A muon is produced by a collision between a cosmic ray and an oxygen nucleus in

Question

A muon is produced by a collision between a cosmic ray and an oxygen nucleus in the upper atmosphere at an altitude of 50 km. It travels vertically downward to the surface of the Earth where it arrives with a total energy of 240 MeV. The rest energy of a muon is 105.7 MeV. What is the kinetic energy of a muon at the surface? What is the velocity of a muon at the surface? (Give your answer as a fraction of the speed of light c.) What is its momentum at the surface? (Give your answer in units of MeV/c. There is no need to enter units.)

Explanation / Answer

Part a

Kinetic erngy = Total Energy - Rest Energy

Put values
KE = 240- 105.7
KE = 134.3MeV

Part b:
KE = 0.5 * m * v^2

Total Energy, E = m*C^2
so mass of muon, m = (240) / (c)^2

Hence Velocity of muon, v = sqrt(2*KE/m)
v = sqrt(2*134.3/(240))*C
v = 1.057 *c

part C :
Momentum, p = M*V = ((240) / (c)^2 ) * 1.06*c
P = 240*1.057/c
P = 253.68Mev/c

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