Problem 1 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: Calculate the de Broglie wavelength

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

Step 3: 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.

Step 4: Calculate the wave number \( k \)

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

Step 5: 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 \( E \):\[E = \frac{(4.50 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} \approx 1.11 \times 10^{-17} \, \text{J}\]

Step 6: 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 \( \omega \):\[\omega = \frac{1.11 \times 10^{-17}}{1.05 \times 10^{-34}} \approx 1.06 \times 10^{17} \, \text{rad/s}\]

Step 7: 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.

Key Concepts

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.