6.3 Solutions
345
Fig. 6.6 (a) Scattering of a
proton with a stationary
electron (b) Momentum
triangle
6.54 The energy released in the decay of pion is
Q = m π c
2
− (m μ c
2
+ m ν c
2 )
= (273 − 207 − 0)m e c
2
= 66 × 0.511 = 33.73 MeV
Energy conservation gives
T μ + T ν = 33.73
(1)
In order to conserve momentum, muon neutrino must move in opposite direction
p ν = p μ
(2)
Multiplying (2) by c and squaring
c
2 p
2 ν = T
2 ν = K μ + 2K μ m μ c
2
(3)
Solving (1) and (3) and using
m μ c
2
= 207m e c
2
= 207 × 0.511 = 105.77 MeV
K μ = 4.08 MeV, K ν = 29.65 MeV
Observe that the lighter particle carries greater energy.
6.55 Let the mass of the final single body be M which moves with a velocity βc.
Momentum conservation gives
m × 0.6c/(1 − 0.6
2 )
1/2
= Mβc/(1 − β
2 )
1/2
Or 3m/4 = Mβ/(1 − β
2 )
( 1 )
Since the total energy is conserved
mc
2
/(1 − 0.6
2 )
1/2
+ mc
2
= Mc
2
/(1 − β
2 )
1/2
(2)
(a) Using (2) in (1), β = 1/3
(b) Using β = 1/3 in (2), M = 2.12 m
345
Fig. 6.6 (a) Scattering of a
proton with a stationary
electron (b) Momentum
triangle
6.54 The energy released in the decay of pion is
Q = m π c
2
− (m μ c
2
+ m ν c
2 )
= (273 − 207 − 0)m e c
2
= 66 × 0.511 = 33.73 MeV
Energy conservation gives
T μ + T ν = 33.73
(1)
In order to conserve momentum, muon neutrino must move in opposite direction
p ν = p μ
(2)
Multiplying (2) by c and squaring
c
2 p
2 ν = T
2 ν = K μ + 2K μ m μ c
2
(3)
Solving (1) and (3) and using
m μ c
2
= 207m e c
2
= 207 × 0.511 = 105.77 MeV
K μ = 4.08 MeV, K ν = 29.65 MeV
Observe that the lighter particle carries greater energy.
6.55 Let the mass of the final single body be M which moves with a velocity βc.
Momentum conservation gives
m × 0.6c/(1 − 0.6
2 )
1/2
= Mβc/(1 − β
2 )
1/2
Or 3m/4 = Mβ/(1 − β
2 )
( 1 )
Since the total energy is conserved
mc
2
/(1 − 0.6
2 )
1/2
+ mc
2
= Mc
2
/(1 − β
2 )
1/2
(2)
(a) Using (2) in (1), β = 1/3
(b) Using β = 1/3 in (2), M = 2.12 m