/*! 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 4 A proton and an antiproton annih... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 44: Problem 4

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon (a) if the p and \(\overline{p}\) are initially at rest and (b) if the p and \(\overline{p}\) collide head-on, each with an initial kinetic energy of 620 MeV.

Short Answer

Expert verified
(a) 938.27 MeV; frequency \(1.48 \times 10^{23}\, \text{Hz}\); wavelength \(2.02 \times 10^{-15}\, \text{m}\). (b) 1558.27 MeV; frequency \(2.47 \times 10^{23}\, \text{Hz}\); wavelength \(1.21 \times 10^{-15}\, \text{m}\).

Step by step solution

01

Analyze Rest Annihilation Energy

When a proton (p) and an antiproton (\(\overline{p}\)) annihilate at rest, their combined rest mass energy is converted into photon energy. The rest mass energy of each particle is given by \(E = mc^2\), where \(m_{\text{p}} = 938.27\, \text{MeV}/c^2\). Thus, \(E = 2 \times 938.27\, \text{MeV} = 1876.54\, \text{MeV}\) total for two photons, so each photon has \(E = 938.27\, \text{MeV}\).
02

Calculate Energy of Photon (at rest)

Each photon's energy is half the rest mass energy of the combined system at annihilation, hence \(E_\gamma = 938.27\, \text{MeV}\).
03

Calculate Frequency of Photon at Rest

Frequency \(f\) is calculated using the formula \(E = hf\), where \(h\) is Planck's constant \(6.626 \times 10^{-34}\, \text{Js}\). Therefore, \(f = \frac{E}{h} = \frac{938.27\times 10^6 \times 1.602 \times 10^{-19}\, \text{J}}{6.626 \times 10^{-34}\, \text{Js}}\).
04

Calculate Wavelength of Photon at Rest

The wavelength \(\lambda\) can be found using \(\lambda = \frac{c}{f}\), where \(c\) is the speed of light \(3 \times 10^8\, \text{m/s}\). Use the frequency found in Step 3 to find \(\lambda\).
05

Analyze Kinetic Energy Contribution (Head-on Collision)

For a head-on collision with kinetic energy, the total energy before annihilation becomes the sum of the rest mass energy and kinetic energy of each particle: \(E = 2 \times (938.27\, \text{MeV} + 620\, \text{MeV})\).
06

Calculate Each Photon's Energy (head-on)

Each photon's energy is \(E_\gamma = \frac{2 \times 1558.27\, \text{MeV}}{2} = 1558.27\, \text{MeV}\) after the kinetic energy is included.
07

Calculate Frequency of Photon (head-on)

The frequency \(f\) is calculated from \(E_\gamma = hf\) as \(f = \frac{1558.27 \times 10^6 \times 1.602 \times 10^{-19}\, \text{J}}{6.626 \times 10^{-34}\, \text{Js}}\).
08

Calculate Wavelength of Photon (head-on)

Using \(\lambda = \frac{c}{f}\) with the new frequency from Step 7, calculate \(\lambda\) for the head-on collision scenario.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photon Energy
Photon energy plays a crucial role in the annihilation of a proton and an antiproton. When these particles annihilate, they transform entirely into energy in the form of photons. This transformation uses the principle given by Einstein's equation, \(E = mc^2\). In this context, \(m\) is the mass of the proton or antiproton, and \(c\) is the speed of light.

  • For a proton or antiproton at rest, each has a rest mass energy of 938.27 MeV.
  • Upon annihilation, both rest mass energies add up, giving a total of 1876.54 MeV, which is shared between the two photons produced.
  • Consequently, each photon in this scenario has an energy of 938.27 MeV.
However, if the particles collide head-on, they possess additional kinetic energy before annihilating, affecting the photons' energy:

  • In a head-on collision with 620 MeV kinetic energy per particle, the total energy for photon creation becomes \(2 \times (938.27 \text{MeV} + 620 \text{MeV}) = 3116.54 \text{MeV}\).
  • Thus, each photon now receives an increased energy of 1558.27 MeV.
Frequency Calculation
Calculating the frequency of a photon involves understanding the relationship between energy and frequency through Planck's equation: \(E = hf\). Here, \(E\) denotes the energy of the photon, \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\), and \(f\) is the frequency we wish to determine.

For photons resulting from proton-antiproton annihilation:

  • At rest, a photon with 938.27 MeV will have its energy converted to joules by multiplying by \(1.602 \times 10^{-19}\) J/MeV.
  • The frequency \(f\) is calculated as \(f = \frac{E}{h}\), where \(E\) is the energy converted to joules.
In a head-on collision scenario, the frequency increases:

  • With each photon's energy at 1558.27 MeV, convert it to joules similarly.
  • Once again, apply \(f = \frac{E}{h}\) to find the higher frequency, as the energy is greater here.
Wavelength Determination
The wavelength of a photon is closely related to its frequency through the equation \(\lambda = \frac{c}{f}\), where \(c\) is the speed of light \((3 \times 10^8 \text{ m/s})\). Understanding the wavelength helps in visualizing the electromagnetic properties of photons resulting from annihilation.

  • For photons with rest energy, once the frequency is computed, \(\lambda\) can be derived to show how the photon's wavelength fits into the electromagnetic spectrum.
  • Longer wavelengths equate to lower frequencies, reflecting lower-energy photons, while shorter wavelengths indicate higher energy.
If there is a kinetic energy influence:
  • The increased frequency from head-on collisions results in shorter wavelengths for those photons. This means that these high-energy photons are more to the 'high-frequency' end of the spectrum.
  • Knowing the wavelength is pivotal in applications, like understanding the type of radiation produced.

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

Determine the electric charge, baryon number, strangeness quantum number, and charm quantum number for the following quark combinations: (a) \(uds\); (b) c\(\overline{u}\); (c) ddd; and (d) d\(\overline{c}\). Explain your reasoning.

The \(\phi\) meson has mass \(1019.4 MeV/c^2\) and a measured energy width of \(4.4 MeV/c^2\). Using the uncertainty principle, estimate the lifetime of the \(\phi\) meson.

One proposed proton decay is \(p^+ \rightarrow e^+ + \pi^0\), which violates both baryon and lepton number conservation, so the proton lifetime is expected to be very long. Suppose the proton half-life were \(1.0 \times 10^{18} y\). (a) Calculate the energy deposited per kilogram of body tissue (in rad) due to the decay of the protons in your body in one year. Model your body as consisting entirely of water. Only the two protons in the hydrogen atoms in each \(H_2O\) molecule would decay in the manner shown; do you see why? Assume that the \(\pi^0\) decays to two \(\gamma\) rays, that the positron annihilates with an electron, and that all the energy produced in the primary decay and these secondary decays remains in your body. (b) Calculate the equivalent dose (in rem) assuming an RBE of 1.0 for all the radiation products, and compare with the 0.1 rem due to the natural background and the 5.0-rem guideline for industrial workers. Based on your calculation, can the proton lifetime be as short as \(1.0 \times 10^{18} y\)?

Suppose that positron-electron annihilations occur on the line 3 cm from the center of the line connecting two detectors. Will the resultant photons be counted as having arrived at these detectors simultaneously? (a) No, because the time difference between their arrivals is 100 ms; (b) no, because the time difference is 200 ms; (c) yes, because the time difference is 0.1 ns; (d) yes, because the time difference is 0.2 ns.

In which of the following reactions or decays is strangeness conserved? In each case, explain your reasoning. (a) \(K^+ \rightarrow \mu^+ + \nu_\mu\); (b) \(n + K^+ \rightarrow p + \pi^0\); (c) \(K^+ + K^- \rightarrow \pi^0 + \pi^0\); (d) \(p + K^- \rightarrow \Lambda^0 + \pi^0\).
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Electrostatics

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

Read Explanation

Engineering Physics

Read Explanation

Fundamentals of 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.