3D scattering of electrons
from nuclei
Finding the distribution of charge
(protons) and matter in the nucleus
[Sec. 3.3 & 3.4 Dunlap]
The Standford Linear
Accelerator, SLAC
Electron scattering at Stanford 1954 - 57
1961 Nobel
Prize winner
Professor Hofstadter’s group worked here at SLAC during the
1960s and were the first to find out about the charge distribution
of protons in the nucleus – using high energy electron scattering.
c
A linear accelerator LINAC
was used to accelerate the
electrons
Electron scattering experiments at SLAC 1954 - 57
e-
Why use electrons?
• Why not alpha’s or protons or neutrons?
• Why not photons?
Alphas, protons or neutrons have two disadvantages
(1) They are STRONGLY INTERACTING – and the strong force between nucleons is so
mathematically complex (not simple 1/r2) that interpreting the scattering data would
be close to impossible.
(2) They are SIZEABLE particles (being made out of quarks). They have spatial
extent – over ~1F. For this reason any diffraction integral would have to include an
integration over the “probe” particle too.
Photons have a practical disadvantage: They could only be produced at this very high
energy at much greater expense. First you would have to produce high energy
electrons, then convert these into high energy positrons – which then you have to
annihilate. And even then your photon flux would be very low. Energy analysis of
photons after scattering would be also very difficult.
Why use electrons?
• Why not alpha’s or protons or neutrons?
• Why not photons?
Electrons are very nice for probing the nucleus because:
(1) They are ELECTRO-MAGNETICALLY INTERACTING – and the electric force takes
a nice precise mathematical form (1/r2)
(2) They are POINT particles (<10-3 F – probably much smaller). [Like quarks they are
considered to be “fundamental” particles (not composites)]
(3) They are most easily produced and accelerated to high energies
Concept of Cross-section
Case for a single nucleus where
particle projectile is deterministic
Case for multiple nuclei where
projectile path is not known.
The effective area is the all important thing – this is the Cross-Section.
Nuclear unit = 1 b = 1 barn = 10-24cm-2 = 10-28m-2 = 100 F2
Rutherford scattering of negatively
charged particles
2


d
1 2
Zze 2
4 
4 

s0 . csc

.
csc

d 16
2  2(4 0 ) m 02 
2
Alpha scattering
2

d 
Ze 2
4 

csc

d  2(4 0 )m 02 
2
Electron scattering
Rutherford scattering of negatively
charged relativistic particles
Known as Mott scattering
2


d 
Ze
4  
2  

csc
.
1

sin



d  2(4 0 )m 02 
2 
c
2
2
2
0
2
Z<<1
Extra relativistic
kinematic factor
2
2


d  Z .c 

4
2
0

csc .1  2 sin 
2 
d  2m0 
2  c
2

2
02
d  Z .c 
4 
2  

 csc .1  2 sin 
d  2 p0 
2  c
2
e2
(4 0 )c

1
137
Fine structure constant
0  c
Which for extreme relativistic electrons becomes:
2
 Z .c 
d
4 
2 

csc
.
cos
d  2 pc 
2
2
2
2
d

(

c
)
2
4 
2 
2
 Z 2.
csc
cos

Z
f
(

)
d Mott
4T 2
2
2
pc  E  T
More forward directed distribution
Mott Scattering
d
1


 c 
 Z 2 2 .  csc4 cos2
d Mott 4
2
2
T 
2
Mott differential scattering
Take the nucleus to have point charge Ze - e being the charge on the proton.
2
2
d
2
2
2  (c)
4 
2 
2
Z .
csc
cos

Z
f
(

)

Z
.
f
(

)
m
m
d Mott
4T 2
2
2
where f m ( ) is theMottscatteringamplitudeat angle per unit charge
If that charge is spread out then an element of charge d(Ze) at a point r will
give rise to a contribution to the amplitude of
d ( )   (r )d . f m ( ).ei
dΨ
Where  is the extra “optical” phase
introduced by wave scattering by the element
of charge at the point r compared to zero
phase for scattering at r=0
r
But the Nucleus is an Extended Object
Wavefront of incident
electron
 ( )
Wavefront of electron
scattered at angle 
NOTE: All points on plane AA’ have the same
phase when seen by observer at 
Can you see why?
FINDING THE PHASE
Wavefront of incident
electron
 ( )
p

  p.r / 

r
Wavefront of electron
scattered at angle 
rcos
p  2 p sin
The extra path length for P2P2’
 2.OX . sin
The phase difference for P2P2’   2

 q.r
2
 2(k ) sin

2

2
2.OX . sin




2  2k .OX . sin   p r cos   p.r
2


THE DIFFRACTION INTEGRAL
Wavefront of incident
electron
p
Wavefront of electron
scattered at angle 

r
Charge in this volume element is:
d ( )
dq   (r ).d   (r ).r 2 sin .dd
The wave amplitude d at  is given by:
d   (r )r 2dr sin dd.eip.r /  . f ( )
Amount of wave
Phase factor
Mott scattering
THE DIFFRACTION INTEGRAL
The wave amplitude d at  is given by:
d   (r )r 2dr sin dd.eip.r /  . f ( )
Amount of wave
Phase factor
Mott scattering
The total amplitude of wave going at angle  is then:
2
 ( )  f ( ) 


  ( r )e


 
ip .r / 
dV  f ( ) FT 3  (r )
 0 0 r 0
Eq (3.15)
The no of particles scattered at angle  is then proportional to:
 ( )  f ( ) [ FT  (r ) ]
2
From which we find:
d
 FT
d

2
3
3
 ( r ) 
d
 [ F (  p /  )] 2
d
2
d
d
d
d
2
f ( )
2
Mott
Eq (3.14)
Mott
Form Factor F(q)
The effect of diffractive interference
d
d
F (p / )
2
Mott
From nucleus
due to wave
interference
p
Fig 3.6 450 MeV e- on 58Ni
E
p  k 
c
E
450MeV
k

c 197MeV .F
 2.28F 1
Additional Maths for a hard edge nucleus
We can get a fairly good look at the form factor for a nucleus by approximating the
nucleus to a sharp edge sphere:
2


 
1
2
ip .r / 
F (p /  ) 

(
r
)
e
dV

Z 0  0 r0
Z
 
2
ip .r cos / 

(
r
).
r
dr
sin

.
e
d

0 0


F (q)  2   (r ).r 2 dr eiqr cos d (cos )

4
Z
0

  (r )
0
40

Zq
0
sin qr 2
.r dr
qr
R
 r sin qr.dr
0
40 1
sin qR  qR cos qR

Zq q 2

q
3  sin qR  qR cos qR 

qR 
( qR) 2


p

0
0
r=R
3.Z
0 
4R 3
Spherical Bessel Function of order 3/2
F (q) 
3  sin qR  qR cos qR 

qR 
(qR) 2

tan qR  qR
q
p


Condition of zeros

2  2k sin 

2
2 p sin
Wavenumber mom transfer
4.5
7.7
11
14
qR
Fig 3.6 450 MeV e- on 58Ni
1.1xR=4.5
R=4.1F
1.8xR=7.7
R=4.3F
2.6xR=11
R=4.2F
Proton distributions
Mass distributions
 (r )   P (r )   N (r )
 N
  P (r ) 1  
 Z
The Woods-Saxon Formula
 (r ) 
0
1  exp  (r  R0 ) / a 
R0=1.2 x A1/3 (F)
a  0.52  0.01
t is width of the surface region of a nucleus; that is, the distance over which
the density drops from 90% of its central value to 10% of its central value
F
Charge distributions can also be obtained by Inverse
Fourier Transformation of the Form Factor F(q)
F (q) 
d
d nucleus
 FT 3  ( r )
d
d Mott
 (r )  FT 3 F ( q )