A pion at rest (mπ = 273me) decays to a muon (mµ = 207me) and an antineutrino (mp ≈ 0). The reaction is written π– → µ– +
(a)
The kinetic energy of the muon.
Answer to Problem 39.61P
The kinetic energy of the muon is
Explanation of Solution
Given info: The mass of pion at rest
Write the equation of kinetic energy of the muon.
Here,
The value of
Substitute
The rest mass energy of an electron
Substitute
Write the equation of conservation of energy.
Here,
Write the equation of conservation of momentum.
Here,
Write the equation of momentum of muon.
Here,
Substitute
Substitute
Substitute
Substitute
Substitute
Conclusion:
Therefore, the kinetic energy of the muon is
(b)
The energy of the antineutrino in electron volts.
Answer to Problem 39.61P
The energy of the antineutrino is
Explanation of Solution
Given info: The mass of pion at rest
Write the equation of energy of the antineutrino.
Substitute
Substitute
Conclusion:
Therefore, the energy of the antineutrino is
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Chapter 39 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
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