9.3 Solutions 505
9.7 If a particle m 1 moving with total energyE 1 collides with the target of mass
m 2 , then the total energy in the CMS will be
E∗=
(
m 12 +m 22 + 2 m 2 E 1) 1 / 2
IfE 1 >>m 1 orm 2 , thenE∗∝√
E 1
9.8 ECM=E∗≈(2mELab)^1 /^2 =(2× 0. 511 × 10 −^3 ×50)^1 /^2
= 0 .226 GeV=226 MeV
σ= 4 πα^2 ^2 c^2 /3(ECM)^2
=(4π/ 1372 )×(197.3MeV-fm)^2 /(3× 2262 )= 1. 7 × 10 −^4 fm^2
= 1. 7 × 10 −^30 cm^2
Macroscopic cross-section per atom
Σ=σN 0 ρ/A= 1. 7 × 10 −^30 × 6. 02 × 1023 × 11. 4 / 207 = 0. 5636 × 10 −^7 cm−^1
Σe=Z.Σatom= 82 × 0. 5636 × 10 −^7 = 4. 62 × 10 −^6 cm−^1
Interaction mean free pathλ= 1 /Σe= 1 / 4. 62 × 10 −^6 = 2. 16 × 105 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
IfEis the positron energy required in the LS, then
E=E∗^2 / 2 me=(1.876)^2 / 2 × 0. 511 × 10 −^3 = 3 ,444 GeV
(b) Energy requirements are drastically reduced
(c) Each beam of energymp−me≈938 MeV, need to be oppositely directed.
9.3.3 Interaction........................................
9.10 In the collision of a particle of massm 1 of total energyE 1 withmat rest, the
centre-of-mass energy is
E∗=(
m 12 +m 22 + 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 reso-
nanceΔ(1,236 MeV) also can not be produced.9.11σ= 10 −^42 E= 10 −^42 × 200 = 2 × 10 −^40 m^2
Number of nucleons/m^3 ,n=ρ/mp= 7 , 900 / 1. 67 × 10 −^27 = 4. 7 × 1030
Σ=nσ= 4. 7 × 1030 × 2 × 10 −^40 = 9. 4 × 10 −^10 m−^1
The mean free pathλ= 1 /Σ= 1 / 9. 4 × 10 −^10 = 1. 06 × 109 m.
If t is the thickness of iron then the probability for interaction
P=t/λ=t/ 1. 06 × 109 = 1 / 109
Required thickness of iron,t≈ 1 .0m.