/*! 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 20 Use the Heisenberg uncertainty p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 20

Use the Heisenberg uncertainty principle to estimate the minimum velocity of a proton or neutron in a \({ }^{208} \mathrm{~Pb}\) nucleus, which has a diameter of about \(13 \mathrm{fm}\) ( \(1 \mathrm{fm}=10^{-15} \mathrm{~m}\) ). Assume that the speed is nonrelativistic, and then check at the end whether this assumption was warranted.(answer check available at lightandmatter.com)

Short Answer

Expert verified
The minimum velocity is approximately \( 4.87 \times 10^{6} \, \text{m/s} \), which is nonrelativistic.

Step by step solution

01

Understand the Uncertainty Principle

The Heisenberg uncertainty principle states that the uncertainty in position \( \Delta x \) and the uncertainty in momentum \( \Delta p \) are related by \( \Delta x \Delta p \geq \frac{h}{4\pi} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ Js} \).
02

Calculate Uncertainty in Position

The diameter of the nucleus is \(13\, \text{fm}\), so the uncertainty in position \( \Delta x \), considering the neutron/proton to be confined within the nucleus, can be taken as around half of this. Hence, \( \Delta x \approx 6.5\, \text{fm} = 6.5 \times 10^{-15}\, \text{m} \).
03

Use the Principle to Find Uncertainty in Momentum

Using the uncertainty principle, we substitute for \( \Delta x \) and solve for \( \Delta p \): \[\Delta p \geq \frac{h}{4\pi \Delta x} = \frac{6.626 \times 10^{-34}}{4\pi \times 6.5 \times 10^{-15}}\approx 8.13 \times 10^{-21} \, \text{kg} \cdot \text{m/s}.\]
04

Compute Minimum Velocity Using Momentum

Momentum is defined as \( p = mv \), so the uncertainty in velocity \( \Delta v \) can be found using \( \Delta p = m \Delta v \), where the mass of a proton/neutron \( m \approx 1.67 \times 10^{-27} \text{kg} \). Solving for \( \Delta v \):\[\Delta v = \frac{\Delta p}{m} = \frac{8.13 \times 10^{-21}}{1.67 \times 10^{-27}} \approx 4.87 \times 10^{6} \, \text{m/s}.\]
05

Check Nonrelativistic Assumption

The assumption is that the velocity is nonrelativistic, meaning \( v \ll c \) (the speed of light, \( 3 \times 10^{8} \, \text{m/s} \)). Here, the computed velocity \( 4.87 \times 10^{6} \, \text{m/s} \) is significantly less than \( c \), confirming that the nonrelativistic assumption is valid.

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 91影视!

Key Concepts

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

Nonrelativistic Speed
Nonrelativistic speed refers to velocities that are much less than the speed of light, symbolized as \( c \), which is approximately \( 3 \times 10^8 \, \text{m/s} \). When the speed of an object is nonrelativistic, it means that the effects of special relativity, like time dilation and length contraction, are negligible.
One way to ensure that a velocity is nonrelativistic is to ensure it is significantly less than \( c \). For instance, in the exercise, the calculated velocity of the proton or neutron, \( 4.87 \times 10^6 \, \text{m/s} \), is far below the speed of light.
This confirms the nonrelativistic assumption, making classical or Newtonian physics sufficient to describe the motion without needing adjustments for relativistic effects.
Using nonrelativistic speeds simplifies calculations, especially in contexts like nuclear physics, where particles often move fast but not close to light-speed.
Momentum Uncertainty
Momentum uncertainty is a critical part of the Heisenberg Uncertainty Principle, which states there's a limit to how precisely we can know both the position and the momentum of a particle.
The momentum uncertainty \( \Delta p \) relates to how accurately we can measure a particle's momentum at a particular time.
According to the uncertainty principle, given by the formula:
  • \( \Delta x \Delta p \geq \frac{h}{4\pi} \)
where \( \Delta x \) is the position uncertainty, \( \Delta p \) is the momentum uncertainty, and \( h \) is Planck's constant.
To find the momentum uncertainty, one inversely considers the precision in locating the particle's position.
In the exercise, the particle is confined within a nucleus with a diameter of 13 femtometers. Thus, the position uncertainty \( \Delta x \) is half that diameter. Using these values, we derive \( \Delta p \), ensuring our measurements respect the natural limits imposed by quantum mechanics.
Planck's Constant
Planck's constant \( h \) is a fundamental constant in physics, with a value of approximately \( 6.626 \times 10^{-34} \, \text{Js} \).
It plays a crucial role in the uncertainty principle, showing the scale at which quantum effects become significant. The constant relates to the size of quantum effects, being the smallest action that can occur in the natural world.
Planck's constant connects the energy of photons to their frequency through the equation:
  • \( E = hf \)
Here, \( f \) is the frequency of the photon.
This constant is essential in the Heisenberg uncertainty equation as well, limiting the precision of simultaneous measurements of position and momentum. In the exercise, Planck's constant is used to determine the product of uncertainties \( \Delta x \Delta p \).
Thus, Planck's constant ensures our understanding of microscopic phenomena remains consistent with observed quantum behavior.

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

We are given some atoms of a certain radioactive isotope, with half-life \(t_{1 / 2}\). We pick one atom at random, and observe it for one half-life, starting at time zero. If it decays during that one-half-life period, we record the time \(t\) at which the decay occurred. If it doesn't, we reset our clock to zero and keep trying until we get an atom that cooperates. The final result is a time \(0 \leq t \leq t_{1 / 2}\), with a distribution that looks like the usual exponential decay curve, but with its tail chopped off. (a) Find the distribution \(D(t)\), with the proper normalization.(answer check available at lightandmatter.com) (b) Find the average value of \(t\). (answer check available at lightandmatter.com) (c) Interpreting your result from part b, how does it compare with \(t_{1 / 2} / 2\) ? Does this make sense? Explain.

Before the quantum theory, experimentalists noted that in many cases, they would find three lines in the spectrum of the same atom that satisfied the following mysterious rule: \(1 / \lambda_{1}=1 / \lambda_{2}+1 / \lambda_{3}\). Explain why this would occur. Do not use reasoning that only works for hydrogen - - - such combinations occur in the spectra of all elements. [Hint: Restate the equation in terms of the energies of photons.]

A blindfolded person fires a gun at a circular target of radius \(b\), and is allowed to continue firing until a shot actually hits it. Any part of the target is equally likely to get hit. We measure the random distance \(r\) from the center of the circle to where the bullet went in. (a) Show that the probability distribution of \(r\) must be of the form \(D(r)=k r\), where \(k\) is some constant. (Of course we have \(D(r)=0\) for \(r>b .)\) (b) Determine \(k\) by requiring \(D\) to be properly normalized.(answer check available at lightandmatter.com) (c) Find the average value of \(r\).(answer check available at lightandmatter.com) (d) Interpreting your result from part c, how does it compare with \(b / 2 ?\) Does this make sense? Explain.

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)

A photon collides with an electron and rebounds from the collision at 180 degrees, i.e., going back along the path on which it came. The rebounding photon has a different energy, and therefore a different frequency and wavelength. Show that, based on conservation of energy and momentum, the difference between the photon's initial and final wavelengths must be \(2 h / m c\), where \(m\) is the mass of the electron. The experimental verification of this type of "pool-ball" behavior by Arthur Compton in 1923 was taken as definitive proof of the particle nature of light. Note that we're not making any nonrelativistic approximations. To keep the algebra simple, you should use natural units - - in fact, it's a good idea to use even-more-naturalthan- natural units, in which we have not just \(c=1\) but also \(h=1\), and \(m=1\) for the mass of the electron. You'll also probably want to use the relativistic relationship \(E^{2}-p^{2}=m^{2}\), which becomes \(E^{2}-p^{2}=1\) for the energy and momentum of the electron in these units.
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Rotational Dynamics

Read Explanation

Electromagnetism

Read Explanation

Measurements

Read Explanation

Space Physics

Read Explanation

Physics of Motion

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.