Transcript Nuclear charge distributions

16.451 Lecture 7: Understanding the cross-section results 25/9/2003
From last class, the cross section for electron scattering from nuclear charge Z:
2
2
2
2
p
2
d

(

)
4
Z


f
 e 
2


 F (q ) 

  2
2  4  
2
2 2 

d

o  (q   )
 (c) 



4 Z2
(c) 4
 e2 
 4  
o

point charge cross-section:
most notably, falls off as q-4
(units should be fm2)
Check units:
2 (cp ) 2
f
q4
 F (q 2 ) 


2
form factor squared
(dimensionless)
[c]  [e 2 4 o ]  MeV.fm; [cp]  MeV; [q]  fm-1
2
1
(
MeV
)
 d 
2
2

(
MeV.fm
)

fm
 d 
-4
fm

 (MeV.fm) 4
1
What does this function look like?
2 (cp ) 2
2
2
d

(

)
4
Z


f
 e 

 


(c) 4  4  o 
q4
 d 
2
2
2
 F (q 2 )   Z 2  d ( )   F (q 2 ) 



 d  o 
First calculate the point charge cross section for Z = 1:
constants:
point charge
2
4  e2 
4
2
9
2



1
.
44

5
.
45

10
(
M
eV.fm
)


(c) 4  4  o 
(197 .5) 4
For the kinematic factors, we need our results from lecture 5 for the momenta,
being of course very careful with the units:
e
e


pi

pf
pf 

pi
1
pi
M

q
(1  cos )
(q) 2  pi2  ( p f ) 2  2 pi p f cos
Point charge cross section, continued...
3
First, let’s consider a heavy nuclear target, so that pf  pi and work out Z2d/d)o
Choose the example of 248 MeV/c electrons scattering from 208Pb as in Krane, fig. 3.2
pf  pi
6.44 fm2
GeV/c
or
(GeV/c) 2
Kinematic relations become:
p f  pi  constant
(q) 2  2 p 2i (1  cos )
(ħq)2
p 2f
d
~ 4
d
q
1  cos  2 sin 2 ( / 2) ...
208Pb,
pi = 248 MeV/c
 (deg)
const.
1
 d 
~

 
4
q
sin 4 ( / 2)
 d  o
Point charge calculation for 248 MeV/c electrons scattering from 208Pb, Z = 82...
4
const. 0.0567 fm 2 /sr
 d 
Z 

 
4
q
sin 4 ( / 2)
 d  o
2
 d 
2
Z2 
 fm /sr
 d  o
NB, the graph blows up at  = 0.
In this limit, the practical cutoff
is given by -2, a scale set by the size
of the atom....
Log scale! The point
charge cross section
drops
like a stone!!!
 (deg)
How are we doing? Compare to data from Krane:
5
 = 10°,
d/d = 105 b/sr
= 10 fm2/sr ...
Point charge calculation:
d(10°)/d = 982 fm2/sr
d
 Z2
d

 d 
2

 F (q )
 d  o
(F(q2))2 at 10° =
10/982 = 0.01 !!!
Also, the graphs are not smooth like 1/q4 -- evidence that the target has finite size!

2
Summary so far:
6
We can predict the cross-section exactly for a pointlike target with nonrelativistic
quantum mechanics.
(This approach is correct for a target particle that has charge but no magnetic
moment, i.e. intrinsic angular momentum of zero. We can’t use this for the proton
without adding some refinements, so along the way we are stopping to look at the
charge distributions of nuclei. Nuclei with (Z, N) both even, such as 208Pb, have
angular momentum zero, so our theory is perfect for this case!)
Recall our basic result from lecture 6:
2
 d ( ) 
2



  point charge cross  section   F (q ) 


d



Where F(q2) is the Fourier transform
of the target charge density:
F (q 2 ) 

e
 
iq  r
 (r ) d 3 r
Measuring d/d and dividing by the point charge result yields a value for F(q2)
How to find the charge distribution?
form factor:
F (q 2 ) 
inverse Fourier transform:

e
 
iq  r
7
 (r ) d 3 r
 (r ) 
1
(2 )
3

e
 
 iq  r
F (q 2 ) d 3 q
In principle, one could measure the form factor, and numerically integrate to
invert the Fourier transform and find (r).
However, this doesn’t work in practice, because the integral has to be done over
a complete range of q from 0 to ∞, and no experiment can ever span an infinite
range of momentum transfer!
(It is bad enough trying to acquire data at large momentum transfer because the
basic cross-section drops like q-4  the rate of scattered particles into a detector
gets too small – see slide 14, lecture 4)
What to do? ...
A practical solution:
8
Experimental data are fitted to a functional form for F(q2); parameters extracted
from the fit are used to invert the transform and deduce (r)....
Example: elastic electron scattering
from gold (A=197, Z = 79). Best fit to
data is given by solid curves.
(Ref: R. Hofstadter, Electron Scattering
& Nuclear Structure, 1963)
Discontinuities are evidence
of diffraction-like behavior,
characteristic of a Fourier
transform, but the edges are
fuzzy!
Illustration of “Diffraction” behavior in F(q2): Uniform sphere
F (q 2 ) 

e
 
iq  r
2
 (r ) d 3 r 
 
d
0


0
F (q )
2

r 2  (r )dr sin  e
2
9
iqr cos
0
  o , r  R ;
  0, r  R
Density:
F (q 2 ) 
3 (sin x  x cos x)
 x  qR 
In contrast, the minima are not as sharp for nuclei...
d
x3
Zeroes never quite get to
zero on a log scale!
Nuclear charge distributions from experiment:
10
Functional form :

o
1 e
(r  R) / a
R  1.1 A1/ 3 fm
a  0.5 fm
Approx. constant
central density
diffuse surface excludes
sharp diffraction minima
Contrast to the proton (recall lecture 4):
11
Electric charge distribution:
Proton Form factor data:
 (r )  e o exp(  M r )
M  4.33 fm 1
r2

fm 3
e
2


Q
p
2

GE (Q )  1 
2 

0.71 GeV 

2
1/ 2

12
 0.80 fm
M