Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 34: Problem 34

Is it possible to determine an electron's velocity accurate to \(\pm 1 \mathrm{m} / \mathrm{s}\) while simultaneously finding its position to within \(\pm 1 \mu \mathrm{m} ?\) What about a proton?

Short Answer

Expert verified
After conducting the calculations, we find that the uncertainty in velocity is much larger than \(\pm 1 m/s\) for both the electron and proton. Thus, it is not possible to simultaneously find the velocity within \(\pm 1 m/s\) and position to within \(\pm 1 \mu m\) for either particle due to the Heisenberg Uncertainty Principle.

Step by step solution

01

Apply the Heisenberg Uncertainty Principle

Heisenberg uncertainty principle can be mathematically represented as: \(\Delta p \Delta x \geq \frac{h}{4\pi}\), where \(\Delta p\) is the uncertainty in momentum, \(\Delta x\) is the uncertainty in position and \(h\) is the Planck constant. First, let's find the minimum momentum for an electron that can be determined while knowing its position with accuracy of \(\pm 1 \mu m\). We know that \(\Delta p = m \Delta v\). Use this relation to replace \(\Delta p\).
02

Solve the equation for electrons and protons

For an electron, substitute the known values \(\Delta x = 1 \mu m = 1*10^-6\) m, \(m_e = 9.11*10^-31\) kg and \(h = 6.626*10^-34\) Js into the inequality and solve for \(\Delta v\). Repeat the procedure for a proton with \(m_p = 1.67*10^-27\) kg.
03

Compare the results for electrons and protons

Compare the velocity uncertainties calculated for the electron and proton to \(\pm 1 m/s\). If the value is larger, then the velocity of the particle cannot be measured up to \(\pm 1 m/s\) while knowing its position up to \(\pm 1 \mu m\).

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

Find \(\lambda_{\text {peak }}\) and \(\lambda_{\text {median }}\) for Earth, considered a \(288-\mathrm{K}\) blackbody.

The stopping potential in a photoelectric experiment is \(1.8 \mathrm{V}\) when the illuminating radiation has wavelength \(365 \mathrm{nm}\). Determine (a) the work function of the emitting surface and (b) the stopping potential for 280 -nm radiation.

The radiance of a blackbody peaks at \(660 \mathrm{nm}\). (a) What's its temperature? (b) How does its radiance at 400 nm compare with that at \(700 \mathrm{nm} ?\)

Find (a) the wavelength and (b) the energy in electronvolts of the photon emitted when a Rydberg hydrogen atom drops from the \(n=180\) level to the \(n=179\) level.

Why does the photoelectric effect suggest that light has particlelike properties?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Engineering Physics

Read Explanation

Fluids

Read Explanation

Turning Points in Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Nuclear 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