402 7 Nuclear Physics – INowR 0 = 1. 44 zZ/T= 1. 44 × 2 × 13 / 4. 5 = 8. 32
Hencermin= 1. 5 × 8. 32 = 12 .48 fm7.25 The minimum distance of approachrminis obtained in the head-on colli-
sion when the initialαparticle energy is entirely converted into the potential
energy.
T=zZe^2 / 4 πε 0 rmin
Puttingrmin=R, the nuclear radius, numerically
T(MeV)= 1. 44 zZ/R(fm)= 1. 44 × 2 × 47 / 7 = 19 .34 MeV
7.26 For the same charges of interacting particles, target thickness and beam inten-
sity and fixed scattering angle, the Rutherford scattering depends inversely as
the square of particle energy. So long as the incident particle is outside the
target nucleus, the Rutherford scattering is expected to be valid. But when
the incident particle touches the nucleus, pure Rutherford scattering would be
invalidated. We can assume that the observed counting rateN 8 is the expected
counting rate for the lowest energy (8 MeV). For higher energyTthe counting
rate is expected to beNE=N 8 (8/T)^2. In the table below are displayed the
calculated counting rates as well as the observed ones.
T(MeV)8 12182226273034
N(Cal) 91,000 40,400 17,970 12,030 8,600 8,000 6,471 5,038
N(Obs) 91,000 40,300 18,000 12,000 8,400 100 12 1.1
Comparison between the calculated and observed counting rates indicates
that Rutherford scattering begins to break down at 26 MeV. Since scattering
angle isθ= 180 ◦, we are concerned with head-on collisions. Hence
R 0 =R= 1. 44 zZ/T= 1. 44 × 2 × 79 / 26 = 8 .75 fm.
Hence the radius of gold nucleus is 8.75 fm.
7.27 When distance of closest approachb=R 0
tan(θ/2)=R 0 / 2 b= 0. 5 =tan 26. 56 ◦
Therefore,θ/ 2 = 26. 56 ◦orθ= 53. 1 ◦
7.28 We work out in the CMS because^7 Li being a light nucleus will recoil in the
encounter. Equating the potential energy at the closest distance of approach
R 0 to the initial kinetic energy,
- 44 zZ/R 0 =μv^2 / 2 =m 1 m 2 v^2 /(m 1 +m 2 )=m 2 T/(m 1 +m 2 )
whereμis the reduced mass. Solving forR 0
R 0 =(1. 44 zZ/T)(1+m 1 /m 2 )=(1.44x2x3/ 0 .5)(1+ 4 /7)= 27 .15 fm
7.29 Givennt= 1. 0 × 1019 nuclei/cm^2
The fractionΔN/N= 1 −
(
πR 02 cot^2 (θ/2).nt)
/ 4 = 1 −(π/4)(1. 44 zZ/T)^2
cot^2 (θ/2).nt = 1 −π(1. 44 × 2 × 79 / 0 .5)^2 cot^2 (30/2)×(1. 0 × 1019 ×
10 −^26 )/ 4 = 0. 82
The factor 10−^26 has been introduced to convert fm into cm^2.