508
9 Particle Physics – I
Q
2
= 2 E i E f (1 − cos θ)
Thus, the approximation is valid when the particle mass is much smaller
than the energies.
9.3.4 Decay
9.20 (a) If a particle A at rest decays into B and C then the total energy of B will be
E B =
m A
2
+ m B
2
− m C
2
/2m A
Inserting the values m A = 494 MeV, m B = 135 MeV and m C =
140 MeV, we find E(π
0 ) = 245.6 MeV.
(b) Kinetic energy of π
0 is E π0 − m π0 = 245.6 − 135 = 110.6 MeV
9.21 If I 0 is the original intensity and I the observed intensity, then
I = I 0 e
−s/vτ
= I 0 e
−as
γ = 1 + T /m = 1 + 100/139 = 1.719
β = (γ
2
− 1)
1/2
/γ = (1.719
2
− 1)
1/2
/1.719 = 0.813
a = 1/vτ → τ = 1/va = 1/β ca = 1/(0.813 × 3 × 10
8
× 9.1 × 10
−2 )
τ = 4.5 × 10
−8 s
Proper life time τ 0 = τ/γ = 4.5 × 10
−8
/1.719 = 2.62 × 10
−8 s
9.22 m K c
2
= 966.7 × 0.511 = 494 MeV
m π c
2
= 273.2 × 0.511 = 139.5 MeV
Energy released in the decay
Q = m K − 3m π = 494.0 − 3 × 139.5 = 75 MeV
Maximum kinetic energy of one pion will occur when the other two pions
go together in the opposite direction, that is energy is shared between one π
and 2 π’s
Non-relativistically
T π (max) = Q × 2m π /(2m π + m π ) = 75 × 2/3 = 50 MeV
Relativistically
T π (max) = E π − m π =
m K
2
+ m π
2
− (2m π )
2
/2m K
− m π
Inserting the values of m K and m π , we find T π (max) = 48.4 MeV. The difference in the two results is 3.3%.
9.23 In the decay A → B + C,
E B =
m A
2
+ m B
2
− m C
2
/2m A
Here B = ν, A = π, C = μ. It follows that
E
ν =
m
2
π − m
2
μ
/2m π (because m ν = 0)
9.24 The maximum energy of neutron is obtained when the neutrino has zero
energy, neutron and muon being emitted in the opposite direction.
The energy released in the decay is
Q = m Σ − (m n + m μ + m ν ) = 1,189 − (939 + 106 + 0) = 144 MeV
Non-relativistically,
T n = Q m μ /(m μ + m n ) = 144 × 106/(106 + 939) = 14.6 MeV
Relativistically,
9 Particle Physics – I
Q
2
= 2 E i E f (1 − cos θ)
Thus, the approximation is valid when the particle mass is much smaller
than the energies.
9.3.4 Decay
9.20 (a) If a particle A at rest decays into B and C then the total energy of B will be
E B =
m A
2
+ m B
2
− m C
2
/2m A
Inserting the values m A = 494 MeV, m B = 135 MeV and m C =
140 MeV, we find E(π
0 ) = 245.6 MeV.
(b) Kinetic energy of π
0 is E π0 − m π0 = 245.6 − 135 = 110.6 MeV
9.21 If I 0 is the original intensity and I the observed intensity, then
I = I 0 e
−s/vτ
= I 0 e
−as
γ = 1 + T /m = 1 + 100/139 = 1.719
β = (γ
2
− 1)
1/2
/γ = (1.719
2
− 1)
1/2
/1.719 = 0.813
a = 1/vτ → τ = 1/va = 1/β ca = 1/(0.813 × 3 × 10
8
× 9.1 × 10
−2 )
τ = 4.5 × 10
−8 s
Proper life time τ 0 = τ/γ = 4.5 × 10
−8
/1.719 = 2.62 × 10
−8 s
9.22 m K c
2
= 966.7 × 0.511 = 494 MeV
m π c
2
= 273.2 × 0.511 = 139.5 MeV
Energy released in the decay
Q = m K − 3m π = 494.0 − 3 × 139.5 = 75 MeV
Maximum kinetic energy of one pion will occur when the other two pions
go together in the opposite direction, that is energy is shared between one π
and 2 π’s
Non-relativistically
T π (max) = Q × 2m π /(2m π + m π ) = 75 × 2/3 = 50 MeV
Relativistically
T π (max) = E π − m π =
m K
2
+ m π
2
− (2m π )
2
/2m K
− m π
Inserting the values of m K and m π , we find T π (max) = 48.4 MeV. The difference in the two results is 3.3%.
9.23 In the decay A → B + C,
E B =
m A
2
+ m B
2
− m C
2
/2m A
Here B = ν, A = π, C = μ. It follows that
E
ν =
m
2
π − m
2
μ
/2m π (because m ν = 0)
9.24 The maximum energy of neutron is obtained when the neutrino has zero
energy, neutron and muon being emitted in the opposite direction.
The energy released in the decay is
Q = m Σ − (m n + m μ + m ν ) = 1,189 − (939 + 106 + 0) = 144 MeV
Non-relativistically,
T n = Q m μ /(m μ + m n ) = 144 × 106/(106 + 939) = 14.6 MeV
Relativistically,