Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The rest energy of an electron is \(0.511 \mathrm{MeV} .\) What momentum (in MeV/c) must an electron have in order that its total energy be 3.00 times its rest energy?

Short Answer

Expert verified
Answer: The momentum of the electron is approximately 1.44 MeV/c.

Step by step solution

01

Write down the relativistic energy-momentum relationship

The relativistic energy-momentum relationship is given by: E² = (mc²)² + (pc)² where E is the total energy, m is the mass, c is the speed of light, and p is the momentum.
02

Express the total energy in terms of rest energy

We are given that the total energy is 3 times the rest energy, so we can write: E = 3mc²
03

Substitute the total energy expression into the energy-momentum relationship

Now, substitute the total energy expression into the energy-momentum relationship: (3mc²)² = (mc²)² + (pc)²
04

Solve the equation for momentum (p)

To find the momentum, we can rearrange the equation and solve for p: (3mc²)² - (mc²)² = (pc)² (8m²c⁴) = (pc)² p = \sqrt{8m²c²}
05

Substitute the values and calculate the momentum

Now we can substitute the given rest energy value and the speed of light to find the momentum in MeV/c: Rest energy (E₀) = 0.511 MeV c = 1 (when working in units of MeV/c) E₀ = mc² m = \frac{E₀}{c²} = \frac{0.511}{(1)²} = 0.511 Now, substitute these values into the momentum equation: p = \sqrt{8(0.511)²} p ≈ 1.44 \, MeV/c The momentum of the electron when its total energy is 3 times its rest energy is approximately 1.44 MeV/c.

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 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

See all solutions

Recommended explanations on Physics Textbooks

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