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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( 40 ) m 02 2 Alpha scattering 2 d Ze 2 4 csc d 2(40 )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(40 )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 (40 )c 1 137 Fine structure constant 0 c Which for extreme relativistic electrons becomes: 2 d Z .c 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 . csc 4 cos 2 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. d d Z . 2 Mott 2 (c) 2 4T 2 csc 4 2 cos 2 2 Z 2 f m ( ) Z . f m ( ) 2 2 where f m ( ) is the Mott scattering amplitude at 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 ( ).e i 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 .dd The wave amplitude d at is given by: d (r )r 2 dr sin dd.e ip.r / . f ( ) Amount of wave Phase factor Mott scattering THE DIFFRACTION INTEGRAL The wave amplitude d at is given by: d (r )r 2 dr sin dd.e ip.r / . f ( ) Amount of wave Phase factor Mott scattering The total amplitude of wave going at angle is then: 2 ( ) f ( ) ( r )e ip.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 3 (r ) ]2 2 From which we find: d FT d 2 3 ( r ) d [ F ( p / )] 2 d 2 d d d d 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 450 MeV k c 197 MeV .F 2.28 F 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 ip.r / F (p / ) (r )e dV Z 0 0 r 0 Z 2 ip.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 40 Zq 0 sin qr 2 .r dr qr R r sin qr.dr 0 40 1 sin qR qR cos qR Zq q 2 q 3 sin qR qR cos qR 2 qR ( qR ) p 0 0 r=R 3.Z 0 4R 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 )