Question

When a hypothetical diatomic molecule having atoms 0.8860 nm apart undergoes a rotational transition from the \(l\) = 2 state to the next lower state, it gives up a photon having energy 8.841 \(\times\) 10\(^-$$^4\) eV. When the molecule undergoes a vibrational transition from one energy state to the next lower energy state, it gives up 0.2560 eV. Find the force constant of this molecule.

Short answer

The force constant is found by relating vibrational frequency to energy and mass; precise values depend on specific atomic mass.

Step-by-step solution

Understand the Problem

We need to determine the force constant of a diatomic molecule based on given rotational and vibrational energy differences. Specifically, we are given the bond length, the energy difference for a rotational transition, and the energy difference for a vibrational transition.

Calculate Rotational Inertia

The rotational energy difference \( \Delta E_r \) is related to the change in rotational states. Using \( \Delta E_r = \frac{\hbar^2}{2I}(l'(l'+1) - l(l+1)) \), where \( I = \mu r^2 \) and \( \mu = \frac{m_1 m_2}{m_1 + m_2} \). But since \( I \) is unknown, we'll set up an expression considering the known energy difference of \( 8.841 \times 10^{-4} \, \text{eV} \) for \( l = 2 \) to \( l = 1 \).

Calculate Reduced Mass

The bond length \( r \) is given as \( 0.8860 \, \text{nm} \), so the reduced mass \( \mu \) needs the assumption of atomic masses. Common diatomic elements have similar mass properties, so using hydrogen molecules as an assumption can give \( \mu = \frac{m_H}{2} = 0.5m_H \). However, exact mass is needed for accurate calculation.

Calculate the Vibrational Frequency

The vibrational energy difference is given as \( 0.2560 \, \text{eV} \). From quantum mechanics, the energy of a vibrational state change is \( \Delta E_v = h u (v+1/2) - h u (v-1/2) = h u \), thus \( u = \frac{0.2560}{h} \), where \( h \) is Planck's constant.

Calculate Force Constant

Using the vibrational frequency, the force constant \( k \) is calculated from \( u = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}} \). Rearrange to solve for \( k \): \[k = (2 \pi u)^2 \mu\]\. Substitute the values for \( u \) and assumed \( \mu \) to find \( k \).

Calculation with Assumed Values

Assume typical molecular mass values for diatomic hydrogen for practical calculation. Convert bond length to meters and energy values to joules for consistency in calculation.

Key concepts

Rotational Transitions

In molecular physics, rotational transitions occur when a molecule changes its rotational energy state. These transitions involve the rotation of the entire molecule around its center of mass. The energy levels for rotational states of a diatomic molecule are quantized and can be described using quantum numbers. The transition from one rotational level to another involves absorbing or emitting a photon whose energy corresponds to the difference between these levels.
For example, in the problem, we have a diatomic molecule transitioning from the rotational state with quantum number \(l = 2\) to \(l = 1\). The energy released in this process, given as \(8.841 \times 10^{-4} \text{ eV}\), corresponds to the energy of the photon emitted. Calculating these differences requires understanding quantized rotational energies, where the general formula for the energy of a rotational state \(l\) is \(E_l = \frac{\hbar^2}{2I}l(l+1)\), with \(\hbar\) being the reduced Planck’s constant and \(I\) the moment of inertia of the molecule.

Vibrational Transitions

Vibrational transitions in molecules occur when the molecule changes its vibrational energy state, which is associated with the periodic motion of atoms within the molecule. Unlike rotational transitions, these are linked to the stretching and compressing of bonds within the molecule.
The vibrational energy levels of a molecule are also quantized. When a molecule transitions between vibrational levels, it may either absorb or release a photon with energy corresponding to the difference between these vibrational levels. In the described exercise, the energy difference for such a transition is given as \(0.2560 \text{ eV}\).
The energy associated with these transitions is often much larger than that for rotational transitions due to the higher energy involved in altering the bond lengths, thereby providing an insight into the molecular force constant and bond strength.

Force Constant

The force constant \(k\) of a molecule is a measure of the strength of the bond between the atoms in a molecule. This constant comes into play when analyzing vibrational transitions, as it relates to how stiff or flexible a molecular bond is. It can be derived from the vibrational frequency of the molecule.
In the exercise, we calculated the vibrational frequency \(u\) using \(u = \frac{0.2560}{h}\), where \(h\) is Planck’s constant. The relationship between the force constant and vibrational frequency is given by the formula:\[k = (2 \pi u)^2 \mu\],where \(\mu\) is the reduced mass of the molecule. A higher force constant indicates a stronger bond, which requires more energy to be compressed or elongated.

Reduced Mass

Reduced mass \(\mu\) is a crucial concept in molecular physics, especially for analyzing both rotational and vibrational spectra of diatomic molecules. The reduced mass simplifies the two-body problem into a one-body problem, allowing easier mathematical handling.
It is defined as:\[\mu = \frac{m_1 m_2}{m_1 + m_2}\],where \(m_1\) and \(m_2\) are the masses of the two atoms in the molecule.
In the problem at hand, having assumed equal atomic masses, the reduced mass simplifies the computation of rotational inertia and vibrational frequency. Correctly estimating \(\mu\) is essential since it directly influences the calculation of both the moment of inertia for rotational transitions and the force constant for vibrational transitions. This highlights its role in accurately modeling molecular behaviors.