1000 Solved Problems in Modern Physics

470 8 Nuclear Physics – II

Ignoring the statistical factor g, at resonance

σt= 7 , 000 × 10 −^24 cm^2 =

λ^2

π

Γs

Γ

(4)

Butλ=

0. 286

√

E

=

0. 286

√

0. 178

= 0. 678 A ̊= 0. 678 × 10 −^8 cm

From (3) and (4) we get

Γs/Γ= 47. 815 × 10 −^5 (5)

Inserting (5) and the value ofσtin (3), we find

σs= 3 .35b

8.71Γ=Γn+Γγ+Γα= 4. 2 + 1. 3 + 2. 7 = 8 .2eV

σ(n,γ)=

λ^2

4 π

ΓγΓn

(E−ER)^2 +Γ

2

4

λ=

0. 286

√

70

= 3. 418 × 10 −^10 cm

ER=60 eV,E=70 eV,Γγ= 1 .3 eV andΓn= 4 .2eV

Ignoring the g - factor, we findσ(n,γ)=1215 b.

σ(n,α)=σ(n,γ).

Γα

Γγ

= 1215 ×

2. 7

1. 3

=2523 b

8.72 Breit–Wigner’s formula is

σtotal=

πλ-^2 ΓsΓg

(E−ER)^2 +Γ

2

4

(1)

For spin zero target nucleus, the statistical factorg = 1. At resonance

energy (1) reduces to

Γs=

πΓσtotal

λ^2

(2)

λ=

0. 286

√

E

A ̊=^0 √.^286

250

= 0. 018 A ̊= 1. 8 × 10 −^10 cm

Substituting,σtotal= 1300 × 10 −^24 cm^2 ,Γ=20 eV andλ= 1. 8 × 10 −^10 cm

in (2), we findΓs= 2 .5eV.

8.3.13 DirectReactions...................................

8.73 Q=[(md+mN)−(mα+mC)]×^931 .44 MeV
=[( 2. 014102 + 14. 003074 )−( 14. 002603 + 12. 0 )]× 931. 44
= 13 .57 MeV