David Ward Lent/Easter Terms 2014 Part II Particle and Nuclear Physics Outline solutions 1. Natural Units; practice in using natural units, and converting between natural and SI units (a) 1 1 λ = =7.15 GeV− =1.41 fm m usinghc ¯ =0.197 GeV fm to change units. (b) We have α 1/137 and √s =91.2 GeV. ≈ 8 2 8 2 11 σ =2.68 10− GeV− =2.68 10− (0.197 fm) =1.04 10− barns = 0.0104 nb ⇒ · · × · (c) Using standard dimensional analysis to insert the missing factors ofh ¯ and c we find: h¯ 4πα2h¯2c2 λ = and σ = mc 3s (noting that α is dimensionless, so equal to 1/137 in any system of units). 2. Semi-Empirical Mass Formula (a) Bookwork. 1 (b) Treat nucleus as uniformly charged sphere of radius R = R0A 3 , charge Ze. Find self- energy by bringing spheres of charge from infinity. Charge of sphere of radius r is 3 2 4 3 Ze(r/R) , and charge of spherical shell between r and r + dr is Ze(4πr dr)/( 3 πR ). So, the self-energy is R 4πr2dr Ze(r/R)3 3Z2e2 Ze 4 3 = Z0 3 πR 4πǫ0r 20πǫ0R Hence, 3e2 aC = =0.72 MeV 20πǫ0R0 (c) The B/A vs A curve peaks around A 56, so for light nuclei fusion can be energetically favourable, while for very heavy nuclei∼ fission is possible. 1 (d) At fixed A the most stable nucleus will be that with lowest mass (not necessarily the ∂M greatest binding energy). So we require =0 ∂Z !A 2aC Z 4aA(A 2Z) (mn mp)+4aA (mp mn)+ 1 − =0 Z = − 1 ⇒ − A 3 − A ⇒ 2aC A− 3 +8aA/A Inserting numerical values for A = 300 gives Z = 114 etc. (e) By analogy with (b), and neglecting the proton-neutron mass difference, we have the gravitational p.e. given by: 2 2 2 2 3GM 3Gm A 5 3Gm n 3 n 37 = 1 = aGA with aG = =5.8 10− MeV − 5R − 5R0A 3 − 5R0 · If neutrons only, N = A and Z = 0, so the binding energy reduces to 2 5 B = a N a N 3 a N + a N 3 V − S − A G which must be > 0 if the ”nucleus” is bound. The aS term is negligible, so we find 2 N 3 > (a a )/a =5 1055. So M > 8 1028 kg which is 0.07 solar masses. A − V G · · 3. The Fermi Gas model Density of states: 3 2 3 1 4√2m R0 g(ǫ)= BAǫ 2 where B = 3πh¯3 Fermi energy is defined by 2 ǫF 3 3 2 2 3N N = g(ǫ)dǫ = BAǫF ǫF = Z0 3 ⇒ 2BA 1 If N = Z = 2 A, neutrons and protons have the same Fermi energy 2 2 3A 3 3 3 ǫF = = = 33.2 MeV 4BA 4B (n.b. non-relativistic) and the corresponding Fermi momentum = √2mǫF = 250 MeV/c. Total KE carried by the neutrons is 5 2 ǫF 2 5 2 3N 3 3 3 3 5 2 3 ǫg(ǫ)dǫ = BAǫF = BA = N Z0 5 5 2BA 5 2BA Adding in the protons similarly, total KE is 2 3 3 3 5 5 (N 3 + Z 3 ) 5 2BA Writing N = 1 A(1 + α) and Z = 1 A(1 α) where α =(N Z)/A, we can expand to second 2 2 − − order in α 5 2 5 5 A 3 10 (N Z) 3 3 (N + Z ) 2+ −2 ≈ 2 " 9 A # 2 So the asymmetry term in the energy is 2 5 3 3 3 A 3 10 (N Z)2 (N Z)2 −2 aA − 5 2BA 2 9 A ≡ A 1 from which we find, after a little algebra, aA = 3 ǫF = 11 MeV. This is about half the fitted value for aA. Pairing energy 1/g(ǫ ) in this model. ∼ F 1 2 3A g(ǫF )= BAǫF = 4ǫF so pairing energy 4ǫF /3A 0.44 MeV, taking A = 100 and ǫF = 33 MeV. For comparison, ∼ ∼3 SEMF fit gives 33.5 MeV/A 4 1 MeV. Again the Fermi gas model is smaller by about a factor of two, because it only takes∼ account of KE, and not of any PE contribution, i.e. the inter-nucleon attraction. 4. Practice in using cross-sections 2 (a) Number of scattering centres per unit area = ρ/MA where ρ = 0.1 kg m− , and MA 5 1 is the mass per nucleus. Rate of scattering = 10 (ρ/MA)σ = 688 s− (since the total 1 cross-section σ = 270.01 b); transmitted intensity = 99312 s− . 1 (b) Rate of fissions = 688 (200/270.01) = 510 s− . × 1 (c) Total rate of elastic scattering = 688 (0.01/270.01) = 0.0255 s− . Area of sphere at 2 × 5 1 2 radius 10 m is 400π m , so flux of neutrons = 0.0255/400π =210− s− m− . 5. Nuclear sizes and Form Factors Start from 2 3 iq r F (q )= d r e · ρ(r) Z For the purpose of integration, take q as the polar axis, hence: 2 2 iqr cos θ′ F (q ) = 2πd(cos θ′) r dr e ρ(r) Z ′ π eiqr cos θ 4π = 2πr2drρ(r) = ∞ r sin qrρ(r)dr Z " iqr #0 q Z0 Then, for qr 1, expand the sin keeping the first two terms: ≪ 4π (qr)3 1 F (q2)= ∞ r qr ρ(r)dr =1 q2R2 q Z0 " − 3! ···# − 6 ··· 15 1 For a 200 MeV electron (ultrarelativistic) E = p = k in natural units, so k =1.02 10 m− . 1 14 1 2 2 2 ×15 q =2k sin θ =1.95 10 m− . So, given F (q ) =0.7 we obtain √R =5 10− m. 2 × | | × Oscillatory structure in F (q2) suggests non-smoothness in ρ(r), i.e. more like a top-hat than a Gaussian form (c.f. Fourier transforms of a Gaussian and a top-hat in 1D.) Best fit is a top-hat with rounded corners (e.g. a Fermi function). 3 6. Mirror Nuclei 1 1 The pair of mirror nuclei have (N,Z)=( 2 (A 1), 2 (A 1)). They have different Coulomb energies; ± ∓ 2 2 aC A +1 A 1 aCA 3αA ∆EC = 1 − = 1 = A 3 " 2 − 2 # A 3 5R 2 1 using aC =3e /(20πǫ0R0)=3α/5R0 and R = R0A 3 . The only other terms which are differ in the SEMF between the mirror nuclei are Z(m +m )+(A Z)m , so M(A, Z +1) M(A, Z)= p e − n − 2me + Emax = ∆EC + mp + me mn ∆EC = Emax +1.8 MeV. Insert numbers to get the − ⇒ 1 answers on the problem sheet. Find R/A 3 is reasonably constant, but higher than the usual value of R0. 7. Scattering in Nuclear Physics Born approximation is based on first order perturbation theory, so it is suitable when the potential is weak. In contrast, the partial wave method is general, but involves the calculation of an infinite number of partial waves. However, if the projectile’s energy is small enough, and/or the potential is sufficiently short range, so that kR 1, then only the ℓ = 0 partial ≪ wave is significant in the neighbourhood of the potential, and the method is quite easy to apply. Of course, if the potential is also weak, then either approach is viable. The question doesn’t imply that V (r) is weak, but we are asked about the low energy limit, so the partial wave approach is the obvious choice. Wave functions are: 2m(V E) 2mV r<b ψ = A sinh qr where q = 0 − 0 s h¯2 ≈ s h¯2 2mE r>b ψ = B sin(kr + δ0) where k = s h¯2 Matching boundary conditions at r = b we have: A sinh qb = B sin(kb + δ0) ; qA cosh qb = kB cos(kb + δ0) Dividing: tanh qb tan(kb + δ ) kb + δ = 0 0 as k, δ 0 q k ≈ k 0 → Hence: δ tanh qb dσ tanh qb 2 f(θ) 0 = b = b − ≡ k q − ⇒ dΩ " q − # 8. Breit-Wigner Formula in Nuclear Scattering We have Γn = 0.0104 eV and Γγ = 0.105 eV, so that Γ = Γn +Γγ = 0.1154 eV, assuming that there are no other decay modes. Hence, σn/σγ =Γn/Γγ, and given σγ = 75 nb, we obtain σn =7.4 kb. The c.m. energy of the neutron is approximately equal to the lab. energy of 2.2 eV, since the 2 11 target is much more massive. We thus infer λ = h /2mE = 1.92 10− m. We then know × everything in the cross-section formula apart fromqg (note that E = E0 at the resonance peak). We find g=0.77 3 , which corresponds to J = 1 for the spin of the compound nucleus. ≈ 4 4 9. Nuclear Shell Model Sequence of energy levels is given in lecture notes. Each has degeneracy (2j + 1). Shell model predicts that the spin and parity is that of the unpaired nucleon. Hence: Z N Unpaired nucleon predicted J P c.f. experiment? 3 1 + 2He 2 1 1s1/2 2 yes 9 3 − 4Be 4 5 1p3/2 2 yes 7 3 − 3Li 34 1p3/2 2 yes 12 + 6 C 66 None 0 yes 13 1 − 7 C67 1p1/2 2 yes 15 1 − 7 N78 1p1/2 2 yes 17 5 + 8 O89 1d5/2 2 yes 23 5 + 11Na 11 12 1d5/2 2 no 131 11 − 54 Xe 77 54 1h11/2 2 no 207 5 − 82 Pb 125 82 2f5/2 2 no 23 The last three disagree with experiment.