470
8 Nuclear Physics – II
Ignoring the statistical factor g, at resonance
σ t = 7,000 × 10
−24 cm
2
=
λ
2
π
.
Γ s
Γ
(4)
But λ =
0.286
√
E
=
0.286
√
0.178
= 0.678 ˚
A = 0.678 × 10
−8 cm
From (3) and (4) we get
Γ s /Γ = 47.815 × 10
−5
(5)
Inserting (5) and the value of σ t in (3), we find
σ s = 3.35b
8.71 Γ = Γ n + Γ γ + Γ α = 4.2 + 1.3 + 2.7 = 8.2 eV
σ (n, γ ) =
λ
2
4π
.
Γ γ Γ n
(E − E R ) 2 +
Γ 2
4
λ =
0.286
√
70
= 3.418 × 10
−10 cm
E R = 60 eV, E = 70 eV, Γ γ = 1.3 eV and Γ n = 4.2 eV
Ignoring the g - factor, we find σ (n, γ ) = 1215 b.
σ (n, α) = σ (n, γ ).
Γ α
Γ γ
= 1215 ×
2.7
1.3
= 2523 b
8.72 Breit–Wigner’s formula is
σ total =
π -
λ
2
Γ s Γg
(E − E R ) 2 +
Γ 2
4
(1)
For spin zero target nucleus, the statistical factor g = 1. At resonance
energy (1) reduces to
Γ s =
πΓσ total
λ 2
(2)
λ =
0.286
√
E
˚
A =
0.286
√
250
= 0.018 ˚
A = 1.8 × 10
−10 cm
Substituting, σ total = 1300×10
−24 cm
2
, Γ = 20 eV and λ = 1.8×10
−10 cm
in (2), we find Γ s = 2.5 eV.
8.3.13 Direct Reactions
8.73 Q = [(m d + m N ) − (m α + m C )] × 931.44 MeV
= [(2.014102 + 14.003074) − (14.002603 + 12.0)] × 931.44
= 13.57 MeV