/*! 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 10 A particle moving in one dimensi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle moving in one dimension (the \(x\) -axis) is described by the wave function \(\psi(x)=\left\\{\begin{array}{ll}{A e^{-b x},} & {\text { for } x \geq 0} \\ {A e^{b x},} & {\text { for } x<0}\end{array}\right.\) where \(b=2.00 \mathrm{m}^{-1}, A>0,\) and the \(+x\) -axis points toward the right. (a) Determine \(A\) so that the wave function is normalized. (b) Sketch the graph of the wave function. (c) Find the probability of finding this particle in each of the following regions: (i) within 50.0 \(\mathrm{cm}\) of the origin, (ii) on the left side of the origin (can you first guess the answer by looking at the graph of the wave function? (iii) between \(x=0.500 \mathrm{m}\) and \(x=1.00 \mathrm{m} .\)

Short Answer

Expert verified
a) \(A = \sqrt{2} \, \mathrm{m}^{-\frac{1}{2}}\); b) Sketch shows exponential growth and decay; c) i) 0.432, ii) 0.5, iii) 0.017.

Step by step solution

01

Understand the Problem

We need to normalize the wave function \( \psi(x) \), graph it, and then find the probability of finding the particle in specified regions along the \( x \)-axis.
02

Normalization Condition

To normalize the wave function, the integral of \( |\psi(x)|^2 \) over all space must equal 1:\[ \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \]Split the integral into two parts for \( x < 0 \) and \( x \geq 0 \): \[ \int_{-\infty}^0 |A e^{b x}|^2 \, dx + \int_0^{\infty} |A e^{-b x}|^2 \, dx = 1 \]
03

Calculate Each Integral

For \( x < 0 \), calculate:\[ \int_{-\infty}^0 (A e^{b x})^2 \, dx = A^2 \int_{-\infty}^0 e^{2b x} \, dx = A^2 \left[ \frac{e^{2b x}}{2b} \right]_{-\infty}^0 = \frac{A^2}{2b} \]For \( x \geq 0 \), calculate:\[ \int_0^{\infty} (A e^{-b x})^2 \, dx = A^2 \int_0^{\infty} e^{-2b x} \, dx = A^2 \left[ -\frac{e^{-2b x}}{2b} \right]_0^{\infty} = \frac{A^2}{2b} \]
04

Solve for Normalization Constant \( A \)

Adding both parts from Step 3 gives:\[ \frac{A^2}{2b} + \frac{A^2}{2b} = 1 \]\[ \frac{A^2}{b} = 1 \]\[ A^2 = b \]\[ A = \sqrt{b} \]Since \( b = 2.00 \, \mathrm{m}^{-1} \), then \( A = \sqrt{2.00} = \sqrt{2} \, \mathrm{m}^{-\frac{1}{2}} \).
05

Sketch the Graph of the Wave Function

Plot the function \( \psi(x) \): it decays exponentially with \( e^{-2x} \) for \( x \geq 0 \) and grows with \( e^{2x} \) for \( x < 0 \). The function is symmetric in terms of its exponential decay characteristics about \( x = 0 \).
06

Probability (i) within 50.0 cm of the Origin

Convert \(50.0 \, \mathrm{cm} = 0.50 \, \mathrm{m}\).Calculate:\[ \int_{-0.50}^{0.50} |\psi(x)|^2 \, dx = \int_{-0.50}^{0} (A e^{2x})^2 \, dx + \int_{0}^{0.50} (A e^{-2x})^2 \, dx \] Evaluating these integrals:\[ \int_{-0.50}^{0} e^{4x} \, dx = \left[ \frac{e^{4x}}{4} \right]_{-0.50}^{0} = \frac{1 - e^{-2}}{4} \]\[ \int_{0}^{0.50} e^{-4x} \, dx = \left[ -\frac{e^{-4x}}{4} \right]_{0}^{0.50} = \frac{1 - e^{-2}}{4} \]Thus, total probability is \( \frac{1 - e^{-2}}{2} \approx 0.432 \).
07

Probability (ii) on the Left Side of the Origin

Calculate:\[ \int_{-\infty}^{0} |\psi(x)|^2 \, dx = \frac{A^2}{2b} = \frac{1}{2} \]This represents the probability of 0.5 because of the symmetric nature of the exponential function around the origin.
08

Probability (iii) between 0.500 m and 1.00 m

Calculate:\[ \int_{0.50}^{1.00} |\psi(x)|^2 \, dx = A^2 \int_{0.50}^{1.00} e^{-4x} \, dx \]Evaluating gives:\[ \left[ -\frac{e^{-4x}}{4} \right]_{0.50}^{1.00} = \frac{e^{-2} - e^{-4}}{4} \approx 0.017 \].

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 a fundamental theory in physics that describes nature at the smallest scales of energy levels of atoms and subatomic particles. It challenges our classical intuition with concepts like superposition and entanglement. In this context, the wave function, often denoted as \( \psi(x) \), is a central concept. This function provides a probabilistic description of a particle's quantum state as it exists in all possible areas. It is a complex-valued function, and its absolute square, \( |\psi(x)|^2 \), gives the probability density of a particle's position concerning the x-axis.
Thus, the wave function must be normalized, meaning the total probability of finding a particle anywhere on the x-axis must equal one. To achieve this, the integral of \( |\psi(x)|^2 \) over the whole space should sum up to one. These fundamental principles are vital as they transition concepts from classical probability to quantum probabilities, where certainty in position vanishes, replaced by likelihoods.
Probability Calculations
In the realm of quantum mechanics, probability calculations involve integrating the squared amplitude of the wave function over a specified region. This task tells us the likelihood of finding a particle within a given interval. For instance, if we have a particle described by a wave function \( \psi(x) \), to find the probability that it exists within a range, say from \( a \) to \( b \), we compute:
\[ P(a \leq x \leq b) = \int_a^b |\psi(x)|^2 \, dx \]The normalization condition guarantees that the total probability across all such ranges sums to one. For practical examples:
  • Integrating over a finite region gives the fraction of probability mass within that region.
  • Understanding these calculations helps in making predictions about particle positions and behaviors that can be verified experimentally.
By using these calculations, physicists can ascertain more about the possible location of particles, testing them against experimental results to validate current theories.
Exponential Functions
Exponential functions frequently appear in quantum mechanics, especially in the representation of wave functions. They inherently describe growth and decay processes, which makes them essential for modeling the behavior of particles in bound and unbound states.For a wave function that involves exponentials, like \( \psi(x) = A e^{-bx} \) for \( x \ge 0 \), we observe certain properties:
  • Exponential decay shows how probabilities diminish or decay from a certain point, aligning with how quantum particles are more likely near specific regions, like potential wells.
  • For \( x < 0 \), where \( \psi(x) = A e^{bx} \) indicates growth - although such a term is less common in physically observable states, it's used here to illustrate mathematical principles.
These functions, with their consistent rates of change, serve as natural solutions to differential equations in quantum scenarios, describing phenomena such as radioactive decay or electronic behavior in fields. When graphed, these curves provide a visual representation of probability distribution, essential in grasping quantum states' locality.

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

Compute \(|\Psi|^{2}\) for \(\Psi=\psi \sin \omega t,\) where \(\psi\) is time independent and \(\omega\) is a real constant. Is this a wave function for a stationary state? Why or why not?

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?

The penetration distance \(\eta\) in a finite potential well is the distance at which the wave function has decreased to 1\(/ e\) of the wave function at the classical turning point: $$\psi(x=L+\eta)=\frac{1}{e} \psi(L)$$ The penetration distance can be shown to be $$\eta=\frac{\hbar}{\sqrt{2 m\left(U_{0}-E\right)}}$$ The probability of finding the particle beyond the penetration distance is nearly zero. (a) Find \(\eta\) for an electron having a kinetic energy of 13 eV in a potential well with \(U_{0}=20 \mathrm{eV} .\) (b) Find \(\eta\) for a 20.0 -MeV proton trapped in a 30.0 -Me \(\mathrm{V}\) -deep potential well.

Consider a particle in a box with rigid walls at \(x=0\) and \(x=L .\) Let the particle be in the ground level. Calculate the probability \(|\psi|^{2} d x\) that the particle will be found in the interval \(x\) to \(x+d x\) for \((\) a \() x=L / 4 ;(\) b) \(x=L / 2 ;(\) c) \(x=3 L / 4\)

(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.

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.