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PHY492: Nuclear & Particle Physics Lecture 4 Nature of the nuclear force Reminder: Investigate www.nndc.bnl.gov Physics of nuclei Topics to be covered •size and shape •mass and binding energy •charge distribution •angular momentum (spin) •symmetries (parity) •magnetic moments 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). 1 V r3 ⎫ ⎬ r A3 VA⎭ •radioactivity •energy levels •reactions Nuclear radius ( r = 1.2 × 10 January 22, 2007 −15 ) m A 1 3 Carl Bromberg - Prof. of Physics 2 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 eV/c2 less than proton mass plus electron mass. When the hydrogen atom is formed, 13.5 eV is released in photons. mH c 2 − m p + me c 2 = −13.5 eV 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 2 − Zm p + Nmn c 2 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. where u = 931.5 MeV, the average Δ = M ( A, Z ) c 2 − A ⋅ u nucleon “mass” in 12C. ( ) B.E. = Δ + A ⋅ u − Zm p + Nmn c 2 January 22, 2007 Carl Bromberg - Prof. of Physics 3 Nuclear charge distribution via electron scattering Electron scattering Relativistic invariants: t,s Momentum transfer squared : t (difference of final & initial momentum)2 2 2 t = ( Eb′ − Eb ) − ( pb′ − pb ) Center of Mass Energy squared : s ee2 4 (effective m c of beam + target) ⎧ p′b , Eb′ ) ( pb , Eb ) Feynman ( ⎪ 2 2⎪ s = ( E b + E t ) − ( pb + pt ) ⎨ diagram ⎪ ( pt , Et ) ⎪⎩ ( pt′, Et′) A For target in the Lab frame 2 t = Et′ − mt c 2 − pt′2 c 2 2 2 2 = Et′ − pt′ c − 2 E ′mt c 2 + mt 2 c 4 ( ( ) ( ) ) = −2mt c 2 E ′ − mt c 2 = −2mt c 2Tt ′ = − pt′2 c 2 January 22, 2007 N.R. A 2 2 t = ( Et′ − Et ) − ( p′t − pt ) In lab, target starts at rest, picks up the transferred momentum − momentum2 of target Carl Bromberg - Prof. of Physics 4 Electron scattering to probe nucleus Rutherford scattering (θ) Electron scattering Momentum transfer2 2 ⎛ ZZ ′e2 ⎞ dσ 1 (θ ) = ⎜ dΩ ⎝ 4E ⎟⎠ sin 4 θ 2 Assume only the direction of beam changes so that E,|p| = E’,|p’| 2 2 t = ( Eb′ − Eb ) − ( pb′ − pb ) c 2 Momentum transfer q q 2 c 2 = −t = 2 p 2 c 2 (1 − cosθ ) = −2 pb pb′ c 2 + 2 pb pb′ c 2 cosθ = −2 p 2 c 2 (1 − cosθ ) q 2 = 2 p 2 (1 − cosθ ) Trigonometry θ 1 − cosθ = 2 2 dq 2 = −2 p 2 d ( cosθ ) sin 2 dΩ = −2π d ( cosθ ) January 22, 2007 v = velocity of electron ≈ c Z =1 Rutherford electron scattering (q) dσ (θ ) = 2 dq ( 4π Z ′e2 v2 ) 2 1 q4 Carl Bromberg - Prof. of Physics 5 Point-like scattering Rutherford scattering ( dσ 4π Z ′e = 2 dq v2q 4 ) 2 2 Mott included effects of electron spin (relativistic quantum mechanics) ⎡ dσ ⎤ ⎤ 2 θ ⎡ dσ θ = 4 cos θ ( ) ( ) ⎢ dq 2 ⎥ ⎢ dq 2 ⎥ 2 ⎣ ⎦ Mott ⎣ ⎦ Rutherford What happens if the charge distribution is not point-like? In other words, the electron approaches the nuclear surface? dσ 2 ⎡ dσ ⎤ = F (q) ⎢ 2 ⎥ 2 dq ⎣ dq ⎦ Mott Nuclear form factor January 22, 2007 Carl Bromberg - Prof. of Physics 6 Form factors -nuclear charge distribution Nuclear charge distribution Nuclear form factor ρ ( r ) = Ze f ( r ) F ( q ) = ∫ d 3r f ( r ) eiqir / Modified differential cross section Fourier transform of distribution dσ 2 ⎡ dσ ⎤ = F (q) ⎢ 2 ⎥ 2 dq ⎣ dq ⎦ Mott ( ) f (r ) F q2 (1 / 4π )δ ( r ) (a 3 (a 2 8π ) exp ( −ar ) 2π ) 32 ⎛ r2 ⎞ exp ⎜ − 2 ⎟ ⎝ 2a ⎠ 3 (r < R) 4π R 3 January 22, 2007 2 1 ⎛ q2 ⎞ ⎜⎝ 1 + 2a 2 2 ⎟⎠ −2 ⎛ −q 2 ⎞ exp ⎜ 2 2 ⎟ ⎝ 2a ⎠ 3 3 ⎡ qR qR qR ⎤ sin − cos ⎥⎦ ( qR )3 ⎢⎣ Carl Bromberg - Prof. of Physics 7 Obtain Rutherford scattering via quantum mechanics Assume: •particles are plane waves (known as the Born Approximation) •target recoil energy is neglected •particles have spin 0 (electrons are fermions - spin 1/2) •point like scattering (charge distribution included later) Fermi’s Golden Rule Probability per unit time to make a transition from 2 2π initial state ψ to any one of many final states ψ’ , P= ψ ′ H ψ ρ ( E′) with a density of final states ρ(Ε’ ) . tricky 1 ipi r / 1 ip ′ i r / e ,ψ′ = e Free particle wave functions: ψ = V V 1 −ip ′ i r / ∗ 3 ∗ ψ′ = e ψ ′ H ψ = d r ⎡⎣ψ ′ H ( r )ψ ⎤⎦ Matrix V element Coulomb potential V (r) = Z ′e r January 22, 2007 ∫ e 3 iqi r / = ∫ d r V( r ) e V Z ′e2 ψ′ H ψ = V momentum transfer q = p′ − p eiqr cos θ / 2 ∫ r r drd ( cosθ ) dφ Carl Bromberg - Prof. of Physics 8 Evaluation of the matrix element Z ′e2 eiqr cos θ / 2 ψ′ H ψ = r drd ( cosθ ) dφ ∫ V r ∞ 1 2π Z ′e2 iqr cos θ / = rdr e d ( cosθ ) ∫ ∫ V 0 −1 ∞ 2π Z ′e2 iqr / −iqr / ⎡ = rdr 2 / qr e − e / 2i ⎤⎦ ( ) ∫ ⎣ V 0 ( ) ∞ 4π Z ′e2 4π Z ′e2 2 ∞ = dr sin ( qr / ) = − ⎡⎣ cos ( qr / ) ⎤⎦ 0 2 ∫ qV 0 qV 4π Z ′e2 2 = q 2V January 22, 2007 Scattering probability per unit time Carl Bromberg - Prof. of Physics 9 Other factors in golden rule Density of final states ρ(E) Probability ---> cross sections 2 2π P= H fi ρ ( E ′ ) dpx dx = 2π (each state) dn ( p′ ) = Vd 3 p′ ( 2π ) 3 = Vp′ 2 dp′dΩ ( 2π ) (scatters/second/beam particle/target particle) 3 dE ′ = v′dp′ V ψ normalization = ; v = v′ v′ beam velocity dn ( p′ ) V p′ 2 dΩ ρ ( E′) = = dE ′ v′ ( 2π )3 2 V V 2π P= H fi ρ ( E ′ ) v v V dσ = P v January 22, 2007 (scatters/second) ( beam/second ) ( targets/unit-area ) Carl Bromberg - Prof. of Physics 10 Fermi’s “Golden Rule” ---> Rutherford scattering 2 V 2π dσ = H fi ρ ( E ′ ) v 2π V ⎡ 4π Z ′e ⎤ ⎡ V p′ 2 ⎤ dσ = ⎢ ⎢ ⎥ 3 ⎥ dΩ 2 v ⎣ q V ⎦ ⎢⎣ v ( 2π ) ⎥⎦ 2 2 2 2 2π ⎡⎣ 4π Z ′e ⎤⎦ p′ 2 ⎛ π dq 2 ⎞ = ( 2π )3 v 2 q 4 ⎜⎝ p′ 2 ⎟⎠ 2 ( dσ 4π Z ′e = 2 dq v2q 4 January 22, 2007 ) 2 2 Trigonometry q 2 = 2 p 2 (1 − cosθ ) dq 2 = −2 p 2 d cosθ dΩ = −2π d cosθ Rutherford scattering Carl Bromberg - Prof. of Physics 11