506
9 Particle Physics – I
9.12 (a) In elastic scattering, total kinetic energy is conserved. Life-time means
mean-life-time i.e., the time in which the population of unstable particles
is reduced by a factor e. The important interactions that the muons undergo
are elastic scattering and the reaction, μ
−
+ p → n + ν μ .
Let the muons travel a distance d metres. Then their intensity will be
reduced due to decay by a factor
I /I 0 = exp(−d/vγ τ 0 )
( 1 )
γ = 1/(1 − v
2
/c
2 )
1/2
= 1/
1 − (10
6
/3 × 10
8 )
2
1/2
= 1.0033
I /I 0 = exp −(d/(1.0033 × 10
6
× 2.2 × 10
−6 ))
= exp −(0.453 d)
( 2 )
The intensity reduction due to interaction will be
I /I 0 = exp(−dΣ)
( 3 )
Σ = σ n = 0.1 × 10
−28
× 2.69 × 10
25
= 2.69 × 10
−4
I /I 0 = exp(−2.69 × 10
−4 ) d
(4)
Comparing (2) and (4), reduction due to decay will be by far greater than
by interaction for any value of d.
(b) Repetition of calculation for T = 4 GeV, γ = 38.7, β ≈ 1, and d =
12,000 m, gives I /I 0 = e
−0.47
= 0.625, so that a large number of muons
survive at the ground level. The distance at which muons of energy 4 GeV
will be reduced by e is given by d = vγ τ 0 = 3×10
8
×38.7×2.2×10
−6
=
2.55 × 10
4 m = 25.5 km.
9.13 The geometrical cross-section is
σ g = π R
2
= π(r 0 A
1/3 )
2
= π r 0
2 A
2/3
= π (1.3 × 10
−13 )
2 (208)
2/3
= 1.86 × 10
−24 cm
2
Number of lead atoms/cm
3
, n = N 0 ρ/A = 6 × 10
23
× 11.3/207 =
3.275 × 10
22
Macroscopic cross-section
Σ = n σ = (3.275 × 10
22 ) (1.86 × 10
−24 ) = 0.06 cm
−1
The interaction length, λ = 1/Σ = 1/0.06 = 16.7 cm
9.14 Number of interactions per second in volume V
I = ΣVI 0
= σ N 0 ρV I 0 /A
= (40 × 10
−27 cm
2 )(6.02 × 10
23
× 0.071/2) × 125 × 2 × 10
3
= 213.7 neutral pions/s are produced.
Each pion decays into two photons. Therefore number of photons produced
per second = 427
9.15 I /I 0 = exp(−Σx)
Σ = nσ = Nρσ/A
9 Particle Physics – I
9.12 (a) In elastic scattering, total kinetic energy is conserved. Life-time means
mean-life-time i.e., the time in which the population of unstable particles
is reduced by a factor e. The important interactions that the muons undergo
are elastic scattering and the reaction, μ
−
+ p → n + ν μ .
Let the muons travel a distance d metres. Then their intensity will be
reduced due to decay by a factor
I /I 0 = exp(−d/vγ τ 0 )
( 1 )
γ = 1/(1 − v
2
/c
2 )
1/2
= 1/
1 − (10
6
/3 × 10
8 )
2
1/2
= 1.0033
I /I 0 = exp −(d/(1.0033 × 10
6
× 2.2 × 10
−6 ))
= exp −(0.453 d)
( 2 )
The intensity reduction due to interaction will be
I /I 0 = exp(−dΣ)
( 3 )
Σ = σ n = 0.1 × 10
−28
× 2.69 × 10
25
= 2.69 × 10
−4
I /I 0 = exp(−2.69 × 10
−4 ) d
(4)
Comparing (2) and (4), reduction due to decay will be by far greater than
by interaction for any value of d.
(b) Repetition of calculation for T = 4 GeV, γ = 38.7, β ≈ 1, and d =
12,000 m, gives I /I 0 = e
−0.47
= 0.625, so that a large number of muons
survive at the ground level. The distance at which muons of energy 4 GeV
will be reduced by e is given by d = vγ τ 0 = 3×10
8
×38.7×2.2×10
−6
=
2.55 × 10
4 m = 25.5 km.
9.13 The geometrical cross-section is
σ g = π R
2
= π(r 0 A
1/3 )
2
= π r 0
2 A
2/3
= π (1.3 × 10
−13 )
2 (208)
2/3
= 1.86 × 10
−24 cm
2
Number of lead atoms/cm
3
, n = N 0 ρ/A = 6 × 10
23
× 11.3/207 =
3.275 × 10
22
Macroscopic cross-section
Σ = n σ = (3.275 × 10
22 ) (1.86 × 10
−24 ) = 0.06 cm
−1
The interaction length, λ = 1/Σ = 1/0.06 = 16.7 cm
9.14 Number of interactions per second in volume V
I = ΣVI 0
= σ N 0 ρV I 0 /A
= (40 × 10
−27 cm
2 )(6.02 × 10
23
× 0.071/2) × 125 × 2 × 10
3
= 213.7 neutral pions/s are produced.
Each pion decays into two photons. Therefore number of photons produced
per second = 427
9.15 I /I 0 = exp(−Σx)
Σ = nσ = Nρσ/A