: \[R = 74.14 \, \mathrm{pm} \times 10^{-12} \, \text{m/pm} \approx 7.414 \times 10^{-11} \, \text{m}\] Now, substitute the values of \(\mu\) and \(R\) into the formula: \[I \approx (8.34 \times 10^{-28})(7.414 \times 10^{-11})^2\] After calculating, we get: \[I \approx 3.64 \times 10^{-47}\, \text{kg.m}^{2}\]

Calculate the angular momentum

To find the angular momentum (\(L\)) of the hydrogen molecule in the \(J=1\) rotational level, we can use the formula: \[L = \sqrt{J(J+1)} \cdot \hbar\] where \(J\) is the rotational quantum number and \(\hbar\) is the reduced Planck's constant \(\left(\hbar = 1.05 \times 10^{-34}\, \text{Js}\right)\). For this problem, \(J=1\). Substitute the values into the formula: \[L \approx \sqrt{1(1+1)} \cdot 1.05 \times 10^{-34}\] After calculating, we get: \[L \approx 1.47 \times 10^{-34}\, \text{kg.m}^{2}\text{s}^{-1}\]

Calculate the energy in the J=1 rotational level

Finally, we need to calculate the energy (\(E\)) in the \(J=1\) rotational level. The formula for the energy of a diatomic molecule in a given rotational level is: \[E_J = \frac{1}{2} \cdot \frac{L^2}{I}\] Substitute the values of \(L\) and \(I\) we already found into the formula: \[E_1 \approx \frac{1}{2} \cdot \frac{(1.47 \times 10^{-34})^2}{3.64 \times 10^{-47}}\] After calculating, we get: \[E_1 \approx 8.6 \times 10^{-24}\, \text{J}\] To summarize our results: - Reduced mass (\(\mu\)): \(\approx 8.34 \times 10^{-28} \, \text{kg}\) - Moment of inertia (\(I\)): \(\approx 3.64 \times 10^{-47} \, \text{kg.m}^{2}\) - Angular momentum (\(L\)): \(\approx 1.47 \times 10^{-34} \, \text{kg.m}^{2}\text{s}^{-1}\) (for \(J=1\)) - Energy in the \(J=1\) rotational level (\(E_1\)): \(\approx 8.6 \times 10^{-24} \, \text{J}\)

Key concepts

Reduced Mass

Understanding the concept of reduced mass is crucial when dealing with rotational spectroscopy. Reduced mass simplifies the calculation of rotational dynamics by transforming a two-body problem into a single-body model.
In a diatomic molecule like \( \mathrm{H}_{2} \), the reduced mass \( \mu \) is calculated using the formula:

• \( \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \)

For \( \mathrm{H}_{2} \), both \( m_1 \) and \( m_2 \) are equivalent, representing the mass of a hydrogen atom. This value is approximately \( 1.67 \times 10^{-27} \, \text{kg} \).
This results in a reduced mass \( \mu \approx 8.34 \times 10^{-28} \, \text{kg} \). The significance of reduced mass goes beyond simple mathematics. It allows the coupling of a molecule's rotational constants with its vibrational and electronic states, providing a fuller picture of molecular behavior during rotation.

Moment of Inertia

The moment of inertia \( I \) plays a pivotal role in determining how a molecule rotates. For diatomic molecules, it's necessary to understand how mass is distributed relative to the axis of rotation.
Calculated via the formula:

• \( I = \mu \cdot R^2 \)

where \( R \) is the bond length, such as \( 74.14 \, \text{pm} \) for \( \mathrm{H}_{2} \). It must be converted to meters to assist in computations, resulting in \( R \approx 7.414 \times 10^{-11} \, \text{m} \).
The calculated \( I \approx 3.64 \times 10^{-47} \, \text{kg.m}^{2} \) reveals the rotational capabilities of the molecule.
The lower the moment of inertia, the more agile the rotation. Hence, small, light diatomic molecules like hydrogen exhibit brisk rotational behavior.

Angular Momentum

Angular momentum \( L \) reflects the rotational motion of a molecule, closely related to its rotational quantum state \( J \). In rotational spectroscopy, \( L \) is calculated using:

• \( L = \sqrt{J(J+1)} \cdot \hbar \)

where \( \hbar \) is the reduced Planck's constant. For \( J=1 \) in \( \mathrm{H}_{2} \), the angular momentum \( L \approx 1.47 \times 10^{-34} \, \text{kg.m}^{2}\text{s}^{-1} \).
This measurement is crucial as it provides insights into the energy states of molecules and their transitions during spectroscopy. Recognizing these values aids in decoding molecular spectra and predicting reaction outcomes.

Rotational Energy Levels

Rotational energy levels are quantized, meaning a molecule can only occupy specific energy states. For a given rotational quantum number \( J \), the energy \( E_J \) is determined by:

• \( E_J = \frac{1}{2} \cdot \frac{L^2}{I} \)

For \( \mathrm{H}_{2} \) at \( J=1 \), \( E_1 \approx 8.6 \times 10^{-24} \, \text{J} \). These values help scientists understand the distribution and magnitude of rotational transitions.
The molecular spectrum features specified peaks correlating to energy levels. By analyzing these, we can study molecular interactions and predict how environmental changes might affect molecular motion. This understanding is fundamental to fields ranging from quantum chemistry to material science.