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Using Erest = m c 2
These questions give practice in using Erest = m c 2 to calculate photon energies and masses of particles created or annihilated.
Energy and mass
The energy of a particle at rest is all due to its mass. This energy is called the rest energy. The rest energy in joules of a particle with mass m measured in kilograms is given by
Erest = m c 2, where c is the speed of light in metres per second. An energy in joules can be converted to electron volts by dividing by 1.60 10 19 J eV 1.
The mass of an electron or positron is 9.11 10 31 kg. The speed of light is 3.00 108 m s 1.
1. Show that the rest energy of an electron is 8.2 10 14 J.
2. Use the answer to question 1 to show that the rest energy of an electron is 0.51 MeV.
3. Write down the rest energy of a positron (antielectron).
4. An electron and a positron which meet annihilate one another. By how much does the rest energy decrease in total? Express the answer in MeV.
5. The annihilation of an electron and a positron at rest produces a pair of identical gamma ray photons travelling in opposite directions. Write down in MeV the energy you expect each photon to have
6. A single photon passing near a nucleus can create an electronpositron pair. Their rest energy comes from the energy of the photon. Write down the smallest photon energy that can produce one such pair.
7. Cosmic rays can send high-energy photons through the atmosphere. What approximately is the maximum number of electronpositron pairs that a 10 GeV photon can create?
8. The isotope EMBED Equation.3 is stable. The light isotope EMBED Equation.3 emits positrons and gamma rays including a photon of energy 1.28 MeV. How can decays of this nucleus result in both annihilation and creation of electronpositron pairs?
9. A photon can create particleantiparticle pairs of greater mass than electrons and positrons. Approximately what energy must a photon have to create a protonantiproton pair? (The mass of a proton is 2000 times the mass of an electron).
10. Why do the photons from the annihilation of an electronpositron pair not themselves go on to create new electronpositron pairs?
Practical advice
These are intended as simple practice calculations on energy changes in particleantiparticle creation and annihilation.
Alternative approaches
You might start in the context of PET scans, which use positron annihilation.
Social and human context
The existence of the creation and annihilation of matter makes for a profound shift in the way the world has to be imagined. Its permanence is undermined.
Answers and worked solutions
1.
EMBED Equation.3
2. Energy in eV is energy in joules divided by 1.60 10 19 J eV 1:
EMBED Equation.3
3. 0.51 MeV since electrons and positrons have identical mass.
4. 1.02 MeV. Two particles, each of rest energy 0.51 MeV, are annihilated.
5. 0.51 MeV each. The two photons share the total 1.02 MeV.
6. 1.02 MeV. The photon has to create the rest energy of two particles, each 0.51 MeV.
7. Approximately 104 pairs. Each pair requires approximately 1 MeV = 1 106 eV and 10 GeV = 10 109 eV is available.
8. The 1.28 MeV photon has enough energy to create an electron positro
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10. Their 0.51 MeV energy is too small (minimum 1.02 MeV).
External references
This activity is taken from Advancing Physics chapter 17, 30S
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