Asked by Andy
The average lifetime of muons at rest is 2.20 us. A laboratory measurement on muons traveling in a beam emerging from a particle accelerator yields an average muon lifetime of 19.668 us. What is the speed of the muons in the laboratory?
u=((1-(2.2e-6/19.688e-6)^2))^(1/2)=0.994c=2.98e8 m/s
ok I've got this part but I'm having trouble with these parts of the problem
B.What is their kinetic energy? (MeV)
C.What is their momentum? (MeV/c - Do not enter a unit) The mass of a muon is 207 times that of an electron.
u=((1-(2.2e-6/19.688e-6)^2))^(1/2)=0.994c=2.98e8 m/s
ok I've got this part but I'm having trouble with these parts of the problem
B.What is their kinetic energy? (MeV)
C.What is their momentum? (MeV/c - Do not enter a unit) The mass of a muon is 207 times that of an electron.
Answers
Answered by
drwls
The ratio by which the apparent lifetime is lengthened, 8.94, equals the "gamma" factor,
1/sqrt[1 - (v/c)^2]
Solve for v/c
b. The kinetic energy (in Joules) is
(gamma-1)*Mo*c^2 = 7.94 Mo c^2
where Mo is the rest mass of the mu-meson. You will have to convert that to MeV.
1 MeV = 1.6*10^-12 J
c. The momentum is gamma*Mo*v.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
You will have to convert that to MeV/c
1 MeV/c = 5.33*10^-21 kg m/s
1/sqrt[1 - (v/c)^2]
Solve for v/c
b. The kinetic energy (in Joules) is
(gamma-1)*Mo*c^2 = 7.94 Mo c^2
where Mo is the rest mass of the mu-meson. You will have to convert that to MeV.
1 MeV = 1.6*10^-12 J
c. The momentum is gamma*Mo*v.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
You will have to convert that to MeV/c
1 MeV/c = 5.33*10^-21 kg m/s
Answered by
Andy
drwls you are awesome!!
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