Question

An oxygen molecule has a vibrational mode that behaves approximately like a simple harmonic oscillator with frequency \(2.99 \cdot 10^{14} \mathrm{rad} / \mathrm{s} .\) Calculate the energy of the ground state and the first two excited states.

Short answer

Question: Calculate the energy for the ground state (n = 0), the first excited state (n = 1), and the second excited state (n = 2) of a simple harmonic oscillator with a given angular frequency ω = 2.99 × 10^14 rad/s. Answer: The energy for the ground state (n = 0) is 1.57 × 10^(-19) J, the energy for the first excited state (n = 1) is 4.7 × 10^(-19) J, and the energy for the second excited state (n = 2) is 7.82 × 10^(-19) J.

Step-by-step solution

Calculate the energy for the ground state (n = 0)

Plug in the values of \(n = 0\), \(\hbar\) and \(\omega\) into the energy equation: \(E_0 = (0+\frac{1}{2})\hbar\omega = \frac{1}{2}\hbar\omega\) Now, insert the given values of \(\hbar = 1.05 \cdot 10^{-34} \mathrm{J\cdot s}\) and \(\omega = 2.99 \cdot 10^{14} \mathrm{rad/s}\) into the equation, and calculate the energy for the ground state: \(E_0 = \frac{1}{2} (1.05 \cdot 10^{-34} \mathrm{J\cdot s}) (2.99 \cdot 10^{14} \mathrm{rad/s})\) \(E_0 = 1.57 \cdot 10^{-19}\;\mathrm{J}\)

Calculate the energy for the first excited state (n = 1)

Plug in the values of \(n = 1\), \(\hbar\), and \(\omega\) into the energy equation: \(E_1 = (1 + \frac{1}{2})\hbar\omega = \frac{3}{2}\hbar\omega\) Now, insert the given values of \(\hbar\) and \(\omega\) into the equation, and calculate the energy for the first excited state: \(E_1 = \frac{3}{2} (1.05 \cdot 10^{-34} \mathrm{J\cdot s}) (2.99 \cdot 10^{14} \mathrm{rad/s})\) \(E_1 = 4.7 \cdot 10^{-19}\;\mathrm{J}\)

Calculate the energy for the second excited state (n = 2)

Plug in the values of \(n = 2\), \(\hbar\), and \(\omega\) into the energy equation: \(E_2 = (2 + \frac{1}{2})\hbar\omega = \frac{5}{2}\hbar\omega\) Now, insert the given values of \(\hbar\), and \(\omega\) into the equation, and calculate the energy for the second excited state: \(E_2 =\frac{5}{2}(1.05\cdot10^{-34}\mathrm{J\cdot s })(2.99\cdot10^{14}\mathrm{rad/s})\) \(E_2 = 7.82 \cdot 10^{-19}\;\mathrm{J}\) In conclusion, the energy for the ground state (n = 0) is \(1.57 \cdot 10^{-19}\;\mathrm{J}\), the energy for the first excited state (n = 1) is \(4.7 \cdot 10^{-19}\;\mathrm{J}\), and the energy for the second excited state (n = 2) is \(7.82 \cdot 10^{-19}\;\mathrm{J}\).

Key concepts

Energy Quantization

In the quantum world, one of the most pivotal concepts is that of energy quantization. This principle suggests that rather than being continuous, energy occurs in discrete packets called quanta. The oxygen molecule vibrational mode, approximated as a quantum harmonic oscillator, clearly demonstrates energy quantization through its distinct energy levels.

When discussing a quantum harmonic oscillator, each energy level is defined by the equation \(E_n = (n + \frac{1}{2})\hbar\omega\), where \(n\) is the quantum number corresponding to different states of vibration, \(\hbar\) is the reduced Planck's constant, and \(\omega\) is the angular frequency of the oscillator. The ground state (\(n = 0\)) has the lowest possible energy but is not zero, reflecting the zero-point energy that arises from the uncertainty principle in quantum mechanics. As \(n\) increases, we observe higher energy levels which correspond to the excited states of the oscillator. The exercise shows a stepwise calculation for \(n = 0\), \(n = 1\), and \(n = 2\), each resulting in quantized energy values specific to those levels.

Vibrational Modes

Vibrational modes refer to the specific patterns in which the components of a molecule move relative to each other. In the case of the oxygen molecule being likened to a harmonic oscillator, its vibrational mode represents the rhythmic movement back and forth from an equilibrium position.

These movements are not arbitrary but follow well-defined modes, each with an associated frequency. Think of it as a guitar string that vibrates at certain frequencies to create distinct musical notes. For a diatomic molecule like oxygen, vibrational modes are relatively straightforward, having one fundamental mode which occurs along the axis connecting the two atoms. Every increase in the vibrational quantum number, \(n\), represents an increment in the vibrational energy, leading to higher vibrational states that still maintain the characteristic nature of simple harmonic motion.

Simple Harmonic Motion

At the heart of understanding vibrational modes for molecules and quantum oscillators is the concept of simple harmonic motion (SHM). SHM is the physics term for what happens when an object oscillates symmetrically around an equilibrium position, like a child's swing swinging back and forth, or the pendulum of a grandfather clock.

In simple harmonic motion, the restoring force that brings the object back to its equilibrium position is directly proportional to the displacement but in the opposite direction. Mathematically, it's described by Hooke's Law, which parallels the quantum oscillator equations, emphasizing the relationship between the oscillation frequency and the energy of the system. The frequency given in the exercise \(\omega = 2.99 \cdot 10^{14} \mathrm{rad/s}\) represents this ideal oscillatory motion, and when plugged into the quantum harmonic oscillator's energy equation, it serves to indicate how energy levels are quantized even under such a classical concept.