1000 Solved Problems in Modern Physics

8.3 Solutions 469

8.68

dσ

dΩ

=

I

I 0 NdΩ

=A+Bcos^2 θ

I 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/Sr

8.69σ(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 s

Therefore σ(total) =

9. 42 × 104

6. 25 × 1010 × 1. 3 × 1019

= 1. 159 × 10 −^25 cm^2

=116 mb

8.3.12 Nuclear Reactions via Compound Nucleus.............

8.70 Breit–Wigner formulae are

σt=π

λ-^2 ΓsΓ.g

(E−ER)^2 +Γ

2

4

(1)

σs=π

λ-^2 Γs^2 .g

(E−ER)^2 +Γ

2

4

(2)

Dividing (2) by (1)

σs

σt

=

Γs

Γ

(3)