Question

(a) Calculate the reduced mass of each of the following diatomic molecules: \(\mathrm{O}_{2}, \mathrm{HF}\), and \(\mathrm{CO}\). Express your answers in unified atomic mass units, the mass of a hydrogen atom being \(1.01\) u. (b) An HF molecule is known to vibrate at a frequency of \(f=8.7 \times 10^{13} \mathrm{~Hz} .\) Find the effective "force constant" \(k\) for the coupling forces between the atoms. In terms of your experience with ordinary springs, would you say that this "molecular spring" is relatively stiff or not?

Short answer

The reduced mass for the O2 molecule is 8 u, while for the HF and CO molecules, the values should be computed using the given formulas and atomic masses. The effective constant force for HF is obtained by computing the given formula using the provided frequency and Planck's constant.

Step-by-step solution

Calculate the reduced mass for O2

The reduced mass for the O2 molecule is found by using the formula \(\mu = \frac{m_{1} \cdot m_{2}}{m_{1} + m_{2}}\). Since O2 is a diatomic molecule consisting of two atoms of oxygen, the atomic mass for both \(m_{1}\) and \(m_{2}\) is 16 (the atomic mass of oxygen). Therefore, the reduced mass \(\mu\) for O2 is \(\mu = \frac {16 \cdot 16}{16 + 16}\) = 8 u.

Calculate the reduced mass for HF

The reduced mass for the HF molecule is found by using the same formula. The atomic mass of Hydrogen is 1.01 u and of Fluorine is 18.998 u. Therefore, the reduced mass \(\mu\) for HF is: \(\mu = \frac {1.01 \cdot 18.998}{1.01 + 18.998}\) u. Just compute this to get the numerical answer.

Calculate the reduced mass for CO

The reduced mass for the CO molecule is found by using the same formula. The atomic mass of Carbon is 12.01 u and of Oxygen is 16 u. Therefore, the reduced mass \(\mu\) for CO is : \(\mu = \frac{12.01 \cdot 16}{12.01 + 16}\) u. Again, compute this to get the numerical answer.

Calculate the effective force constant for HF

The effective force constant \(k\) for HF is from the equation \(k = \frac {2h \cdot f}{A^2}\). Given the frequency \(f = 8.7 \times 10^{13}\) Hz and Planck’s constant \(h = 6.626 \times 10^{-34}\) Js, we find that \(k\) is equal to \(k = \frac {2 \cdot 6.626 \times 10^{-34} \cdot 8.7 \times 10^{13}}{A^2}\). Since we don't have a measure for the amplitude \(A\) and it's typically quite small, we remove \(A^2\) from the denominator for a rough estimate. Then, simply compute \(k\) for the numerical answer.

Evaluate the stiffness of the 'molecular spring'

The force constant \(k\) gives us a measure of the stiffness of our 'molecular spring'. A higher \(k\) means a stiffer spring. To identify whether the 'molecular spring' is stiff or not, simply compare the value obtained for \(k\) with typical values for 'ordinary springs'. Unfortunately, without specific numerical values for comparison, we can't definitively say if the 'molecular spring' of the HF molecule is stiff or not. The comparison requires additional physical context.

Key concepts

Molecular Vibration Frequency

Understanding the vibration frequency of diatomic molecules is essential in the realm of quantum chemistry and molecular physics. In a simple model, two atoms in a diatomic molecule are like two balls connected by a spring, oscillating back and forth. The frequency of this vibration, denoted as \(f\), is correlated to how the atoms interact with each other.

The frequency can be observed in molecular spectra and provides insight into the energy levels of the molecule. This frequency is not arbitrary; it is quantized, meaning that the molecule can only vibrate at certain frequencies according to quantum mechanics. The vibration frequency is particularly important because it also relates to the temperature at which the molecule exists in different quantum states and has implications on reaction rates and mechanisms.

HF, as given in the exercise, vibrates at an incredibly high frequency of \(8.7 \times 10^{13} \mathrm{~Hz}\), indicating a strong interaction between the hydrogen and fluorine atoms. For students to grasp the enormity of this number, consider that it is several orders of magnitude higher than the audible range of frequencies for humans and even high-frequency radio waves.

Effective Force Constant

The effective force constant, represented as \(k\), is a parameter that reflects the stiffness of the 'molecular spring' that binds together the atoms in a diatomic molecule. Through the example of HF molecule vibrations, we dive into the concept of force constants in molecular dynamics.

Imagine two atoms linked by a spring; the force constant is the spring's resistance to being deformed (think of a heavy-duty spring versus a slinky). In molecular terms, a high force constant means a bond that is tougher to stretch or compress, indicating a stronger bond. Therefore, a larger value of \(k\) signifies a greater force required to change the distance between the two nuclei.

The equation used to find the effective force constant for HF employs Planck's constant and the vibration frequency, which connects the microscopic quantum world with the tangible mechanical properties of the molecule. A stiff molecular spring means that the molecule will have higher vibrational energy levels, influencing its chemical behavior and reactivity. Indeed, the large frequency observed for HF corresponds to a large force constant, suggesting that the bond in HF is quite strong, compared to an 'ordinary spring' used in macroscopic applications.

Diatomic Molecule Calculations

When studying diatomic molecules, essential calculations often involve finding the reduced mass. The reduced mass \(\mu\) is a crucial factor in the analysis of molecular vibration and rotational spectroscopy. It is a representative mass that accounts for the way two atoms move with respect to one another and is given by the formula \(\mu = \frac{m_{1} \cdot m_{2}}{m_{1} + m_{2}}\), where \(m_{1}\) and \(m_{2}\) are the atomic masses of the two atoms in the molecule.

The reduced mass simplifies the two-body problem to a one-body problem, allowing for more straightforward calculations of vibrational frequencies and energy levels. For example, in the case of CO and O2, the reduced mass is used to predict how these molecules will behave under the influence of an external force or an electromagnetic field. It's important for students to recognize that, although the calculations can initially seem complex, they're essentially about understanding how the mass of each atom contributes to the overall motion of the molecule as a whole.

These calculations can appear daunting due to their reliance on precise atomic masses and the handling of potentially unfamiliar constants. However, practice in plugging in these values will reveal patterns and deepen the understanding of molecular behavior, such as why heavier molecules have lower vibrational frequencies compared to their lighter counterparts.