1000 Solved Problems in Modern Physics

400 7 Nuclear Physics – I

Therefore, Mean free pathλ= 1 /Σ= 0 .2cm

Fraction of scattered particles=Probability of scattering through more than

900 ist/λ= 10 −^5 / 0. 2 = 5 × 10 −^5.

7.19 rmin=R 20

[

1 +

(

+^4 b

2

R^20

) 12 ]

R 0 = 1. 44 zZ/T= 1. 44 × 1 × 50 /(1× 10 −^3 )= 7. 2 × 104 fm

bp=L=

b=/p=c/cp= 197 .3MeVfm/ 1 × 10 −^3 MeV= 1. 973 × 105 fm

b/R 0 = 1. 973 × 105 /(7. 2 × 104 )= 2. 74

rmin=^72.^2 × 104

[

1 +

(

1 + 4 × 2. 742

)^12 ]

= 2. 365 × 105 fm

7.20 The mean free pathλ= 1 /Σ= 1 /N.dσ (1)

whereNis the number of atoms per cm^3

N=N 0 ρ/A= 6. 02 × 1023 × 19. 3 / 197 = 5. 9 × 1022 (2)

dσ=(1. 44 zZ/ 4 T)^2 dΩ/sin^4 (θ/2) (3)

Putz= 1 ,Z= 79 ,θ= 60 ◦,dΩ=area/(distance)^2 = 0. 5 / 102 = 0. 005

Using these values in (3),dσ= 2 .588 fm^2

= 2. 588 × 10 −^26 cm^2 (4)

Using (2) and (4) in (1) we findλ=655 cm,

The probability of scattering at 60◦,

P=t/λ (5)

wheretis the foil thickness

t=pλ=(655× 1 / 5 × 106 )cm= 1. 31 μm.

7.21 The counting rate is dependent on the factor

(zZ/T)^2 (N 0 /A)(ρt/sin^4 (θ/2))

In the problemz,T,(ρt) andθare unchanged. Hence, counting rate with

platinum/counting rate with silver

=(Zpt^2 /Apt)/(ZAg^2 /AAg)= 1. 52

Substituting the known values:ZAg= 47 ,AAg= 108 .87 andApt=195, the

above equation can be solved to yieldZpt= 77 .55 or 78.

7.22 Let the particle of massm 1 , chargez, velocityvand kinetic energyTcollide
elastically with an electron of massm 2 =m. The electron velocity before
and after the collision isv 2 ∗=vcin the CMS. The velocityv 2 ∗is combined
vectorially with the CMS velocityvcto yield the LS velocityv 2 at angleφ
with the incident direction