Question

Calculate the frequency and wavelength of the radiation absorbed when a quantum harmonic oscillator with a frequency of \(3.15 \times 10^{13} \mathrm{s}^{-1}\) makes a transition from the \(n=2\) to the \(n=3\) state.

Short answer

The frequency of the absorbed photon is \(\nu = \frac{\hbar (2\pi f)(1)}{h} = \frac{ (1.054 \times 10^{-34} Js)(2\pi \cdot 3.15 \times 10^{13} s^{-1})(1)}{6.626 \times 10^{-34} Js} \approx 1.58 \times 10^{13} s^{-1}\). The wavelength of the absorbed photon is \(\lambda = \frac{c}{\frac{\hbar (2\pi f)(1)}{h}} = \frac{ 3.00 \times 10^8 m/s}{1.58 \times 10^{13} s^{-1}} \approx 1.90 \times 10^{-5} m\).

Step-by-step solution

Calculate the energy difference between the two states

In a quantum harmonic oscillator, the energy levels are given by: \(E_{n} = \hbar \omega (n+\frac{1}{2})\) where \(\hbar\) is the reduced Planck constant, \(\omega\) is the angular frequency of the oscillator, and \(n\) is the quantum number of the energy level. The angular frequency \(\omega\) can be calculated from the frequency \(f\) given in the problem: \(\omega = 2\pi f\) The energy difference between the initial state (\(n=2\)) and the final state (\(n=3\)) is: \(\Delta E = E_{3} - E_{2} = \hbar \omega (3+ \frac{1}{2}) - \hbar\omega (2+ \frac{1}{2})\)

Calculate the frequency of the absorbed photon

According to the conservation of energy, the energy of the absorbed photon must equal the energy difference between the two states: \(\Delta E = h\nu\) where \(h\) is the Planck constant and \(\nu\) is the frequency of the absorbed photon. Thus, the frequency of the absorbed photon is: \(\nu = \frac{\Delta E}{h}\) We can substitute the expression for \(\Delta E\) from Step 1 and the relation between \(\omega\) and \(f\) to find the frequency of the absorbed photon: \(\nu = \frac{\hbar \omega (1)}{h} = \frac{\hbar (2\pi f)(1)}{h}\)

Calculate the wavelength of the absorbed photon

The wavelength \(\lambda\) of the absorbed photon can be obtained using the speed of light \(c\) and the frequency \(\nu\): \(\lambda = \frac{c}{\nu}\) Substitute the expression for \(\nu\) from Step 2 to find the wavelength of the absorbed photon: \(\lambda = \frac{c}{\frac{\hbar (2\pi f)(1)}{h}}\) Now, we can plug in the given values for the frequency of the oscillator (\(3.15 \times 10^{13} s^{-1}\)), the speed of light (\(3.00 \times 10^8 m/s\)), and the constants \(\hbar\) and \(h\) to calculate the frequency and wavelength of the absorbed photon. Note: The reduced Planck constant \(\hbar\) is approximately \(1.054 \times 10^{-34} Js\), and the Planck constant \(h\) is approximately \(6.626 \times 10^{-34} Js\).

Key concepts

Energy Levels in a Quantum Harmonic Oscillator

In the realm of quantum mechanics, the concept of energy levels is integral to understanding phenomena like photon absorption. Within a quantum harmonic oscillator, energy levels are quantized, meaning they can only have specific values. Unlike classical systems where energy can vary continuously, this quantum system follows the formula: \[E_{n} = \hbar \omega \left(n + \frac{1}{2}\right)\] Here,

• \(E_{n}\) is the energy of the state with quantum number \(n\).
• \(\hbar\) is the reduced Planck constant, a fundamental constant equal to about \(1.054 \times 10^{-34} \text{J s}\).
• \(\omega\) is the angular frequency of the oscillator and is related to its frequency by \(\omega = 2\pi f\).

Each added quantum number \(n\) represents a higher energy state. When a transition occurs between these states, such as from \(n=2\) to \(n=3\), energy is either absorbed or emitted. The energy difference \(\Delta E\) between these states is crucial for photon absorption and is calculated by subtracting the energies of these levels: \[\Delta E = \hbar \omega \left(3 + \frac{1}{2}\right) - \hbar \omega \left(2 + \frac{1}{2}\right)\] This difference tells us how much energy a photon must have to facilitate this transition.

Photon Absorption in Quantum Systems

Photon absorption occurs when a photon of light transfers its energy to a quantum particle, such as an electron in an atomic or molecular system. For a quantum harmonic oscillator, this process is closely tied to transitions between quantized energy levels. When a photon is absorbed, its energy must exactly match the difference \(\Delta E\) between these two energy levels. This is described by the equation: \[\Delta E = hu\] where

• \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \text{J s}\).
• \(u\) is the frequency of the absorbed photon.

Therefore, by knowing the energy difference \(\Delta E\), we can find the frequency \(u\) of the absorbed photon by rearranging the formula: \[u = \frac{\Delta E}{h}\] This relationship underscores the quantum nature of light and matter interactions, revolving around exact energy matches between photons and atomic or molecular transitions.

Frequency and Wavelength Calculation

Calculating the frequency and wavelength of radiation involves understanding the relationship between these properties of electromagnetic waves. Once the frequency \(u\) is determined from photon absorption, the wavelength \(\lambda\) can be calculated using the speed of light \(c\), as given by the equation: \[\lambda = \frac{c}{u}\] where

• \(\lambda\) is the wavelength of the photon.
• \(c\) is the speed of light, approximately \(3.00 \times 10^8 \text{m/s}\).

This formula shows that the wavelength is inversely proportional to the frequency: as frequency increases, the wavelength decreases, and vice versa.For instance, substituting the expression \(u = \frac{\hbar \omega (1)}{h}\) obtained earlier into this equation allows us to calculate the wavelength of the absorbed photon: \[\lambda = \frac{c}{\frac{\hbar (2\pi f)(1)}{h}}\] By inserting given constants and values specific to the problem, such as the frequency of the oscillator and the values of \(\hbar\) and \(h\), one can derive precise measures of the photon's frequency and wavelength.