8.3 Solutions
469
8.68
dσ
dΩ
=
I
I 0 N dΩ
= A + B cos
2
θ
I 0 = 8 × 10
12
/m
2
− s
N = number of target atoms intercepting the beam
=
N 0 ρt
A
=
6.02 × 10
23
× 2.7 × 10
−3
27
= 6.02 × 10
19
dΩ =
0.01
6 2 = 2.78 × 10
−4
A + Bcos
2 30
◦
=
50
8 × 10 8 × 6.02 × 10 19 × 2.78 × 10 −4 = 3.73 × 10
−24
A + Bcos
2 45
◦
=
40
8 × 10 8 × 6.02 × 10 19 × 2.78 × 10 −4 = 2.99 × 10
−24
Solving the above equations we find
A = 1.57 b/Sr
B = 2.88 b/Sr
8.69 σ (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 × 4π/2 × 10
−3
= 9.42 × 10
4
/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 × 10
10
/cm
2 s
Therefore σ (total) =
9.42 × 10
4
6.25 × 10 10 × 1.3 × 10 19 = 1.159 × 10
−25 cm
2
= 116 mb
8.3.12 Nuclear Reactions via Compound Nucleus
8.70 Breit–Wigner formulae are
σ t = π
-
λ
2
Γ s Γ.g
(E − E R ) 2 +
Γ 2
4
(1)
σ s = π
-
λ
2
Γ s
2
.g
(E − E R ) 2 +
Γ 2
4
(2)
Dividing (2) by (1)
σ s
σ t
=
Γ s
Γ
(3)