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Question:The mass of a muon is 207 times the electron mass; the average lifetime of muons at rest is 2.20μs. In a certain experiment, muons moving through a laboratory are measured to have an average lifetime of 6.90μs. For the moving muons, what are (a) β , (b) K, and (c) p (in MeV/c)?

Short Answer

Expert verified

Answer

(a) The speed factor for muon is 0.948.

(b) The kinetic energy of muon is 227MeV.

(c) The momentum of muon is 315MeV/c.

Step by step solution

01

Identification of given data

The average life time of muons at rest is t0=2.20μs

The average life time of moving muons is t=6,90μs

The mass of the muon is m=207me

The total energy of a particle has two components. One is the rest energy of particle and other is the kinetic energy of the particle due to the motion of particle.

02

Determination of speed factor of muon

(a)

The speed factor of muon is given as:

β=1-t0t2

Substitute all the values in equation.

β=1-2.20μs6.90μs2β=0.948

Therefore, the speed factor for muon is 0.948.

03

Determination of kinetic energy of muon

(b)

The kinetic energy of muon is given as:

K=mc211-β2-1K=207mec211-β2-1K=207mec211-β2-1\

Here, me is the mass of electron and its value is 9.109×10-31kg, c is the speed of light and its value is 3×108m/s

Substitute these values in equation (3).

K=2079.109×10-31kg3×108m/s211-0.9482-1K=3.635×10-11JK=3.635×10-11J1MeV1.6×10-13JK=227MeV

Therefore, the kinetic energy of muon is 227MeV.

04

Determination of momentum of muon

(c)

The momentum of muon is given as:

p=β1-β2mcp=β1-β2207mecp=207β1-β2mec

Here, me is the mass of electron and its energy equivalent is 0.511MeV/c2.

Substitute all the values in above equation.

p=2070.9481-0.94820.511MeV/c2cp=315MeV/c

Therefore, the momentum of muon is 315MeV/c.

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