8.3 Solutions 469
8.68
dσ
dΩ=
I
I 0 NdΩ=A+Bcos^2 θI 0 =^8 ×^1012 /m^2 −s
N=number of target atoms intercepting the beam=N 0 ρt
A=
6. 02 × 1023 × 2. 7 × 10 −^3
27
= 6. 02 × 1019
dΩ=0. 01
62
= 2. 78 × 10 −^4
A+Bcos^230 ◦=50
8 × 108 × 6. 02 × 1019 × 2. 78 × 10 −^4
= 3. 73 × 10 −^24
A+Bcos^245 ◦=40
8 × 108 × 6. 02 × 1019 × 2. 78 × 10 −^4
= 2. 99 × 10 −^24
Solving the above equations we find
A= 1 .57 b/Sr
B= 2 .88 b/Sr8.69σ(total)=
Total number of particles scattered/sec
(beam intensity) (number of target particles within the beam)
As the scattering is assumed to be isotropic total number of particles scattered
=(Observed number) (4π/dΩ)= 15 × 4 π/ 2 × 10 −^3 = 9. 42 × 104 /s
Beam intensity, that is number of beam particles passing through unit area
per second=
beam current
charge on each proton=
10 × 10 −^9 A
1. 6 × 10 −^19 C
= 6. 25 × 1010 /cm^2 sTherefore σ(total) =9. 42 × 104
6. 25 × 1010 × 1. 3 × 1019
= 1. 159 × 10 −^25 cm^2
=116 mb8.3.12 Nuclear Reactions via Compound Nucleus.............
8.70 Breit–Wigner formulae are
σt=πλ-^2 ΓsΓ.g
(E−ER)^2 +Γ
2
4(1)
σs=πλ-^2 Γs^2 .g
(E−ER)^2 +Γ
2
4(2)
Dividing (2) by (1)
σs
σt=
Γs
Γ