Chapter 42: Problem 13
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
Step by step solution
Understand the Problem
Calculate Rotational Inertia
Calculate Reduced Mass
Calculate the Vibrational Frequency
Calculate Force Constant
Calculation with Assumed Values
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rotational Transitions
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
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
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
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.