Nuclear Systematics and Rutherford scattering
Terminology
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Atomic number (Z) is the number of protons in the nucleus of an atom, and
also the number of electrons in a neutral atom
Nucleon: proton (Z) or neutron (N)
Nuclide: nucleus uniquely specified by the values of N & Z
Mass number (A) is the total number of nucleons in a nucleus (A=Z+N)
Isotopes: nuclides with the same protons (Z) e.g. 235U and 238U
Isotones: nuclides with the same neutrons (N) e.g. 2H (d) and 3He
Isobars: nuclides with the same A
Atomic mass unit (u): one-twelfth of the mass of a neutral atom of 12C (six
protons, six neutrons, and six electrons). 1 u = 1.66 x 10–27 kg = 931.5 MeV/c2
Atomic mass is the mass of a neutral atom and includes the masses of
protons, neutrons, and electrons as well as all the binding energy.
Nuclear mass is the mass of the nucleus and includes the masses of the
protons and neutrons as well as the nuclear binding energy, but does not
include the mass of the atomic electrons or electronic binding energy.
Radioisotopes: members of a family of unstable nuclides with a common value
of Z
Units
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SI units are fine for macroscopic objects like
footballs but are very inconvenient for nuclei and
particles  use natural units.
Energy: 1 eV = energy gained by electron in being
accelerated by 1V.
Mass: MeV/c2 (or GeV/c2)
1 MeV/c2 = 1.78X10-30 kg. 1 GeV/c2 = 1.78X10-27 kg.
Or use Atomic Mass Unit defined by mass of 12C= 12 u
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Momentum: MeV/c (or GeV/c)
1 eV/c = e/c kg m s-1
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Cross sections:
1 barn =10-28 m2
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Length: fermi (fm)
1 fm = 10-15 m.
By Tony Weidberg
Properties of nucleons
http://www.phys.unsw.edu.au/PHYS3050
Chart of the Nuclides
Proton number
Z=N
* Unstable nuclei on either side of
stable ones
* Small Z tendency for Z=N
* Large Z characterized by N>Z
Neutron number
Behaviour of nucleus is determined by the combinations of protons and neutrons
Behaviour of nucleus is determined by strong and electromagnetic interactions
At first sight we would expect the more neutrons the more strongly bound
the nucleus – but in fact there is a tendency for Z = N
Further from Z=N, the more unstable the nuclide becomes
The number of unstable nuclei is around 2000 but is always increasing.
* Tendency for even Z – even N to be the most stable nuclei
* Even – Odd, and Odd – Even configurations are equally likely
* Almost no Odd Z – Odd N are stable, and these are interesting
small nuclei such as 2 H , 6 Li , 10 B , 14C
1
1
3
3
5
5
7
7
Rutherford classified radioactivity
Awarded Nobel prize in chemistry 1908
“for investigations into the disintegration of
the elements and the chemistry of
radioactive substances”
Together with Geiger and Marsden
scattered alpha particles from atomic
nuclei and produced the theory of
“Rutherford Scattering”.
Postulated the existence of the neutron
Ernest Rutherford (1871-1937)
The Geiger – Marsden experiment
The Geiger – Marsden Experiment
glass
214Po
Au
microscope
ZnS
Lead lining
Their data
The Rutherford Cross – Section Formula
∆p
p
p
p  2 p. sin  / 2
vo
m
s
Secret of deriving the formula quickly is to express the momentum transfer in
two ways. The first way you can see in the top parallelogram diagram.
p  2m 0 sin

2
(1)
The Rutherford Cross – Section Formula
φ
The second way is by integrating the force on the alpha across the
trajectory: since we know by Newton’s law and Coulomb’s law:

 dp
Zze 2
F

rˆ
2
dt (4 0 )r
p  
t2
t1
Zze 2
Zze 2 t2
dt
cos.dt 
cos. 2
2

t
(4 0 )r
(4 0 ) 1
r
p  
t2
t1
Zze 2
Zze 2 t2
dt
cos.dt 
cos. 2
2

(4 0 )r
(4 0 ) t1
r
Eqn 3.5
At first sight this integral looks impossible because both  and r are both functions of t.
However the conservation of angular momentum helps:
d
m 0b  mr
dt
Eqn 3.1
2
From which one sees that:
dt
1

.d
2
r
 0b
So that (3.5) becomes
2  2 
Zze 2
p 
cos  .d
  



(4 0 ) 0b 2 2
Zze2

2
cos
(40 )0b
2
Now we can equate this with the first method of finding Δp

Zze 2

p  2m0 . sin  2
. cos
2
(4 0 )b
2
This allows us to get the impact parameter b as a function of θ
Zze 2
 1
Zze 2
 1

b
cot

cot

s
cot
0
(4 0 )m02
2 2 (4 0 ) 12 m02
2 2
2
Where S0 is the distance of closest approach for head on collision
1
m 02
2
Coulomb potential
Zze 2
V (r ) 
(4 0 )r
so
* All particles scattered by more than some value of  must have impact
parameters less than b. So that cross-section for scattering into any angle
greater than  must be:
1

Eqn(3.9)
  b 2   s02 cot 2
4
2
b
Particles fly off into solid angle given by:
d  2 sin d
The differential scattering cross-section is defined as:
d
d d
d 1
1
2
2  



s
cot
0
d
d d  d  4
2  2 sin 
1

1
1
1
 s02 .2 cot .
. .
4
2 sin 2  2 4 sin  cos 
2
2
2
1 2

1 2
1

s0 csc 4 
s0

16
2 16
sin 4
2

Distance of closest approach
d
0
o
o
2
2
b
Zze


2
2
1
1
 vo 
2 m vo  2 m
(4 0 )d
d 
The distance of closest approach “d” will be determined by:
2
2
b
Zze


2
2
1
1
m
v

m
v


o
o 
2
2
(4 0 )d
d 
Using:
Zze 2
s0 
(4 0 ) 2 . 12 m v02
1
 
& b  s0 cot 
2
2
2
2
2 b 
2  s0 
1
1
1
mv

mv

mv
0
0 
 2 0 
2
2
d 
d
2
 b   s0 
1    
d   d 
2
2
2
d d b d d 1
 
                .cot 2    0
2
 s0   s0   s0   s0   s0  4
Solution of this quadratic gives:
d 1
 
 1  cosec  
s0 2 
 2 
we get -
Failure of the Rutherford Formula
Increasing energy and constant angle
Increasing angle and constant energy
Failure of the formula occurs because the distance of closest approach is less than the
diameter of the nucleus. This can happen if (a) the angle of scatter is large or (b) the
energy of the particle is large enough. With Alpha particles from radioactive sources this
is difficult. But with those from accelerators it becomes possible to touch the nucleus
and find out its size because the distance of closest approach is given by:
1 

d  s0 1  cosec 
2 
2