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Chapter 27: Problem 29

In proton-antiproton annihilation, a proton and an antiproton (a negatively charged particle with the mass of a proton) collide and disappear, producing electromagnetic radiation. If each particle has a mass of \(1.67 \times 10^{-27} \mathrm{~kg}\) and they are at rest just before the annihilation, find the total energy of the radiation. Give your answers in joules and in electronvolts.

Short Answer

Expert verified
The total energy is \( 3.00 \times 10^{-10} \) joules or \( 1.87 \times 10^{9} \) electronvolts.

Step by step solution

01

Understand the Concept

In proton-antiproton annihilation, two particles with mass collide and convert their entire rest mass into energy. This is a direct application of Einstein's mass-energy equivalence principle, described by the equation \( E = mc^2 \).
02

Calculate Total Mass

Since the proton and antiproton have the same mass, the total mass involved in the annihilation is the sum of their masses: \[ 2m = 2 imes 1.67 imes 10^{-27} ext{ kg} \] This simplifies to: \[ 3.34 imes 10^{-27} ext{ kg} \]
03

Calculate Energy in Joules

Use the formula \( E = mc^2 \) to calculate the energy:\[ E = (3.34 imes 10^{-27} ext{ kg}) imes (3.00 imes 10^8 ext{ m/s})^2 \]\[ E = 3.00 imes 10^{-10} ext{ J} \]
04

Convert Energy to Electronvolts

One electronvolt (eV) is the amount of kinetic energy gained or lost by an electron as it moves across an electric potential difference of 1 volt. It is equal to \( 1.602 imes 10^{-19} ext{ J} \). To find the energy in electronvolts, use:\[ E (eV) = \frac{3.00 imes 10^{-10} ext{ J}}{1.602 imes 10^{-19} ext{ J/eV}} \]\[ E (eV) = 1.87 imes 10^{9} ext{ eV} \]
05

Organize Final Results

The energy of the radiation produced by the annihilation is \( 3.00 imes 10^{-10} \) joules or approximately \( 1.87 imes 10^{9} \) electronvolts.

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Key Concepts

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

Mass-Energy Equivalence
The idea of mass-energy equivalence is fundamental to understanding proton-antiproton annihilation. This principle was introduced by Albert Einstein and is expressed in his famous equation, \( E = mc^2 \). It tells us that mass and energy are interchangeable; essentially, mass can be converted into energy.In the context of proton-antiproton annihilation, the particles' mass completely converts into energy. This demonstrates how a seemingly small mass can produce a large amount of energy when multiplied by the speed of light squared, \( c^2 \). This step is vital in understanding why such interactions result in substantial energy release.
Rest Mass Energy
Rest mass energy refers to the energy inherent in an object due to its mass, independent of its motion. It's the energy equivalent of a particle's rest mass and can be calculated using the same mass-energy equivalence formula \( E = mc^2 \).For proton-antiproton annihilation, the rest mass energy of both particles is calculated first, as they are at rest just before the collision. This rest mass energy is the foundation for the subsequent energy calculations, helping us determine the total energy released during annihilation.
Energy Conversion
Energy conversion in proton-antiproton annihilation shows how mass transforms into energy. When the proton and antiproton meet, their rest mass isn't lost but converted into pure energy. This conversion is a direct result of particle interactions at the most fundamental level. It clearly outlines that energy in any form (including electromagnetic radiation produced during annihilation) originates from mass through natural processes dictated by physics laws. This concept underscores the broader implications of energy transformations we see in the universe.
Joules to Electronvolts Conversion
To fully appreciate the result of proton-antiproton annihilation, we often need to convert energy from joules to electronvolts. This conversion is crucial because electronvolts (eV) are more commonly used in particle physics to express such high-energy interactions.The conversion factor between these units is \(1 \text{eV} = 1.602 \times 10^{-19} \text{J}\). To convert the annihilation energy from joules to electronvolts, divide the energy in joules by this factor. This unit transformation allows us to express and understand the energy scale of nuclear and subatomic processes more intuitively.

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Most popular questions from this chapter

(a) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of \(0.980 c ?\) (b) What is the kinetic energy of the electron at this speed? Express your answer in joules and in electronvolts.

A particle has a rest mass of \(6.64 \times 10^{-27} \mathrm{~kg}\) and a momentum of \(2.10 \times 10^{-18} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). (a) What is the total energy (kinetic plus rest energy) of the particle? (b) What is the kinetic energy of the particle? (c) What is the ratio of the kinetic energy to the rest energy of the particle?

In certain radioactive beta decay processes (more about these in Chapter 30 ), the beta particle (an electron) leaves the atomic nucleus with a speed of \(99.95 \%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at \(75.00 \%\) the speed of light, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving, (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts (a) and (b), find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

Two events are observed in a frame of reference \(S\) to occur at the same space point, the second occurring \(1.80 \mathrm{~s}\) after the first. In a second frame \(S^{\prime}\) moving relative to \(S\), the second event is observed to occur \(2.35 \mathrm{~s}\) after the first. What is the difference between the positions of the two events as measured in \(S^{\prime} ?\)

A nuclear physicist measures the momentum and corresponding relativistic energy of an atomic particle. A table of her data is shown here. $$\begin{array}{ll}\hline \text { Momentum }(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}) & \text { Energy }(\mathrm{J}) \\\\\hline 1.42 \times 10^{-18} & 7.36 \times 10^{-10} \\\2.00 \times 10^{-18} & 8.50 \times 10^{-10} \\\2.45 \times 10^{-18} & 9.51 \times 10^{-10} \\\2.83 \times 10^{-18} & 1.04 \times 10^{-9} \\\3.17 \times 10^{-18} & 1.13 \times 10^{-9} \\\\\hline\end{array}$$ Make a plot of \(E^{2}\) as a function of \(p^{2} c^{2}\). Using a linear "best fit" to the data and Equation \(27.24,\) determine the mass of the particle. Express your answer in amu. What is the particle?
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