PHY492: Nuclear & Particle Physics. Lecture 4 Nature of the nuclear force. Reminder: Investigate

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1 PHY49: Nuclear & Particle Physics Lecture 4 Nature of the nuclear force Reminder: Investigate

2 Topics to be covered size and shape mass and binding energy charge distribution angular momentum (spin) symmetries (parity) magnetic moments radioactivity energy levels reactions Physics of nuclei Nuclear size Binding energy per nucleon is < 1% of the nucleon mass. The protons and neutrons in the nucleus retain their particle properties. Assume nuclei are spherical and have a constant density (not compressed). r 3 A Nuclear radius r A 1 3 r = ( m) A 1 3 January, 007 Carl Bromberg - Prof. of Physics

3 Nuclear mass and binding energy The mass of a bound system is always less than the mass of its component parts. For example, the mass of the hydrogen atom is 13.5 e/c less than proton mass plus electron mass. When the hydrogen atom is formed, 13.5 e is released in photons. m H c ( m p + m e )c = 13.5 e electron binding energy in hydrogen What is the sign of the binding energy? D&F: the binding energy of a bound system is negative. So be it. Nuclear binding energy B.E. = M ( A, Z )c ( Zm p + Nm n )c where N = A Z B.E. A binding energy per nucleon, ~ the energy to remove one nucleon Sometimes mass excess Δ is given in the tables. Δ = M ( A, Z )c where u = Me, the average A u nucleon mass in 1 C. B.E. = Δ + A u ( Zm p + Nm n )c January, 007 Carl Bromberg - Prof. of Physics 3

4 Nuclear charge distribution via electron scattering Electron scattering Center of Mass Energy squared : s (effective m c 4 of beam + target) t = E t m t c Relativistic invariants: t,s ( ) p t c ( E t p t c ) E m t c + m t c 4 ( ) = m t c T t = = For target in the Lab frame = m t c E m t c s = ( E b + E t ) ( p b + p t ) N.R. Momentum transfer squared : t (difference of final & initial momentum) p t c In lab, target starts at rest, picks up the transferred momentum Feynman diagram momentum of target January, 007 Carl Bromberg - Prof. of Physics 4 t = E b E b e - e - p b, ( p b, E b ) p t, E t A ( ) ( ) ( ) A ( ) ( ) t = E t E t E b ( ) ( pt, ) E t p b p b p t p t

5 Electron scattering to probe nucleus Rutherford scattering (θ) dω θ Z e ( ) = Z 4E q c = t = p c ( 1 cosθ) q = p ( 1 cosθ) θ sin = 1 cosθ dq = p d ( cosθ) dω = πd ( cosθ) 1 sin 4 θ Momentum transfer q Trigonometry Electron scattering Assume only the direction of beam changes so that E, p = E, p ( ) ( ) c t = E b E b p b p b = p b p b c + p b p b c cosθ ( ) = p c 1 cosθ Z = 1 Rutherford electron scattering (q) ( Z ) e ( dq θ) = 4π 1 v q 4 Momentum transfer v = velocity of electron c January, 007 Carl Bromberg - Prof. of Physics 5

6 Point-like scattering Rutherford scattering ( Z ) e dq = 4π v q 4 Mott included effects of electron spin (relativistic quantum mechanics) dq ( θ) Mott = 4 cos θ dq ( θ) Rutherford What happens if the charge distribution is not point-like? In other words, the electron approaches the nuclear surface? dq = F q ( ) dq Mott Nuclear form factor January, 007 Carl Bromberg - Prof. of Physics 6

7 Form factors -nuclear charge distribution Nuclear charge distribution ( ) = Ze f ( r) F( q) = d 3 ρ r Modified differential cross section dq = F q ( ) dq f ( r) F( q ) ( 1 / 4π )δ ( r) 1 ( a 3 8π )exp ar ( a π ) 3 exp r ( ) 1+ q a a 3 4π R (r < R) qr Mott q exp a sin qr ( ) 3 qr cos qr Nuclear form factor r f r iqir / ( )e Fourier transform of distribution January, 007 Carl Bromberg - Prof. of Physics 7

8 Obtain Rutherford scattering via quantum mechanics Assume: Fermi s Golden Rule P = π particles are plane waves (known as the Born Approximation) target recoil energy is neglected particles have spin 0 (electrons are fermions - spin 1/) point like scattering (charge distribution included later) ( ) ψ H ψ ρ E Free particle wave functions: Matrix element Coulomb potential (r) = Z e r Probability per unit time to make a transition from initial state ψ to any one of many final states ψ, with a density of final states ρ(ε ). ψ H ψ = d 3 r ψ H ( r)ψ ψ = 1 ei pir /, ψ = 1 ei p ir / = e d 3 r ( r) e i qir / Z e ψ H ψ = momentum transfer q = p p iqr cosθ / e tricky January, 007 Carl Bromberg - Prof. of Physics 8 ψ = 1 e i p ir / r r drd ( cosθ)dφ

9 Evaluation of the matrix element Z e ψ H ψ = = π Z e iqr cosθ / e r rdr e r drd ( cosθ)dφ iqr cosθ / d ( cosθ) = π Z e 0 rdr ( / qr) e iqr / e ( iqr / ) / i = 4π Z e q = 4π Z e q dr sin ( qr / ) = 4π Z e q 0 Scattering probability per unit time ( ) cos qr / 0 January, 007 Carl Bromberg - Prof. of Physics 9

10 Other factors in golden rule Density of final states ρ(e) Probability ---> cross sections dp x dx = π (each state) ( ) = d 3 ( π) = 3 ( π) 3 dn p d E = v d p ρ E p ( ) = dn ( p ) d E = v p d p dω p dω ( π) 3 = v P P = π H fi v P = v π H fi ρ ( E ) (scatters/second/beam particle/target particle) v = ψ normalization beam velocity ; v = v ρ ( E ) (scatters/second) ( beam/second) targets/unit-area ( ) January, 007 Carl Bromberg - Prof. of Physics 10

11 Fermi s Golden Rule ---> Rutherford scattering = v π H fi ρ ( E ) = π v = 4π Z e q v π 4π Z e p πdq ( π ) 3 v q 4 p ( Z ) e dq = 4π v q 4 p ( π) 3 dω Rutherford scattering Trigonometry q = p ( 1 cosθ) dq = p d cosθ dω = πd cosθ January, 007 Carl Bromberg - Prof. of Physics 11