9.3 Solutions
505
9.7 If a particle m 1 moving with total energy E 1 collides with the target of mass
m 2 , then the total energy in the CMS will be
E
∗
=
m 1
2
+ m 2
2
+ 2 m 2 E 1
1/2
If E 1 >> m 1 or m 2 , then E
∗
∝
√
E 1
9.8 E CM = E
∗
≈ (2mE Lab )
1/2
= (2 × 0.511 × 10
−3
× 50)
1/2
= 0.226 GeV = 226 MeV
σ = 4πα
2
2 c
2
/3(E CM )
2
= (4π/137
2 ) × (197.3 MeV-fm)
2
/(3 × 226
2 ) = 1.7 × 10
−4 fm
2
= 1.7 × 10
−30 cm
2
Macroscopic cross-section per atom
Σ = σ N 0 ρ/A = 1.7 × 10
−30
× 6.02 × 10
23
× 11.4/207 = 0.5636 × 10
−7 cm
−1
Σ e = Z .Σ atom = 82 × 0.5636 × 10
−7
= 4.62 × 10
−6 cm
−1
Interaction mean free path λ = 1/Σ e = 1/4.62 × 10
−6
= 2.16 × 10
5 cm
= 2.16 km
9.9 (a) Minimum energy required in the C.M.S. is
E
∗
= M + M = 2 × 0.938 = 1.876 GeV
If E is the positron energy required in the LS, then
E = E
∗2
/2m e = (1.876)
2
/2 × 0.511 × 10
−3
= 3,444 GeV
(b) Energy requirements are drastically reduced
(c) Each beam of energy m p − m e ≈ 938 MeV, need to be oppositely directed.
9.3.3 Interaction
9.10 In the collision of a particle of mass m 1 of total energy E 1 with m at rest, the
centre-of-mass energy is
E
∗
=
m 1
2
+ m 2
2
+ 2 m 2 E 1
1/2
(1)
Here, m 1 = 938 MeV, m 2 = 938 MeV, E 1 = 938 + 200 = 1,138 MeV
Using these values in (1), we find E
∗
= 1973 MeV
Useful energy for particle production
= E
∗
− (m 1 + m 2 ) = 1, 973 − (938 + 938) = 97 MeV
Pion can not be produced as the threshold energy is 290 MeV. Muons and
electron – positron pairs are not produced in strong interactions. The resonance Δ(1, 236 MeV) also can not be produced.
9.11 σ = 10
−42 E = 10
−42
× 200 = 2 × 10
−40 m
2
Number of nucleons/m
3
, n = ρ/m p = 7, 900/1.67 × 10
−27
= 4.7 × 10
30
Σ = nσ = 4.7 × 10
30
× 2 × 10
−40
= 9.4 × 10
−10 m
−1
The mean free path λ = 1/Σ = 1/9.4 × 10
−10
= 1.06 × 10
9 m.
If t is the thickness of iron then the probability for interaction
P = t/λ = t/1.06 × 10
9
= 1/10
9
Required thickness of iron, t ≈ 1.0 m.
505
9.7 If a particle m 1 moving with total energy E 1 collides with the target of mass
m 2 , then the total energy in the CMS will be
E
∗
=
m 1
2
+ m 2
2
+ 2 m 2 E 1
1/2
If E 1 >> m 1 or m 2 , then E
∗
∝
√
E 1
9.8 E CM = E
∗
≈ (2mE Lab )
1/2
= (2 × 0.511 × 10
−3
× 50)
1/2
= 0.226 GeV = 226 MeV
σ = 4πα
2
2 c
2
/3(E CM )
2
= (4π/137
2 ) × (197.3 MeV-fm)
2
/(3 × 226
2 ) = 1.7 × 10
−4 fm
2
= 1.7 × 10
−30 cm
2
Macroscopic cross-section per atom
Σ = σ N 0 ρ/A = 1.7 × 10
−30
× 6.02 × 10
23
× 11.4/207 = 0.5636 × 10
−7 cm
−1
Σ e = Z .Σ atom = 82 × 0.5636 × 10
−7
= 4.62 × 10
−6 cm
−1
Interaction mean free path λ = 1/Σ e = 1/4.62 × 10
−6
= 2.16 × 10
5 cm
= 2.16 km
9.9 (a) Minimum energy required in the C.M.S. is
E
∗
= M + M = 2 × 0.938 = 1.876 GeV
If E is the positron energy required in the LS, then
E = E
∗2
/2m e = (1.876)
2
/2 × 0.511 × 10
−3
= 3,444 GeV
(b) Energy requirements are drastically reduced
(c) Each beam of energy m p − m e ≈ 938 MeV, need to be oppositely directed.
9.3.3 Interaction
9.10 In the collision of a particle of mass m 1 of total energy E 1 with m at rest, the
centre-of-mass energy is
E
∗
=
m 1
2
+ m 2
2
+ 2 m 2 E 1
1/2
(1)
Here, m 1 = 938 MeV, m 2 = 938 MeV, E 1 = 938 + 200 = 1,138 MeV
Using these values in (1), we find E
∗
= 1973 MeV
Useful energy for particle production
= E
∗
− (m 1 + m 2 ) = 1, 973 − (938 + 938) = 97 MeV
Pion can not be produced as the threshold energy is 290 MeV. Muons and
electron – positron pairs are not produced in strong interactions. The resonance Δ(1, 236 MeV) also can not be produced.
9.11 σ = 10
−42 E = 10
−42
× 200 = 2 × 10
−40 m
2
Number of nucleons/m
3
, n = ρ/m p = 7, 900/1.67 × 10
−27
= 4.7 × 10
30
Σ = nσ = 4.7 × 10
30
× 2 × 10
−40
= 9.4 × 10
−10 m
−1
The mean free path λ = 1/Σ = 1/9.4 × 10
−10
= 1.06 × 10
9 m.
If t is the thickness of iron then the probability for interaction
P = t/λ = t/1.06 × 10
9
= 1/10
9
Required thickness of iron, t ≈ 1.0 m.