/*! 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 1 What is the de Broglie wavelengt... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 1

What is the de Broglie wavelength of a basketball of mass \(0.50 \mathrm{kg}\) when it is moving at \(10 \mathrm{m} / \mathrm{s} ?\) Why don't we see diffraction effects when a basketball passes through the circular aperture of the hoop?

Short Answer

Expert verified
Answer: The diffraction effects become noticeable when the size of the aperture (i.e., the hoop) is comparable to the de Broglie wavelength of the particle. In this case, the de Broglie wavelength of the basketball is \(1.33 \times 10^{-34}\,\mathrm{m}\), while the diameter of a standard basketball hoop is \(0.46\,\mathrm{m}\). The wavelength of the basketball is many orders of magnitude smaller than the diameter of the hoop, which means the diffraction effects are negligible and not noticeable when the basketball passes through the hoop.

Step by step solution

01

Understanding de Broglie's wavelength formula

The de Broglie wavelength is determined by the following formula: $$ \lambda = \frac{h}{mv} $$ where \(\lambda\) is the de Broglie wavelength, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\)), \(m\) is the mass of the particle, and \(v\) is the velocity of the particle.
02

Calculating the wavelength of the basketball

We are given the mass \(m = 0.50\,\mathrm{kg}\) and the velocity \(v = 10\,\mathrm{m/s}\) of the basketball. We now use the de Broglie's wavelength formula to calculate the wavelength : $$ \lambda = \frac{h}{mv} $$ $$ \lambda = \frac{6.626 \times 10^{-34}\, \mathrm{J} \cdot \mathrm{s}}{(0.50\, \mathrm{kg})(10\, \mathrm{m/s})} $$ $$ \lambda \approx 1.33 \times 10^{-34}\, \mathrm{m} $$
03

Understanding why diffraction effects aren't seen with the basketball

The diffraction effects become noticeable when the size of the aperture (i.e., the hoop) is comparable to the wavelength of the particle. The diameter of a standard basketball hoop is \(0.46\,\mathrm{m}\), while the de Broglie wavelength of the basketball is \(1.33 \times 10^{-34}\,\mathrm{m}\). We can see that the wavelength is many orders of magnitude smaller than the diameter of the hoop. Consequently, the diffraction effects are negligible and not noticeable when the basketball passes through the hoop.

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!

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 length \(1.0 \mathrm{nm} .\) What is the magnitude of its momentum in the \(n=4\) state?
An electron in an atom has an angular momentum quantum number of \(2 .\) (a) What is the magnitude of the angular momentum of this electron in terms of \(\hbar ?\) (b) What are the possible values for the \(z\) -components of this electron's angular momentum? (c) Draw a diagram showing possible orientations of the angular momentum vector \(\overrightarrow{\mathbf{L}}\) relative to the z-axis. Indicate the angles with respect to the z-axis.
An electron is confined to a one-dimensional box of length \(L\) (a) Sketch the wave function for the third excited state. (b) What is the energy of the third excited state? (c) The potential energy can't really be infinite outside of the box. Suppose that \(U(x)=+U_{0}\) outside the box, where \(U_{0}\) is large but finite. Sketch the wave function for the third excited state of the electron in the finite box. (d) Is the energy of the third excited state for the finite box less than, greater than, or equal to the value calculated in part (b)? Explain your reasoning. [Hint: Compare the wavelengths inside the box.] (e) Give a rough estimate of the number of bound states for the electron in the finite box in terms of \(L\) and \(U_{0}\).
At a baseball game, a radar gun measures the speed of a 144-g baseball to be \(137.32 \pm 0.10 \mathrm{km} / \mathrm{h} .\) (a) What is the minimum uncertainty of the position of the baseball? (b) If the speed of a proton is measured to the same precision, what is the minimum uncertainty in its position?
A double-slit interference experiment is performed with 2.0-ev photons. The same pair of slits is then used for an experiment with electrons. What is the kinetic energy of the electrons if the interference pattern is the same as for the photons (i.e., the spacing between maxima is the same)?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Fluids

Read Explanation

Electrostatics

Read Explanation

Solid State Physics

Read Explanation

Dynamics

Read Explanation

Further Mechanics and Thermal 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.