Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Problem 4

What is the magnitude of the momentum of an electron with a de Broglie wavelength of \(0.40 \mathrm{nm} ?\)

Short Answer

Expert verified
Answer: The magnitude of the electron's momentum is approximately \(1.6565 \times 10^{-24} \mathrm{kg \cdot m/s}\).

Step by step solution

01

Understand the de Broglie wavelength formula

The de Broglie wavelength formula relates the wavelength (\(λ\)) of a particle to its momentum (\(p\)). The formula is given by: \(\lambda = \frac{h}{p}\), where \(h\) is the Planck constant.
02

Recall the Planck constant

The Planck constant (\(h\)) is a fundamental constant in quantum mechanics, with the approximate value of \(6.626 \times 10^{-34} \mathrm{Js}\).
03

Express the given wavelength in meters

The wavelength provided in the exercise is \(0.40 \mathrm{nm}\). Before plugging it into the de Broglie formula, we need to convert it to meters. To convert nanometers to meters, we use the following conversion factor: \(1 \mathrm{nm} = 10^{-9} \mathrm{m}\). So, \(0.40 \mathrm{nm} = 0.40 \times 10^{-9} \mathrm{m}\).
04

Calculate the momentum of the electron

Now, we can plug the given wavelength and the Planck constant into the de Broglie formula to find the momentum of the electron: $$\lambda = \frac{h}{p}$$ $$p = \frac{h}{\lambda}$$ $$p = \frac{6.626 \times 10^{-34} \mathrm{Js}}{0.40 \times 10^{-9} \mathrm{m}}$$ $$p = \frac{6.626 \times 10^{-34}}{0.40 \times 10^{-9}}$$ Evaluating the expression gives us the momentum of the electron: $$p \approx 1.6565 \times 10^{-24} \mathrm{kg \cdot m/s}$$ Thus, the magnitude of the electron's momentum is approximately \(1.6565 \times 10^{-24} \mathrm{kg \cdot m/s}\).

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

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?
The neutrons produced in fission reactors have a wide range of kinetic energies. After the neutrons make several collisions with atoms, they give up their excess kinetic energy and are left with the same average kinetic energy as the atoms, which is \(\frac{3}{2} k_{\mathrm{B}} T .\) If the temperature of the reactor core is \(T=400.0 \mathrm{K},\) find (a) the average kinetic energy of the thermal neutrons, and (b) the de Broglie wavelength of a neutron with this kinetic energy.
What is the maximum possible value of the angular momentum for an outer electron in the ground state of a bromine atom?
The omega particle \((\Omega)\) decays on average about \(0.1 \mathrm{ns}\) after it is created. Its rest energy is 1672 MeV. Estimate the fractional uncertainty in the \(\Omega\) 's rest energy \(\left(\Delta E_{0} / E_{0}\right)\) [Hint: Use the energy-time uncertainty principle, Eq. \((28-3) .]\)
In the Davisson-Germer experiment (Section \(28.2),\) the electrons were accelerated through a \(54.0-\mathrm{V}\) potential difference before striking the target. (a) Find the de Broglie wavelength of the electrons. (b) Bragg plane spacings for nickel were known at the time; they had been determined through x-ray diffraction studies. The largest plane spacing (which gives the largest intensity diffraction maxima) in nickel is \(0.091 \mathrm{nm} .\) Using Bragg's law [Eq. ( \(25-15\) )], find the Bragg angle for the first-order maximum using the de Broglie wavelength of the electrons. (c) Does this agree with the observed maximum at a scattering angle of \(130^{\circ} ?\) [Hint: The scattering angle and the Bragg angle are not the same. Make a sketch to show the relationship between the two angles.]
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Nuclear Physics

Read Explanation

Electrostatics

Read Explanation

Fields in Physics

Read Explanation

Work Energy and Power

Read Explanation

Circular Motion and Gravitation

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