UNIVERSITY OF SOUTHAMPTON PHYS3002W1 SOLUTIONS FOR EXAMINATION PAPER 2012-2013 NUCLEI AND PARTICLES Duration: 120 min This paper contains 0 questions. Copyright c University of Southampton Page 1 of 13 2 PHYS3002W1 Section A A1 Solution (Seen in the notes) What is meant by “Magic Numbers”? Give three pieces of experimental evidence for the existence of Magic Numbers. Nuclides with atomic number or neutron number equal to a Magic Number are particularly tightly bound. [ 1 ] Evidence for Magic Numbers (any three marks [1]): • Large separation energy of last proton or neutron. • Low neutron absorption cross-section • High excitation energy above ground state • Larger numbers of stable isotopes if Z is magic, and larger numbers of isotones if N is magic. • larger natural abundances of elements with Z magic A2 Solution (partly new, partly seen in the notes and problem sheets) For a nuclide P with atomic number, ZP and atomic mass number AP, write down the relation between nuclear mass, the proton mass, the neutron mass, and the binding energy, BE. In case of α-decay P D + α, derive the relation between P → E c the Q-value of the released energy, binding energy of the parent particle, BP, binding energy of the daughter particle, E E BD, and the binding energy of the α-particle, Bα . For a given nucleide P, the binding energy is the difference between the mass of P the protons (mp) and neutrons (mn) summed together and the nuclear mass (M ): BE = ZPm + (AP ZP)m MP [ 1 ] P p − n − The Q-value is the difference between the mass of the initial and final state particles. [ 1 ] Copyright c University of Southampton Page 2 of 13 3 PHYS3002W1 Since the number of the initial and final state protons (and neutrons) is the same, the Q-value becomes equal to the difference of just binding energies: Q = M M M = BE + BE BE P − D − α D α − P [ 2 ] . A3 Solution (seen in the notes) The deuteron is a bound state of a proton and a neutron. There are no bound states of two protons or two neutrons. Use this information and consideration of the symmetry properties of the deuteron wavefunction to deduce the spin of the deuteron in its ground-state. The wavefunction for deuteron may be written as ΨD = Ψ(r)χS χI which is the product of spatial, spin and Isospin parts of the wavefunction. The ΨD must be overall antisymmetric under interchange of the two nucleons. [ 1 ] The absence of two-proton or two-neutron bound states implies that the deuteron is an isosinglet, I = 0. Therefore the isospin part of the wavefunction, χI is anti- symmetric. [ 1 ] For a two-body system in the ground state, the orbital angular momentum, ℓ is zero, so the spatial part of the wavefunction Ψ(r) is symmetric which is defined by ( 1)ℓ factor. [ 1 ] − This means that the spin part of the wavefunction, χS , must be symmetric, which implies that the ground state has a spin S = 1. A4 Solution (new problem) The production cross section of the Higgs boson (H) in the gluon-gluon fusion process at the Large Hadron collider is σgg H = 20 pb for the Higgs boson mass 125 GeV. The decay branching ratio of the Higgs boson into two photons is → 3 Br(H γγ) = 2 10− . What is the total integrated luminosity is required to measure the cross section of gg H γγ → × → → process with the accuracy 10%? For simplicity you can neglect background processes. TURN OVER Copyright c University of Southampton Page 3 of 13 4 PHYS3002W1 The cross section of the gg H γγ is given by → → [ ] σgg H γγ = σgg H Br(H γγ) 1 → → → × → The number of events, N, integrated luminosity L and the cross section σgg H γγ → → are related as follows: [ 1 ] N = L σgg H γγ × → → To measure the cross section of the Higgs boson production in the gg H γγ → → channel with the accuracy ǫ = 0.1 (10%) we should observe N = 100 such events since an accuracy and a number of events are related as √N/N = ǫ [ 1 ] Therefore the requested total integrated luminosity is equal to 3 1 1 L = N/σgg H γγ = 100/(20 2 10− ) = 2500 pb− = 2.5 fb− → → × × [ 1 ] A5 Solution (analogous seen in the problem sheets) Given that first two shells in the Shell Model are 1s,1p and 1d shells, determine spin and parities of the following nuclei in 15 8 6 their ground state: 8 O, 4Be, 3Li. In case of several possible states, list all possibilities. 15 8 O: Z=8, N=7. There are seven neutrons and eight protons. All protons are paired, so spin and parity are defined by unpaired neutron. Neutrons fill shells as follows ( with j = l + 1/2 fills before j = l 1/2 due to specific features of the spin- − 2 4 1 orbit interactions): (1s1/2) (1p3/2) (1p1/2) Therefore there is one unpaired neutron in 1p state forming state with parity ( 1)l=1 and j = 1/2 : (1/2) . [ 1 ] 1/2 − − 8 4Be: Z=4, N=4. Neutrons and protons are paired, so spin of this nucleide is zero and parity is positive: (0)+. [ 1 ] 6 3Li: Z=3, N=3. There are 3 neutrons and protons filling shells as follows: 2 1 (1s1/2) (1p3/2) . Therefore there is one unpaired proton in 1p3/2 state and one unpaired neutron in the same state which can form (0)+, (1)+, (2)+ and (3)+ states. The + parity is defined by ( 1)1+1 since each nucleon is in 1p state. [ 1 ] − Copyright c University of Southampton Page 4 of 13 5 PHYS3002W1 B1 Solution a) What is meant by a “form-factor”? How does a differential cross-section depend on the form factor? [ 2 ] b) Explain qualitatively why the scattering cross-section for scattering of sufficiently high energy electrons off a nucleus displays diffraction maxima and minima, and how this diffraction pattern can be used to obtain information about the charge distribution in the nucleus. [ 5 ] c) Electrons of energy 1 GeV are scattered off the nuclei, 64Cu, 108Ag and 197Au. The first diffraction minima are observed at scattering angles 7.40, 6.20 and 5.00 for the three nuclei respectively. Write down the relation between the momentum transferred in the scattering process and the scattering angle. Hence make an estimate of the radii of the three nuclei and comment on the dependence of the radii on the atomic mass numbers. [ 11 ] d) What further information about a nucleus can be obtained from scattering using neutrons rather than electrons? [ 2 ] a) A form factor, F(θ) is a factor that multiplies the scattering amplitude in order to account for the non-zero size of the nucleus. [ 1 ] The cross section is proportional to F(θ) 2 [ 1 ] | | b) Diffraction occurs in a scattering experiment when the de Broglie wavelength of the projectile is of the order of the nuclear radius. [ 1 ] The part of the wavefront that passes at a distance r from the centre of the nucleus and is scattered through an angle has a phase difference qr/h¯ relative to the part of the wavefront that passes though the centre, where q is the momentum change of the scattered particle. The total amplitude is the sum of all such parts of the wave- front and maxima and minima are observed for values of q (which depends on the scattering angle ) for which the phases of the parts of the wavefront sum construc- tively or destructively. [ 2 ] The scattering amplitude for a particular part of the wavefront that passes at a dis- tance r from the centre of the nucleus is proportional of the charge density (in the case of electron scattering) or the total nucleon density (in the case of scattering by strongly interacting particles) and so the Diffraction pattern can be used to de- duce information about the charge or nucleon distribution within the nucleus. (Bonus TURN OVER Copyright c University of Southampton Page 5 of 13 6 PHYS3002W1 mark for mentioning that the form-factor is proportional to the Fourier transform of the charge distribution). [ 2 ] c) For electrons of energy 1 GeV, the momentum p is (neglecting the electron mass) 1 GeV/c. The momentum transferred, q is given by q = 2p sin(θ/2). [ 1 ] So for the angles 7.40, 6.20 and 5.00 the corresponding values of q are q = 2000 sin(7.40/2) = 129.MeV/c q = 2000 sin(6.20/2) = 108.MeV/c q = 2000 sin(5.00/2) = 87.2MeV/c [ 3 ] To a good approximation the first minimum occurs when qR/h¯ = π, where R is the nuclear radius. [ 1 ] Therefore for these three nuclei, the radii are given by hc¯ R = π = 3.14 197/129. = 4.8fm qc × hc¯ R = π = 3.14 197/108. = 5.7fm qc × hc¯ R = π = 3.14 197/87.2 = 7.0fm qc × [ 3 ] Copyright c University of Southampton Page 6 of 13 7 PHYS3002W1 The radius increases as the atomic mass number, A, increases. This is expected as we expect the volume of the nucleus to be proportional to A so that the radius is proportional to A1/3. [ 1 ] The radii are approximately in the ratio 641/3 : 1081/3 : 1971/3 = 4.8:5.7:7.0 [ 2 ] d) Neutrons do not probe the electromagnetic forces within the nucleus, but they do probe the strong forces, so that the Diffraction pattern can be used to obtain information about the distribution of nucleons within the nucleus.