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Chapter 26: Problem 52

In proton-antiproton annihilation, a proton and an antiproton (which has the same mass as the proton, but carries a negative charge) interact, and both masses are completely converted to electromagnetic radiation. Assuming that the particles are moving toward one another at a speed of 0.80 c relative to the laboratory reference frame before they annihilate, determine the total energy released in the form of radiation according to a laboratory technician.

Short Answer

Expert verified
3128 MeV.

Step by step solution

01

Identify the Rest Energy

Both a proton and an antiproton have the same rest energy. The rest energy of a proton is calculated using the formula: \( E_0 = mc^2 \), where \( m = 1.67 \times 10^{-27} \text{ kg} \) and \( c = 3 \times 10^8 \text{ m/s} \). Substituting the values, we get: \( E_0 = (1.67 \times 10^{-27} \text{ kg})(3 \times 10^8 \text{ m/s})^2 = 1.503 \times 10^{-10} \text{ J} \).
02

Convert Rest Energy to MeV

The energy in joules can be converted to electron volts using the conversion factor: \( 1 \text{ J} = 6.242 \times 10^{12} \text{ MeV} \). So, \( E_0 = 1.503 \times 10^{-10} \text{ J} \times 6.242 \times 10^{12} \text{ MeV/J} = 938 \text{ MeV} \).
03

Determine the Total Rest Energy of System

Since there are two particles involved (proton and antiproton), the total rest energy of the system before annihilation is: \( 2 \times 938 \text{ MeV} = 1876 \text{ MeV} \).
04

Calculate Total Energy Using Relativistic Energy Formula

The relativistic energy formula is \( E = \gamma m c^2 \), where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \) and \( v = 0.80c \). Calculating \( \gamma \), we get \( \gamma = \frac{1}{\sqrt{1 - (0.80)^2}} \approx 1.667 \). Thus, the energy of one moving proton is \( E = 1.667 \times 938 \text{ MeV} \approx 1564 \text{ MeV} \). For two particles, total energy is \( 2 \times 1564 \text{ MeV} = 3128 \text{ MeV} \).
05

Calculate the Total Energy Released

The total energy released in the form of radiation will be equal to the total energy just before annihilation, which is 3128 MeV, since both masses are completely converted to electromagnetic radiation.

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Key Concepts

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

Relativistic Energy
When dealing with particles traveling at significant fractions of the speed of light, as in the proton-antiproton annihilation scenario, one must take into account relativistic effects. These effects are crucial for accurately calculating the total energy of such fast-moving particles. Relativistic energy can be described by the equation:
  • \(E = \gamma mc^2\)
  • Here, \(\gamma\) is the Lorentz factor, \(m\) is the rest mass, and \(c\) is the speed of light. \(\gamma\) accounts for the increase in energy due to the particle's velocity.
For a particle moving at a speed of 0.80c, \(\gamma\) approximately equals 1.667 because:
  • \(\gamma = \frac{1}{\sqrt{1-v^2/c^2}}\)
  • Substitute \(v = 0.80c\) into the equation to find \(\gamma\).
Using this value, you can calculate the relativistic energy of a proton well beyond its rest energy, preparing for its annihilation with an antiproton.
Rest Energy
Rest energy is a concept that derives from Einstein's famous equation \(E_0 = mc^2\). It refers to the inherent energy contained within a particle when it is at rest. In the context of proton-antiproton annihilation:
  • The rest mass of a proton (or an antiproton, since they're identical in mass) is approximately \(1.67 \times 10^{-27} \text{ kg}\).
  • For a proton, the energy yielded by this mass at rest equals \(1.503 \times 10^{-10} \text{ J}\).
However, since both particles in interaction have rest energy, their total rest energy before annihilation becomes doubled. This means you must think of it as an initial energy reserve before converting to electromagnetic radiation.
Conversion to MeV
Energy in physics and especially in particle physics is often expressed not in joules (J) but in electronvolts (eV) due to the small magnitudes involved. Conversion to mega-electronvolts (MeV), where 1 MeV = \(10^6\) eV, simplifies dealing with such small numbers. When calculating the rest energy for one proton:
  • First calculate it in joules \(1.503 \times 10^{-10} \text{ J}\).
  • Convert this energy to MeV using the relation: \(1 \text{ J} = 6.242 \times 10^{12} \text{ eV}\).
  • The result is approximately 938 MeV for one proton, often used in nuclear and high-energy physics.
For two particles involved, the collective rest energy is simply double, 1876 MeV. This conversion helps communicate and understand the substantial energies particles possess even while at rest.
Electromagnetic Radiation
In particle interactions like proton-antiproton annihilation, the ultimate transformation of matter results in the emission of electromagnetic radiation. This radiation represents the total energy formerly existing as rest mass and kinetic energy of the particles.
  • Following annihilation, matter entirely converts to energy in accordance with Einstein's mass-energy equivalence principle \(E = mc^2\).
  • In this scenario, the energy output released is equivalent to 3128 MeV.
The electromagnetic radiation emitted might typically include high-energy photons known as gamma rays. Understanding this conversion underscores the profound principle that mass can indeed shed its form and exit the universe as pure energy in electromagnetic waves.

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