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An electron is moving as a free particle in the \(-x\) -direction with momentum that has magnitude \(4.50 \times 10^{-24} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) What is the one-dimensional time-dependent wave function of the electron?

Short Answer

Expert verified
\( \psi(x,t) = A e^{i(4.27 \times 10^{10} x - 1.06 \times 10^{17} t)} \)

Step by step solution

01

Write down the de Broglie wavelength formula

The de Broglie wavelength of a particle is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \, \text{Js} \) and \( p \) is momentum. Let's calculate the wavelength of the electron.
02

Calculate the de Broglie wavelength

Given the momentum \( p = 4.50 \times 10^{-24} \, \text{kg} \cdot \text{m/s} \). Substituting into the formula:\[\lambda = \frac{6.626 \times 10^{-34}}{4.50 \times 10^{-24}} \approx 1.47 \times 10^{-10} \, \text{m}\]
03

Identify the time-dependent wave function formula

The one-dimensional time-dependent wave function \( \psi(x,t) \) for a free particle is given by:\[\psi(x, t) = A e^{i(kx - \omega t)}\]where \( k = \frac{2\pi}{\lambda} \) is the wave number and \( \omega = \frac{E}{\hbar} \) is the angular frequency.
04

Calculate the wave number \( k \)

Substitute \( \lambda = 1.47 \times 10^{-10} \, \text{m} \) into the equation for \( k \):\[k = \frac{2\pi}{1.47 \times 10^{-10}} \approx 4.27 \times 10^{10} \, \text{m}^{-1}\]
05

Calculate the energy \( E \) of the electron

The energy \( E \) can be calculated using \( E = \frac{p^2}{2m} \), where \( m \) is the mass of the electron \( 9.11 \times 10^{-31} \, \text{kg} \). First, find \( E \):\[E = \frac{(4.50 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} \approx 1.11 \times 10^{-17} \, \text{J}\]
06

Calculate the angular frequency \( \omega \)

The angular frequency is \( \omega = \frac{E}{\hbar} \), where \( \hbar = \frac{h}{2\pi} \). First, calculate \( \hbar \): \[\hbar = \frac{6.626 \times 10^{-34}}{2\pi} \approx 1.05 \times 10^{-34} \, \text{Js}\]Then, find \( \omega \):\[\omega = \frac{1.11 \times 10^{-17}}{1.05 \times 10^{-34}} \approx 1.06 \times 10^{17} \, \text{rad/s}\]
07

Write the time-dependent wave function

Substitute the values of \( k \) and \( \omega \) into the wave function formula. The complete time-dependent wave function is:\[\psi(x, t) = A e^{i(4.27 \times 10^{10} \cdot x - 1.06 \times 10^{17} \cdot t)}\]Here, \( A \) is the amplitude, which can be left as a constant as it's not specified.

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Key Concepts

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

de Broglie wavelength
The concept of the de Broglie wavelength is central to understanding the wave-particle duality in quantum mechanics. According to de Broglie's hypothesis, every moving particle or object has an associated wavelength, which signifies its wave-like nature. This wavelength is given by the formula:\[ \lambda = \frac{h}{p} \]Here, \( \lambda \) is the de Broglie wavelength, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \), and \( p \) is the momentum of the particle.
For small-scale particles like electrons, this wavelength is significant. It explains phenomena that cannot be accounted for by classical mechanics, such as electron diffraction.
  • The smaller the momentum, the larger the wavelength, highlighting more pronounced wave characteristics.
  • This concept is essential because it bridges classical and quantum physics.
In our exercise scenario, the electron has a momentum of \( 4.50 \times 10^{-24} \, \text{kg} \cdot \text{m/s} \). Substituting into the formula, we find the de Broglie wavelength to be approximately \( 1.47 \times 10^{-10} \, \text{m} \), showcasing the wave-like nature of the electron.
Wave Function
The wave function is a fundamental concept in quantum mechanics, representing the quantum state of a particle or system. It is typically denoted by the Greek letter \( \psi \), and provides a probability amplitude for the position, velocity, and other properties of a particle.
In one-dimensional space, the time-dependent wave function for a free particle can be expressed as:\[ \psi(x, t) = A e^{i(kx - \omega t)} \]Here, \( A \) refers to the amplitude, which gives information about the probability density, \( k \) is the wave number related to the wavelength, and \( \omega \) is the angular frequency.
  • The wave number \( k \) is calculated using \( k = \frac{2\pi}{\lambda} \), indicating how many complete wave cycles exist within a unit distance.
  • \( \omega \) determines how the wave function changes with time, as can be seen from the equation.
This function conveys all the possible outcomes, and their likelihoods, associated with the quantum state of the particle. Despite its purely mathematical nature, the wave function is invaluable for predicting physical behavior on the quantum scale.
Angular Frequency
Angular frequency, represented by \( \omega \), reflects how quickly a particle's wave function oscillates over time. It is a key parameter in understanding wave-like behavior in the quantum realm. The angular frequency is defined as:\[ \omega = \frac{E}{\hbar} \]where \( E \) is the energy of the particle, and \( \hbar \) is the reduced Planck's constant \( \left( \frac{6.626 \times 10^{-34}}{2\pi} \right) \approx 1.05 \times 10^{-34} \, \text{Js} \).
In context, for an electron given the energy calculated from its momentum, \( E = 1.11 \times 10^{-17} \, \text{J} \), the angular frequency is computed to be approximately \( 1.06 \times 10^{17} \, \text{rad/s} \).
  • Angular frequency is crucial in the formulation of the wave function, dictating how a system evolves over time.
  • It helps in understanding interference patterns and the oscillation phenomena.
This metric enhances the ability to describe quantum states dynamically, providing insights into how energy manifests in wave properties.

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Most popular questions from this chapter

An electron is bound in a square well with a depth equal to six times the ground-level energy \(E_{1-\mathrm{LDW}}\) of an infinite well of the same width. The longest-wavelength photon that is absorbed by the electron has a wavelength of 400.0 \(\mathrm{nm} .\) Determine the width of the well.

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 \(n . )\)

Normalization of the Wave Function. Consider a particle moving in one dimension, which we shall call the \(x\) -axis. (a) What does it mean for the wave function of this particle to be normalized? (b) Is the wave function \(\psi(x)=e^{a x},\) where \(a\) is a positive real number, normalized? Could this be a valid wave function? (c) If the particle described by the wave function \(\psi(x)=A e^{b x},\) where \(A\) and \(b\) are positive real numbers, is confined to the range \(x \geq 0\) , determine \(A\) (including its units) so that the wave function is normalized.

(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 \(\mathrm{nm}\) . (b) The electron makes a transition from the \(n=1\) to \(n=4\) level by absorbing a photon. Calculate the wave-length of this photon.

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?

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