Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 80

A particle in a box \([0 ; L]\) is in the third excited state. What are its most probable positions?

Short Answer

Expert verified
The most probable positions for a particle in a box of length L in its third excited state are \(x = \frac{L}{8}, \frac{3L}{8}, \frac{5L}{8},\) and \(\frac{7L}{8}\).

Step by step solution

01

Identify the wave function for the third excited state

The wave function for a particle in a box in its nth state can be given by the formula:\[ \psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) \] Since we are looking for the third excited state, we will set n = 4 (third excited state corresponds to the fourth energy level, with n = 1 being the ground state). Therefore, we have:\[ \psi_4(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{4\pi x}{L}\right) \]
02

Calculate the probability density function

The probability density function is determined by taking the absolute square of the wave function: \[P(x) = |\psi_4(x)|^2 \] Therefore, we have:\[ P(x) = \left(\sqrt{\frac{2}{L}}\sin\left(\frac{4\pi x}{L}\right)\right)^2 = \frac{2}{L}\sin^2\left(\frac{4\pi x}{L}\right) \]
03

Find the maxima of the probability density function

To find the maxima of the probability density function, we'll first find the derivative of the function with respect to x:\[ \frac{dP(x)}{dx} = \frac{d}{dx} \left(\frac{2}{L}\sin^2\left(\frac{4\pi x}{L}\right)\right) \] Using the chain rule, we find:\[ \frac{dP(x)}{dx} = \frac{16\pi}{L^2}\sin\left(\frac{4\pi x}{L}\right)\cos\left(\frac{4\pi x}{L}\right) \] Now, we set the derivative equal to zero to find the critical points, taking into account that the particle is restricted to the interval [0, L]:\[ \frac{16\pi}{L^2}\sin\left(\frac{4\pi x}{L}\right)\cos\left(\frac{4\pi x}{L}\right) = 0 \]
04

Solve for the most probable positions

To solve for x, we can see that at certain points, either the sine or cosine terms within the derivative are equal to zero. Thus, we can write:\[ \sin\left(\frac{4\pi x}{L}\right) = 0 \textrm{ or } \cos\left(\frac{4\pi x}{L}\right) = 0 \] Upon solving the above equations, we find the following most probable positions for x:\[ x = \frac{L}{8}, \frac{3L}{8}, \frac{5L}{8}, \textrm{ and } \frac{7L}{8} \] These are the most probable positions for the particle in its third excited state.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

An electron is confined to a box of width \(0.25 \mathrm{nm}\). (a) Draw an energy-level diagram representing the first five states of the electron. (b) Calculate the wavelengths of the emitted photons when the electron makes transitions between the fourth and the second excited states, between the second excited state and the ground state, and between the third and the second excited states.

A simple model of a radioactive nuclear decay assumes that \(\alpha\) -particles are trapped inside a well of nuclear potential that walls are the barriers of a finite width \(2.0 \mathrm{fm}\) and height \(30.0 \mathrm{MeV}\). Find the tunneling probability across the potential barrier of the wall for \(\alpha\) -particles having kinetic energy (a) \(29.0 \mathrm{MeV}\) and \((\mathrm{b}) 20.0 \mathrm{MeV} .\) The mass of the \(\alpha\) -particle is \(m=6.64 \times 10^{-27} \mathrm{kg}\)

What is the meaning of the expression "expectation value?" Explain.

Find the expectation value of the kinetic energy for the particle in the state, \(\Psi(x, t)=A e^{i(k x-\omega t)} .\) What conclusion can you draw from your solution?

Can a quantum particle 'escape' from an infinite potential well like that in a box? Why? Why not?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Modern Physics

Read Explanation

Kinematics Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Famous Physicists

Read Explanation

Fundamentals of Physics

Read Explanation
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.

Sign-up for free