/*! 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 45 Suppose that an electron, in one... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that an electron, in one dimension, is confined to a certain region of space so that its wavefunction is given by$$\Psi=\left\\{\begin{array}{ll} 0 & \text { if } x<0 \\\A \sin (2 \pi x / L) & \text { if } 0 \leq x \leq L \\\ 0 & \text { if } x>L\end{array}\right.$$ Determine the constant \(A\) from normalization.(answer check available at lightandmatter.com)

Short Answer

Expert verified
The constant \( A \) is \( \sqrt{\frac{2}{L}} \).

Step by step solution

01

Understand Normalization

The wavefunction must be normalized, meaning that the probability of finding the electron anywhere in space is 1. For the wavefunction \( \Psi \), this condition is expressed mathematically as: \( \int_{-\infty}^{\infty} |\Psi|^2 \, dx = 1 \). Since \( \Psi \) is zero outside the interval \( [0, L] \), the integral simplifies to \( \int_{0}^{L} |\Psi|^2 \, dx = 1 \).
02

Substitute the Wavefunction

Inside the region \( [0, L] \), we have \( \Psi(x) = A \sin (2 \pi x / L) \). Substituting this into the normalization condition, we get: \( \int_{0}^{L} |A \sin(2 \pi x / L)|^2 \, dx = 1 \). The absolute value square is simply \( A^2 \sin^2(2 \pi x / L) \).
03

Simplify the Integral

The integral becomes: \( A^2 \int_{0}^{L} \sin^2(2 \pi x / L) \, dx = 1 \). Use the trigonometric identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \) to simplify the integral: \( \int_{0}^{L} \sin^2(2 \pi x / L) \, dx = \frac{1}{2} \int_{0}^{L} (1 - \cos(4 \pi x / L)) \, dx \).
04

Evaluate the Integral

Evaluate the integral \( \frac{1}{2} \int_{0}^{L} (1 - \cos(4 \pi x / L)) \, dx \). This splits into two integrals: \( \frac{1}{2} \left( \int_{0}^{L} 1 \, dx - \int_{0}^{L} \cos(4 \pi x / L) \, dx \right) \). The first integral is simply \( L \), and for the second, \( \int_{0}^{L} \cos(4 \pi x / L) \, dx = \left[ \frac{L}{4 \pi} \sin(4 \pi x / L) \right]_{0}^{L} = 0 \).
05

Solve for Constant \(A\)

Substitute the evaluated integrals back: \( A^2 \frac{1}{2} (L - 0) = 1 \) simplifies to \( A^2 \frac{L}{2} = 1 \). Solving for \( A \), we find \( A^2 = \frac{2}{L} \), giving \( A = \sqrt{\frac{2}{L}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior and interaction of particles at a microscopic scale, such as electrons, photons, and atoms. It differs from classical physics in several fundamental ways. One of its key concepts is that particles exhibit both wave-like and particle-like properties, known as wave-particle duality.
In quantum mechanics:
  • The state of a particle is described by a mathematical function called a wavefunction.
  • Observables, like position or momentum, do not have definite values until measured.
  • The act of measurement affects the system being measured, leading to quantum uncertainty.
These ideas revolutionized our understanding of the physical world, providing insights into atomic and subatomic processes. Quantum mechanics is essential for explaining a wide range of phenomena, including the behavior of semiconductors in electronics and the fundamental interactions within atoms.
Wavefunction
A wavefunction, often denoted as \( \Psi \), is a central concept in quantum mechanics. It is a complex-valued function that provides a complete description of a quantum system. However, the precise physical meaning of the wavefunction is subject to interpretation, with the most common interpretation being the Copenhagen interpretation.
The wavefunction has several key features:
  • It can be a function of position and time: \( \Psi(x, t) \).
  • The probability of finding a particle at a particular position is given by the square of the wavefunction's magnitude: \( |\Psi(x)|^2 \).
  • Wavefunctions must be normalized to ensure that the total probability is 1.
In our exercise, the wavefunction for an electron is given by a piecewise function which includes a trigonometric term \( A \sin(2 \pi x / L) \). This function describes an electron confined within a specific region of space.
Normalization Integral
Normalization is a crucial process in quantum mechanics to ensure that the total probability of finding a particle within all of space is 1. For a wavefunction \( \Psi \), this is mathematically represented by the integral:
\[ \int_{-\infty}^{\infty} |\Psi|^2 \, dx = 1 \]
However, if the wavefunction is non-zero only within a specific interval, the integral simplifies to:
\[ \int_{a}^{b} |\Psi|^2 \, dx = 1 \]
In the example given, the wavefunction is non-zero only between \(0\) and \(L\), allowing us to focus on this range for the normalization integral.
It is from this normalization condition that we find the constant \(A\), ensuring the wavefunction appropriately accounts for the probability of finding the electron within this confined region of space.
Trigonometric Identity Simplification
Simplifying wavefunctions often involves using trigonometric identities to make calculations easier. One useful identity is:
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
This identity helps in transforming the integral of \( \sin^2 \) functions into terms that are easier to evaluate. In the context of the exercise, using this identity converts the integral:
\[ \int_{0}^{L} \sin^2(2 \pi x / L) \, dx \]
into
\[ \frac{1}{2} \left( \int_{0}^{L} 1 \, dx - \int_{0}^{L} \cos(4 \pi x / L) \, dx \right) \]
This simplification reveals that the second integral becomes zero due to the periodic nature of the cosine function over its defined interval. By recognizing and applying these identities, we can solve complex integrals more efficiently when dealing with quantum mechanical problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the following, \(x\) and \(y\) are variables, while \(u\) and \(v\) are constants. Compute (a) \(\partial(u x \ln (v y)) / \partial x\), (b) \(\partial(u x \ln (v y)) / \partial y\). (answer check available at lightandmatter.com)

Show that a wavefunction of the form \(\Psi=e^{b y} \sin a x\) is a possible solution of the Schrödinger equation in two dimensions, with a constant potential. Can we tell whether it would apply to a classically allowed region, or a classically forbidden one?

Devise a method for testing experimentally the hypothesis that a gambler's chance of winning at craps is independent of her previous record of wins and losses. If you don't invoke the definition of statistical independence, then you haven't proposed a test.

Assume that the kinetic energy of an electron the \(n=1\) state of a hydrogen atom is on the same order of magnitude as the absolute value of its total energy, and estimate a typical speed at which it would be moving. (It cannot really have a single, definite speed, because its kinetic and interaction energy trade off at different distances from the proton, but this is just a rough estimate of a typical speed.) Based on this speed, were we justified in assuming that the electron could be described nonrelativistically?

In the photoelectric effect, electrons are observed with virtually no time delay \((\sim 10 \mathrm{~ns})\), even when the light source is very weak. (A weak light source does however only produce a small number of ejected electrons.) The purpose of this problem is to show that the lack of a significant time delay contradicted the classical wave theory of light, so throughout this problem you should put yourself in the shoes of a classical physicist and pretend you don't know about photons at all. At that time, it was thought that the electron might have a radius on the order of \(10^{-15} \mathrm{~m}\). (Recent experiments have shown that if the electron has any finite size at all, it is far smaller.) (a) Estimate the power that would be soaked up by a single electron in a beam of light with an intensity of \(1 \mathrm{~mW} / \mathrm{m}^{2}\).(answer check available at lightandmatter.com) (b) The energy, \(E_{s}\), required for the electron to escape through the surface of the cathode is on the order of \(10^{-19} \mathrm{~J}\). Find how long it would take the electron to absorb this amount of energy, and explain why your result constitutes strong evidence that there is something wrong with the classical theory.(answer check available at lightandmatter.com)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.