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Chapter 40: Problem 42

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?

Short Answer

Expert verified
The photon's wavelength is approximately 111 nm.

Step by step solution

01

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, \( E_n = (n + \frac{1}{2}) \times 5.60 \).
02

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.
03

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.
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider the wave packet defined by $$\psi(x)=\int_{0}^{\infty} B(k) \cos k x d k$$ Let \(B(k)=e^{-\alpha^{2} k^{2}} .\) (a) The function \(B(k)\) has its maximum value at \(k=0 .\) Let \(k_{\mathrm{h}}\) be the value of \(k\) at which \(B(k)\) has fallen to half its maximum value, and define the width of \(B(k)\) as \(w_{k}=k_{\mathrm{h}}\) . In terms of \(\alpha,\) what is \(w_{k} ?\) (b) Use integral tables to evaluate the integral that gives \(\psi(x) .\) For what value of \(x\) is \(\psi(x)\) maximum? (c) Define the width of \(\psi(x)\) as \(w_{x}=x_{\mathrm{h}},\) where \(x_{\mathrm{h}}\) is the positive value of \(x\) at which \(\psi(x)\) has fallen to half its maximum value. Calculate \(w_{x}\) in terms of \(\alpha .\) (d) The momentum \(p\) is equal to \(h k / 2 \pi\) so the width of \(B\) in momentum is \(w_{p}=h w_{k} / 2 \pi .\) Calculate the product \(w_{p} w_{x}\) and compare to the Heisenberg uncertainty principle.

A harmonic oscillator absorbs a photon of wavelength \(8.65 \times 10^{-6} \mathrm{m}\) when it undergoes a transition from the ground state to the first excited state. What is the ground-state energy, in electron volts, of the oscillator?

Chemists use infrared absorption spectra to identify chemicals in a sample. In one sample, a chemist finds that light of wavelength 5.8\(\mu \mathrm{m}\) is absorbed when a molecule makes a transition from its ground harmonic oscillator level to its first excited level. (a) Find the energy of this transition. (b) If the molecule can be treated as a harmonic oscillator with mass \(5.6 \times 10^{-26} \mathrm{kg},\) find the force constant.

An electron is in a box of width \(3.0 \times 10^{-10} \mathrm{m} .\) What are the de Broglie wavelength and the magnitude of the momentum of the electron if it is in (a) the \(n=1\) level; (b) the \(n=2\) level; (c) the \(n=3\) level? In each case how does the wavelength compare to the width of the box?

When a hydrogen atom undergoes a transition from the \(n=2\) to the \(n=1\) level, a photon with \(\lambda=122 \mathrm{nm}\) is emitted. (a) If the atom is modeled as an electron in a one-dimensional box, what is the width of the box in order for the \(n=2\) to \(n=1\) transition to correspond to emission of a photon of this energy? (b) For a box with the width calculated in part (a), what is the ground-state energy? How does this correspond to the ground-state energy of a hydrogen atom? (c) Do you think a one-dimensional box is a good model for a hydrogen atom? Explain. (Hint: Compare the spacing between adjacent energy levels as a function of \(n . )\)
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