Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 30: Problem 12

A proton in Fermilab's Tevatron is accelerated through a potential difference of 2.5 MV during each revolution around the ring of radius \(1.0 \mathrm{km} .\) In order to reach an energy of 1 TeV, how many revolutions must the proton make? How far has it traveled?

Short Answer

Expert verified
Answer: The proton travels 2.5 x 10^9 meters to reach an energy of 1 TeV.

Step by step solution

01

Calculate the energy gained by the proton in each revolution

To find the energy gained by the proton in each revolution, we can use the following formula: \(E = qV\). where \(E\) = energy gained by the proton, \(q\) = charge of the proton (\(1.6\times10^{-19} \,\text{C}\)), \(V\) = potential difference (\(2.5\times10^{6}\,\text{V}\)). \(E = (1.6\times10^{-19} \,\text{C})(2.5\times10^{6}\, \text{V}) = 4.0\times10^{-13}\, \text{J}\). So, the proton gains \(4.0\times10^{-13}\,\mathrm{J}\) of energy in each revolution.
02

Calculate the number of revolutions needed to reach 1 TeV

The energy required to reach 1 TeV (tera-electronvolt) is: \(1\,\mathrm{TeV} = 1\times10^{12}\,\mathrm{eV} = 1\times10^{12}\times1.6\times10^{-19}\,\mathrm{J} = 1.6\times10^{-7}\,\mathrm{J}\). Now, we can find the number of revolutions needed using the following formula: Number of revolutions = \(\frac{\text{Total energy required}}{\text{Energy gained per revolution}}\) Number of revolutions = \(\frac{1.6\times10^{-7}\,\text{J}}{4.0\times10^{-13}\, \text{J}} = 4.0\times10^{5}\) So, the proton needs to make \(4.0\times10^{5}\) revolutions to reach 1 TeV.
03

Calculate the distance travelled by the proton

To calculate the distance travelled by the proton, we will use the following formula: Distance travelled = Number of revolutions \(\times\) circumference of the ring First, let's find the circumference of the ring. Circumference = \(2\pi r\) where \(r\) is the radius (\(1.0\,\text{km} = 1000\,\text{m}\)). Circumference = \(2\pi(1000\,\text{m}) \approx 6283.2\,\text{m}\) Now, we can find the distance travelled: Distance travelled = \(4.0\times10^{5} \times 6283.2\,\text{m} = 2.5\times10^{9}\,\text{m}\) The proton travels \(2.5\times10^{9}\,\text{m}\) to reach an energy of 1 TeV.

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

Three types of sigma baryons can be created in accelerator collisions. Their quark contents are given by uus, uds, and dds, respectively. What are the electric charges of each of these sigma particles, respectively?
At the Stanford Linear Accelerator, electrons and positrons collide together at very high energies to create other elementary particles. Suppose an electron and a positron, each with rest energies of \(0.511 \mathrm{MeV},\) collide to create a proton (rest energy \(938 \mathrm{MeV}\) ), an electrically neutral kaon \((498 \mathrm{MeV}),\) and a negatively charged sigma baryon \((1197 \mathrm{MeV}) .\) The reaction can be written as: $$\mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{p}^{+}+\mathrm{K}^{0}+\Sigma^{-}$$ (a) What is the minimum kinetic energy the electron and positron must have to make this reaction go? Assume they each have the same energy. (b) The sigma can decay in the reaction \(\Sigma^{-} \rightarrow \mathrm{n}+\pi^{-}\) with rest energies of \(940 \mathrm{MeV}\) (neutron) and \(140 \mathrm{MeV}\) (pion). What is the kinetic energy of each decay particle if the sigma decays at rest?
According to Figure \(30.2,\) higher energies correspond with times that are closer to the origin of the universe, so particle accelerators at higher energies probe conditions that existed shortly after the Big Bang. At Fermilab's Tevatron, protons and antiprotons are accelerated to kinetic energies of approximately 1 TeV. Estimate the time after the Big Bang that corresponds to protonantiproton collisions in the Tevatron.
A proton of mass \(0.938 \mathrm{GeV} / c^{2}\) and an antiproton, at rest relative to an observer, annihilate each other as described by \(\mathrm{p}+\overline{\mathrm{p}} \rightarrow \pi^{-}+\pi^{+} .\) What are the kinetic energies of the two pions, each of which has mass $0.14 \mathrm{GeV} / \mathrm{c}^{2} ?$
What is the de Broglie wavelength of a proton with kinetic energy $1.0 \mathrm{TeV} ?$
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Mechanics and Materials

Read Explanation

Quantum Physics

Read Explanation

Fluids

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Famous Physicists

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