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

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.

1961 Nobel Prize winner

cA 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

Alpha scattering

Electron scattering

2csc.

)4(22csc.

16

1 4

2

200

242

0

m

Zzes

d

d

2csc

)4(24

2

200

2

m

Ze

d

d

Rutherford scattering of negatively charged relativistic particles

2sin1.

2csc

2

. 22

204

2

20

cm

cZ

d

d

137

1

)4( 0

2

c

e

Fine structure constant

2sin1.

2csc

2

. 22

204

2

0

cp

cZ

d

d

Which for extreme relativistic electrons becomes:

2cos.

2csc

2

. 24

2

pc

cZ

d

d c0

2sin1.

2csc

)4(22

2

204

2

200

2

cm

Ze

d

d

Extra relativistic kinematic factor

Z<<1

22242

222 )(

2cos

2csc

4

)(.

fZT

cZ

d

d

Mott

TEpc

More forward directed distribution

Known as Mott scattering

Mott Scattering

2cos

2csc.

4

1 242

22

T

cZ

d

d

Mott

Mott differential scatteringTake the nucleus to have point charge Ze - e being the charge on the proton.

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

im efdrd ).(.)()(

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

dΨ

chargeunit per angleat amplitude scatteringMott theis )( where

)(.)(2

cos2

csc4

)(.

222242

222

m

mmMott

f

fZfZT

cZ

d

d

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?

Wavefront of incident electron

Wavefront of electron scattered at angle

)(

The extra path length for P2P2’ 2

sin..2

OX

2sin)(2

2sin2

kpp

p

The phase difference for P2P2’

rq

rprp

OXkOX

.

.cos

2sin..22

sin..22

/.rp

r

FINDING THE PHASE

rcos

Wavefront of incident electron

Wavefront of electron scattered at angle

)(dp

r

THE DIFFRACTION INTEGRAL

Charge in this volume element is: ddrrdrdq .sin).().( 2

The wave amplitude d at is given by: )(..sin)( /.2 fedddrrrd rpi

Amount of wave

Phase factor Mott scattering

THE DIFFRACTION INTEGRAL The wave amplitude d at is given by: )(..sin)( /.2 fedddrrrd rpi

Amount of wave Phase factor Mott scattering

The no of particles scattered at angle is then proportional to:

2322])([)()( rFTf

From which we find: 2)(f

Form Factor F(q)

Eq (3.14)

The total amplitude of wave going at angle is then:

2

0 0 0

3/. )()()()()(r

rpi rFTfdVerf

Eq (3.15)

Mott

Mott

d

dpF

d

d

d

drFT

d

d

2

23

)]/([

)(

The effect of diffractive interference

p

dd

2)/( pF Mott

From nucleus due to wave interference

Fig 3.6 450 MeV e- on 58Ni

128.2

.197

450

F

FMeV

MeV

c

Ek

c

Ekp

Additional Maths for a hard edge nucleusWe can get a fairly good look at the form factor for a nucleus by approximating the nucleus to a sharp edge sphere:

dedrrr

ZdVer

ZpF

r

rpirpi

2

0 0 0 0 0

/cos.2/. .sin).(2

)(1

)/(

p

q )(cos).(2)(0 0

cos2

dedrrrqF iqr

2

20

0

0

2

0

)(

cossin3

cossin14

.sin4

.sin

)(4

qR

qRqRqR

qR

qRqRqRqZq

drqrrZq

drrqr

qrr

ZR

0

r=R0

30 4

.3

R

Z

2sin22

sin2

tan

)(

cossin3)(

2

kpp

q

qRqR

qR

qRqRqR

qRqF

Condition of zeros

4.5 7.7 11 14

Spherical Bessel Function of order 3/2

qR

Wavenumber mom transfer

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

Z

Nr

rrr

P

NP

1)(

)()()(

The Woods-Saxon Formula

0

0

( )1 exp ( ) /

rr R a

R0=1.2 x A1/3 (F)

0.52 0.01a F

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

)()(

)()(

3

3

qFFTr

rFT

dddd

qF

Mott

nucleus

Charge distributions can also be obtained by Inverse Fourier Transformation of the Form Factor F(q)