1000 Solved Problems in Modern Physics

7.3 Solutions 399

σ=

∫

σ(θ)dΩ=

∫π

0

(R^2 /4). 2 πsinθdθ=πR^2

This is called geometric cross-section.

7.3.2 RutherfordScattering...............................

7.12 Rutherford law for scattered particles in the CMS is

σ(θ∗)=dσ(θ∗)/dΩ∗=

1

16

(

zZe^2

T

) 2

1

sin^4

(θ∗

2

)

Now sin (θ∗/2)=sin((π−φ∗)/2)=cos(φ∗/2)=cosφ

dΩ∗= 2 πsinθ∗dθ∗= 2 πsin(π−φ∗)dφ∗

= 2 πsin 2φ.2dφ= 8 πsinφcosφdφ=4 cosφdΩ(φ)

Therefore dσ(φ)/dΩ(φ)=(zZe^2 /T)^2. 1 /cos^3 φ

7.13 The Rutherford scattering cross-section for scattering between angleθ 1 and
θ 2 is given byσ(θ 1 ,θ 2 )=

(

πR 02 / 4

)

[cot^2 (θ 1 /2)−cot^2 (θ 2 /2)]

whereR 0 =zZe^2 / 4 πε 0 T= 1. 44 zZ/T

Therefore,σ(60

◦, 90 ◦)

σ(90◦, 120 ◦)=(cot

230 ◦−cot 245 ◦)/(cot 245 ◦−cot 260 ◦)= 3 / 1

7.14 Since the atomic number and foil thickness and angles are the same, the prob-
ability for scattering will be inversely proportional to the square of energy
and directly proportional to this square of the charge of the particle. Hence the
probability for scattering of 5 MeVα-particles will be^2

2

102 ×

52

12 ×^10

− (^3) = 10 − 3
7.15 The minimum requirement is that the particle should be able to penetrate the
nucleus of radiusR. Thus
T= 1. 44 zZ/R(fm)= 1. 44 × 2 × 60 / 10 ≈ 17 .3MeV
7.16 tan(θ∗/2)=r(collision)/ 2 b
forb=^1 / 2 r(collision),tan (θ∗/2)= 1 =tan 45◦
Therefore,θ∗= 90 ◦
7.17 Givend=d 20 (1+cosec (θ/2))
Hered=rmin,d 0 =R 0
Therefore,R 0 = 1. 44 zZ/T= 1. 44 × 1 × 82 / 11. 8 =10 fm
Rmin=(10/2)(1+cosec 26◦)= 16 .4fm
7.18 Givenθ=2tan−^1 (a/ 2 b)
Therefore,tan (θ/2)=R 0 /^2 b (1)
R 0 = 1. 44 zZ/T= 1. 44 × 1 × 50 / 0. 5 =144 fm
From (1)
b=(R 0 /2) cot (θ/2)=(144 cot 45^0 )/ 2 =72 fm
Where we have usedθ= 90 ◦
σ(90◦, 180 ◦)=πb^2 =π(72)^2 = 16 ,278 fm^2 = 1. 628 × 10 −^22 cm^2
Macroscopic cross-section
Σ=σN 0 ρ/A= 1. 628 × 10 −^22 × 6. 02 × 1023 × 5 / 100 = 4 .9cm−^1