Question

An experimental measurement of the energy levels of a hydrogen molecule, \(\mathrm{H}_{2}\), shows that the energy levels are evenly spaced and separated by about \(9 \cdot 10^{-20} \mathrm{~J}\). A reasonable model of one of the hydrogen atoms would then seem to be that of a hydrogen atom in a simple harmonic oscillator potential. Assuming that the hydrogen atom is attached by a spring with a spring constant \(k\) to the molecule, what is the spring constant \(k\) ?

Short answer

Answer: The spring constant for the oscillating hydrogen atom is approximately \(1.21 \times 10^{-11} \mathrm{N/m}\).

Step-by-step solution

Formula for the energy levels of a simple harmonic oscillator

From quantum mechanics, we know that the energy levels of a simple harmonic oscillator are given by the formula: \(E_n = (n + \frac{1}{2}) \hbar \omega\) Here, \(E_n\) is the energy of the \(n\)-th level, \(\hbar\) is the reduced Planck constant (approximately \(1.054 \times 10^{-34} \mathrm{J\cdot s}\)), and \(\omega\) is the angular frequency.

Determine the energy level difference

According to the given information, the energy levels are evenly spaced and separated by \(9 \times 10^{-20} \mathrm{J}\). Let's use this information to write the energy level difference for two consecutive levels: \(\Delta E = E_{n+1} - E_n = \hbar \omega\) By substituting the given value, we get: \(9 \times 10^{-20} \mathrm{J} = \hbar \omega\) Now, we can solve for the angular frequency, \(\omega\): \(\omega = \frac{9 \times 10^{-20} \mathrm{J}}{\hbar} = \frac{9 \times 10^{-20} \mathrm{J}}{1.054 \times 10^{-34} \mathrm{J\cdot s}} = 8.53 \times 10^{14} \mathrm{s^{-1}}\)

Express the angular frequency in terms of spring constant and mass

The angular frequency can be expressed in terms of the mass (\(m\)) of a hydrogen atom and the spring constant (\(k\)) using the following formula: \(\omega = \sqrt{\frac{k}{m}}\) Here, \(m\) is the mass of a hydrogen atom (approximately \(1.67 \times 10^{-27} \mathrm{kg}\)). Since we found the angular frequency in Step 2, we can now solve for the spring constant, \(k\).

Calculate the spring constant

By rearranging the formula for angular frequency and substituting the known values, we can calculate the spring constant: \(k = m \omega^2 = (1.67 \times 10^{-27} \mathrm{kg})(8.53 \times 10^{14} \mathrm{s^{-1}})^2 = 1.21 \times 10^{-11} \mathrm{N/m}\) The spring constant for the hydrogen atom in the hydrogen molecule is approximately \(1.21 \times 10^{-11} \mathrm{N/m}\).

Key concepts

Simple Harmonic Oscillator

A simple harmonic oscillator is a fundamental concept in physics that depicts a system where an object is subject to a restoring force proportional to its displacement from an equilibrium position. This could be visualized as a mass attached to a spring, which, when displaced, experiences a force pushing it back towards its starting position. In quantum mechanics, a quantum harmonic oscillator follows similar behavior, with the position of a particle, such as an atom, oscillating about a fixed point.

In the context of our problem, we can think of a hydrogen atom in a molecule behaving like a simple harmonic oscillator when it vibrates. The evenly spaced energy levels indicated in the exercise are a hallmark of the quantized nature of such systems. Instead of having a continuum of energies like in classical systems, a quantum harmonic oscillator has discretely spaced energy levels represented by the equation \(E_n = (n + \frac{1}{2}) \hbar \omega\), where \(n\) denotes the quantum number of the state, \(\hbar\) is the reduced Planck’s constant, and \(\omega\) is the quantum angular frequency of the oscillator.

Energy Levels of Hydrogen

In quantum mechanics, the energy levels of the hydrogen atom are of particular interest because they reveal the quantized nature of atomic energies. It’s significant to note that our exercise deals with the energy levels of a hydrogen molecule \(\mathrm{H}_{2}\), not just a single atom, and the simple harmonic oscillator model is used to describe one of the hydrogen atoms within the molecule.

The energy levels of a hydrogen atom are determined by the electron's possible states around the nucleus – each state has a specific energy level, and electrons can transition between these levels by absorbing or releasing energy. In the context of our exercise, the spacing of \(9 \cdot 10^{-20} \mathrm{~J}\) between the energy levels is an abstraction based on behavior similar to that of electrons in a hydrogen atom, rather than precise electronic transitions. This gives us a simplified model for analyzing the vibrational motion within the molecule as a whole.

Spring Constant Calculation

The spring constant, denoted as \(k\), is a measure of the stiffness of a spring. In the context of our exercise, the spring represents the force that binds the hydrogen atom to the rest of the molecule, resisting its movement. The stiffer this 'spring', the more force is required to displace the atom, leading to a higher value of \(k\).

To calculate the spring constant, you usually need two pieces of information: the force applied and the displacement it caused. Alternatively, in a quantum system like our hydrogen molecule, you can determine \(k\) using the quantum angular frequency \(\omega\) (found in the exercise with mass \(m\)) through the relationship \(\omega = \sqrt{\frac{k}{m}}\). In the exercise, after finding the angular frequency \(\omega\), we rearrange the formula to solve for \(k\) and calculate its value, as shown in the solution steps. The result indicates how 'stiff' the force is that's keeping the atom in vibrational motion.

Quantum Angular Frequency

Quantum angular frequency, symbolized by \(\omega\), is an important parameter in quantum mechanics as it relates the energy of a system to its vibrational or rotational motion. Specifically, for a quantum harmonic oscillator, \(\omega\) determines the spacing of the energy levels, and consequently, it directly factors into the energy of each quantized state using the relationship \(E_n = (n + \frac{1}{2}) \hbar \omega\).

In the exercise, upon identifying the energy difference \(\Delta E\) between consecutive levels, we use it to determine \(\omega\). This provides a bridge from the macroscopic measurements down to the microscopic properties of the system. Knowing \(\omega\), we gain insight into the motion of the hydrogen atom as part of the molecule, which, in this case, oscillates in a quantized manner similar to the motion of a weight on a spring, but on the subatomic scale.