/*! 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 15 A neutral pion (mass \(0.135 \ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 30: Problem 15

A neutral pion (mass \(0.135 \mathrm{GeV} / \mathrm{c}^{2}\) ) decays via the electromagnetic interaction into two photons: $\pi^{0} \rightarrow \gamma+\gamma$ What is the energy of each photon, assuming the pion was at rest?

Short Answer

Expert verified
Answer: The energy of each photon emitted in the decay process is \(0.0675 \, \mathrm{GeV}\).

Step by step solution

01

Note Down The Given Information

We are given: 1. Mass of neutral pion: \(0.135 \,\mathrm{GeV}/\mathrm{c}^2\) 2. Decay process: \(\pi^0 \rightarrow \gamma + \gamma\)
02

Recap Conservation of Energy and Momentum

In this decay process, we need to make sure that both energy and momentum are conserved. In other words, the total energy before the decay should be equal to the total energy after the decay, and the total momentum before the decay should be equal to the total momentum after the decay.
03

Apply Conservation of Energy

The energy of the neutral pion is equal to its mass-energy, which means that the sum of the energies of the two photons must be equal to this mass-energy. Since the pion is initially at rest, we know that the energy of each photon must be half of the total energy. So, the energy of each photon is: \(E_\gamma = \frac{1}{2} E_{\pi^0}\), where \(E_{\pi^0}\) is the energy of the neutral pion.
04

Convert Mass-Energy to Energy

We can convert the mass-energy of the pion to energy using the famous equation of Einstein, \(E = mc^2\). Here, c is the speed of light, which will be squared. For the given mass of the pion, it follows \(E_{\pi^0} = (0.135 \,\mathrm{GeV}/\mathrm{c}^2)(c^2) = 0.135 \,\mathrm{GeV}\)
05

Calculate the Energy of Each Photon

Now we can find the energy of each photon using the expression from Step 3: \(E_\gamma = \frac{1}{2} E_{\pi^0} = \frac{1}{2} (0.135 \,\mathrm{GeV}) = 0.0675 \,\mathrm{GeV}\) So, the energy of each photon emitted in the decay process is \(0.0675 \, \mathrm{GeV}\).

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

What is the de Broglie wavelength of an electron with kinetic energy $7.0 \mathrm{TeV} ?$
The \(K^{0}\) meson can decay to two pions: \(K^{0} \rightarrow \pi^{+}+\pi^{-}\) The rest energies of the particles are: \(K^{0}=497.7 \mathrm{MeV}\) \(\pi^{+}=\pi^{-}=139.6 \mathrm{MeV} .\) If the \(K^{0}\) is at rest before it decays, what are the kinetic energies of the \(\pi^{+}\) and the \(\pi^{-}\) after the decay?
In the Cornell Electron Storage Ring, electrons and positrons circulate in opposite directions with kinetic energies of 6.0 GeV each. When an electron collides with a positron and the two annihilate, one possible (though unlikely) outcome is the production of one or more proton-antiproton pairs. What is the maximum possible number of proton-antiproton pairs that could be formed?
When a proton and an antiproton annihilate, the annihilation products are usually pions. (a) Suppose three pions are produced. What combination(s) of \(\pi^{+}, \pi^{-},\) and \(\pi^{0}\) are possible? (b) Suppose five pions are produced. What combination(s) of \(\pi^{+}, \pi^{-},\) and \(\pi^{0}\) are possible? (c) What is the maximum number of pions that could be produced if the kinetic energies of the proton and antiproton are negligibly small? The mass of a charged pion is \(0.140 \mathrm{GeV} / c^{2}\) and the mass of a neutral pion is \(0.135 \mathrm{GeV} / c^{2}.\)
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?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Fluids

Read Explanation

Atoms and Radioactivity

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.