/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 The ground-state energy of a har... [FREE SOLUTION] | 91影视

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Chapter 40: Problem 37

The ground-state energy of a harmonic oscillator is 5.60 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 wavelength of the photon is 220 nm.

Step by step solution

01

Understand the Energy Levels

The energy levels of a harmonic oscillator in quantum mechanics are given by the formula \( E_n = \left( n + \frac{1}{2} \right) h u \), where \( n \) is the quantum number, \( h \) is Planck's constant, and \( u \) is the frequency of the oscillator. The ground-state energy \( E_0 \) is given as 5.60 eV.
02

Calculate Frequency from Ground-State Energy

Given the ground-state energy \( E_0 = 5.60 \text{ eV} \), we know that \( E_0 = \frac{1}{2} h u \). Solving for \( u \), we have \( u = \frac{2 \times 5.60}{h} \). Since there are constants involved, this will remain symbolic for now.
03

Calculate Energy for Levels n = 3 and n = 2

For the level \( n = 3 \), the energy \( E_3 = \left( 3 + \frac{1}{2} \right) h u = 3.5 h u \). For the level \( n = 2 \), the energy \( E_2 = \left( 2 + \frac{1}{2} \right) h u = 2.5 h u \).
04

Determine Energy Difference between Levels

The photon emitted corresponds to the energy difference between these levels: \( \Delta E = E_3 - E_2 = (3.5 h u - 2.5 h u) = h u \). Given in terms of the ground state energy, \( \Delta E = 5.60 \text{ eV} \).
05

Relate Energy to Wavelength

The energy of a photon is related to its wavelength \( \lambda \) by \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light. Solve for \( \lambda \) giving \( \lambda = \frac{hc}{E} \).
06

Calculate Wavelength

Using \( E = 5.60 \times 1.6 \times 10^{-19} \text{ J} \) (converting eV to J), \( h = 6.626 \times 10^{-34} \text{ J} \cdot\text{s} \), and \( c = 3.00 \times 10^8 \text{ m/s} \), plug the values into \( \lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{5.60 \times 1.6 \times 10^{-19}} \).
07

Final Calculation and Result

Perform the calculation to find \( \lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{8.96 \times 10^{-19}} \approx 2.20 \times 10^{-7} \text{ m} \), which converts to \( 220 \text{ nm} \).

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Key Concepts

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

Understanding Energy Levels in a Quantum Harmonic Oscillator
In quantum mechanics, the concept of energy levels is crucial, especially in systems like the quantum harmonic oscillator. An oscillator has discrete energy levels marked by a quantum number, denoted as \( n \). For a quantum harmonic oscillator, each energy level can be calculated using the formula \( E_n = \left(n + \frac{1}{2}\right) h u \). Here, \( h \) is Planck鈥檚 constant, and \( u \) represents the frequency of the oscillator.
  • The ground state corresponds to \( n = 0 \), where the energy is given by \( E_0 = \frac{1}{2} h u \).
  • The subsequent levels \( n = 1, 2, 3, \ldots \) add more energy to the system, with each level differing by a specific quantized amount.
This quantization is fundamental in quantum mechanics, informing us that energy is not continuous but comes in discrete amounts, which is a significant departure from classical mechanics.
Photon Emission During Energy Transitions
Photon emission occurs when an oscillator transitions between energy levels, releasing a photon as it drops to a lower energy state. Here鈥檚 what happens during such an event:
  • An atom or oscillator initially at a higher energy level, like \( n = 3 \), loses energy by emitting a photon.
  • The photon carries away the energy difference between this higher state and the lower state it transitions to, such as \( n = 2 \). The energy of this photon \( \Delta E \) is calculated as the difference between the two levels: \( \Delta E = E_3 - E_2 \).
  • This process is key in many applications, such as lasers and fluorescence.
Understanding this mechanism helps explain fundamental features of light and matter interaction in a quantized system.
Calculating Wavelength from Energy Differences
The wavelength of a photon emitted can be directly connected to the energy difference between quantum levels. This relationship is expressed with the formula: \( E = \frac{hc}{\lambda} \), where \( c \) is the speed of light. Solving for wavelength \( \lambda \), you get: \[ \lambda = \frac{hc}{E} \].
Calculation Steps:
  • Convert the photon energy from electronvolts (eV) to joules (J) for consistency with the SI units used in Planck鈥檚 constant and speed of light.
  • Insert values: \( h = 6.626 \times 10^{-34} \text{ Js} \), \( c = 3.00 \times 10^8 \text{ m/s} \), and the converted \( E \).
  • Compute the wavelength, which indicates how far the photon will travel in one cycle.
By calculating the wavelength, we can predict properties of the emitted photon, critical for practical applications like quantum computing and optical technologies.
Core Principles of Quantum Mechanics
Quantum mechanics is a branch of physics that deals with phenomena at very small scales, like atoms and subatomic particles. It fundamentally changes how we view energy, states, and interactions. Here are some of its underlying principles:
  • Quantization: Energy and matter exist in discrete quantities, not continuous flows. This means systems like harmonic oscillators have specific, quantized energy levels.
  • Wave-Particle Duality: Particles can exhibit properties of both waves and discrete particles, influencing how we predict their behavior, such as in photon emissions.
  • Uncertainty Principle: Some aspects of a system cannot be precisely known simultaneously, like position and momentum, impacting measurements.
These principles show how quantum mechanics breaks away from classical physics, providing a new framework for understanding microscopic phenomena.
Exploring Ground-State Energy of Oscillators
Ground-state energy refers to the lowest energy level that a quantum mechanical system, such as a harmonic oscillator, can have.
  • This ground state energy, represented as \( E_0 \), is never zero due to the Heisenberg uncertainty principle, which dictates that a system cannot have both a precise position and momentum simultaneously.
  • In our scenario, this is given as 5.60 eV, where the system is at its most stable and energy-efficient state without any external energy adding to it.
  • This energy serves as a baseline for calculating higher energy levels: each increase in the quantum number \( n \) adds additional quanta of energy \( hu \) to \( E_0 \).
Understanding ground-state energy is vital for predicting the behavior of quantum systems under minimal conditions, affecting everything from chemical reactions to fundamental physics research.

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Most popular questions from this chapter

A particle is described by a wave function \(\psi(x) = Ae^{-\alpha x^2}\), where A and \(\alpha\) are real, positive constants. If the value of \(\alpha\) is increased, what effect does this have on (a) the particle's uncertainty in position and (b) the particle's uncertainty in momentum? Explain your answers.

When an electron in a one-dimensional box makes a transition from the \(n\) = 1 energy level to the \(n\) = 2 level, it absorbs a photon of wavelength 426 nm. What is the wavelength of that photon when the electron undergoes a transition (a) from the \(n\) = 2 to the \(n\) = 3 energy level and (b) from the n = 1 to the \(n\) = 3 energy level? (c) What is the width \(L\) of the box?

When a hydrogen atom undergoes a transition from the \(n\) = 2 to the \(n\) = 1 level, a photon with \(\lambda\) = 122 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.)

(a) Find the lowest energy level for a particle in a box if the particle is a billiard ball (\(m\) = 0.20 kg) and the box has a width of 1.3 m, the size of a billiard table. (Assume that the billiard ball slides without friction rather than rolls; that is, ignore the \(rotational\) kinetic energy.) (b) Since the energy in part (a) is all kinetic, to what speed does this correspond? How much time would it take at this speed for the ball to move from one side of the table to the other? (c) What is the difference in energy between the \(n\) = 2 and \(n\) = 1 levels? (d) Are quantum-mechanical effects important for the game of billiards?

An electron in a one-dimensional box has ground-state energy 2.00 eV. What is the wavelength of the photon absorbed when the electron makes a transition to the second excited state?
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