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

Why use electrons? Why not alpha’s or protons or neutrons? Why not photons? Alphas, protons or neutrons have two disadvantages 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. 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: They are ELECTRO-MAGNETICALLY INTERACTING – and the electric force takes a nice precise mathematical form (1/r2) They are POINT particles (<10-3 F – probably much smaller). [Like quarks they are considered to be “fundamental” particles (not composites)] 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

Rutherford scattering of negatively charged relativistic particles Known as Mott scattering Extra relativistic kinematic factor Z<<1 Fine structure constant Which for extreme relativistic electrons becomes: More forward directed distribution

Mott Scattering

Mott differential scattering Take 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 r 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

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 Wavefront of electron scattered at angle  rcos The extra path length for P2P2’ The phase difference for P2P2’

THE DIFFRACTION INTEGRAL Wavefront of incident electron p Wavefront of electron scattered at angle  Charge in this volume element is: The wave amplitude d at  is given by: Amount of wave Phase factor Mott scattering

[ ] THE DIFFRACTION INTEGRAL ( ) d s s d = FT r ( r ) d W d W d s d s The wave amplitude d at  is given by: Amount of wave Phase factor Mott scattering The total amplitude of wave going at angle  is then: Eq (3.15) The no of particles scattered at angle  is then proportional to: From which we find: d s [ ] s ( ) d = FT 3 r 2 ( r ) d W d W Mott d s d s = [ F ( D p / h )] 2 Eq (3.14) d W d W Mott Form Factor F(q)

The effect of diffractive interference Mott From nucleus due to wave interference p

Fig 3.6 450 MeV e- on 58Ni

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: r=R

Spherical Bessel Function of order 3/2 Condition of zeros Wavenumber mom transfer 7.7 qR 4.5 11 14

Fig 3.6 450 MeV e- on 58Ni 1.1xR=4.5 R=4.1F 1.8xR=7.7 R=4.3F

Proton distributions

Mass distributions

The Woods-Saxon Formula R0=1.2 x A1/3 (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

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