Problem 42 - University Physics with Modern Physics

Question

The ground-state energy of a harmonic oscillator is 5.60 \(\mathrm{eV} .\) If the oscillator undergoes a transition from its \(n=3\) to \(n=2\) level by emitting a photon, what is the wavelength of the photon?

Step-by-Step Solution

Answer

The photon's wavelength is approximately 111 nm.

Step 1: Understanding the Energy Levels

The energy levels for a quantum harmonic oscillator are given by the formula \( E_n = \left(n + \frac{1}{2}\right)hu \), where \( n \) is the quantum number, \( h \) is Planck's constant, and \( u \) is the frequency. However, we know \( E_0 = 5.60 \mathrm{eV} \). So, \( E_n = (n + \frac{1}{2}) \times 5.60 \).

Step 2: Calculate Energy of n=3 Level

Substitute \( n = 3 \) into the energy level equation: \( E_3 = (3 + \frac{1}{2}) \times 5.60 \) eV. Simplifying gives \( E_3 = 4.5 \times 5.60 = 25.2 \) eV.

Step 3: Calculate Energy of n=2 Level

Substitute \( n = 2 \) into the energy level equation: \( E_2 = (2 + \frac{1}{2}) \times 5.60 \) eV. Simplifying gives \( E_2 = 2.5 \times 5.60 = 14.0 \) eV.

Step 4: Determine the Energy Difference

The photon energy emitted during the transition from \( n=3 \) to \( n=2 \) is the difference in these energy levels: \( \Delta E = E_3 - E_2 = 25.2 - 14.0 = 11.2 \) eV.

Step 5: Convert Energy to Wavelength

Use the photon energy-wavelength relation: \( \lambda = \frac{hc}{\Delta E} \), where \( h = 4.135667696 \times 10^{-15} \mathrm{eV \cdot s} \) and \( c = 3.00 \times 10^8 \mathrm{m/s} \). Thus, \( \lambda = \frac{4.135667696 \times 10^{-15} \times 3.00 \times 10^8}{11.2} \approx 1.11 \times 10^{-7} \) meters or 111 nm.

Key Concepts

Quantum Mechanics

Quantum mechanics is a fundamental theory in physics that provides a comprehensive framework for understanding the physical properties of nature at the smallest scales, particularly the behavior of atoms and subatomic particles. Unlike classical mechanics, quantum mechanics incorporates the concept of waves and particles coexisting, leading to unique phenomena like wave-particle duality and quantization of energy levels. This field of study is essential for explaining how particles behave at atomic and subatomic levels, and it revolutionized our understanding of phenomena such as photon emission and the behavior of harmonic oscillators.

Energy Levels

In quantum mechanics, energy levels refer to the discrete values of energy that a quantum system, such as an electron around an atom or a harmonic oscillator, may have. These levels are quantized, meaning particles can only exist at specific energy levels, and transitions between levels involve absorbing or emitting energy. For harmonic oscillators, the energy is quantized as \( E_n = \left(n + \frac{1}{2}\right)hu \), where \( n \) is the quantum number, representing the state of the system. The ground state is the lowest energy level, and excited states are higher energy levels. Transitions between levels are quantized as energy differences.

Photon Emission

Photon emission occurs when a quantum system, such as an atom or molecule, transitions from a higher energy state to a lower one, releasing energy in the form of a photon. This process is fundamental in atomic physics and is responsible for phenomena such as light emission from stars and lasers. The energy of the emitted photon corresponds to the difference in energy between the two quantum states. The formula \( \Delta E = E_3 - E_2 \) helps calculate the energy difference during a transition. In this context, for the harmonic oscillator example, when moving from \( n=3 \) to \( n=2 \), a photon with energy 11.2 eV is emitted.

Wavelength Calculation

Wavelength calculation involves finding the wavelength of a photon emitted or absorbed during a transition between energy levels. The relationship between energy and wavelength for a photon is given by \( \lambda = \frac{hc}{\Delta E} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \Delta E \) is the energy difference between the two states. This formula highlights the inverse relationship between energy and wavelength: higher energy photons have shorter wavelengths. In the example provided, the calculated energy of 11.2 eV corresponds to a wavelength of approximately 111 nm, illustrating the ultraviolet range of light.