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UNIVERSITY OF SOUTHAMPTON PHYS3002W1 SOLUTIONS FOR EXAMINATION PAPER 2012-2013 NUCLEI AND PARTICLES Duration: 120 min This paper contains 0 questions. c University of Southampton Copyright 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 [1] particularly tightly bound. 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, Z P and atomic mass number AP , write down the relation between nuclear mass, the proton mass, the neutron mass, and the binding energy, BEP . In case of α-decay P → D + α, derive the relation between c the Q-value of the released energy, binding energy of the parent particle, BEP , binding energy of the daughter particle, BED , and the binding energy of the α-particle, BαE . For a given nucleide P, the binding energy is the difference between the mass of the protons (m p ) and neutrons (mn ) summed together and the nuclear mass ( M P): [1] BEP = Z P m p + (AP − Z P )mn − M P The Q-value is the difference between the mass of the initial and final state [1] particles. c University of Southampton Copyright 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 = MP − MD − Mα = BED + BαE − BEP . [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 [1] must be overall antisymmetric under interchange of the two nucleons. 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- [1] symmetric. 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 Br(H → γγ) = 2 × 10−3 . 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 c University of Southampton Copyright Page 3 of 13 4 PHYS3002W1 The cross section of the gg → H → γγ is given by [1] σgg→H→γγ = σgg→H × Br(H → γγ) 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 √ [1] N/N = ǫ Therefore the requested total integrated luminosity is equal to L = N/σgg→H→γγ = 100/(20 × 2 × 10−3) = 2500 pb−1 = 2.5 fb−1 [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 their ground state: 15 8 O: 15 8 6 8 O, 4 Be, 3 Li. In case of several possible states, list all possibilities. 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 spinorbit interactions): (1s1/2)2 (1p3/2)4 (1p1/2)1 Therefore there is one unpaired neutron in 1p1/2 state forming state with parity (−1)l=1 and j = 1/2 : (1/2)−. 8 4 Be: [1] Z=4, N=4. Neutrons and protons are paired, so spin of this nucleide is zero and parity is positive: (0)+. 6 3 Li: [1] Z=3, N=3. There are 3 neutrons and protons filling shells as follows: (1s1/2)2(1p3/2)1 . 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. c University of Southampton Copyright [1] Page 4 of 13 5 PHYS3002W1 B1 Solution [2] a) What is meant by a “form-factor”? How does a differential cross-section depend on the form factor? 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 [5] the charge distribution in the nucleus. c) Electrons of energy 1 GeV are scattered off the nuclei, 64 Cu, 108 Ag and 197 Au. 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 [ 11 ] three nuclei and comment on the dependence of the radii on the atomic mass numbers. 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 [1] the projectile is of the order of the nuclear radius. 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/h̄ 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 wavefront 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- [2] tively or destructively. The scattering amplitude for a particular part of the wavefront that passes at a distance 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 deduce information about the charge or nucleon distribution within the nucleus. (Bonus TURN OVER c University of Southampton Copyright Page 5 of 13 6 PHYS3002W1 mark for mentioning that the form-factor is proportional to the Fourier transform of [2] the charge distribution). c) For electrons of energy 1 GeV, the momentum p is (neglecting the electron mass) [1] 1 GeV/c. The momentum transferred, q is given by q = 2p sin(θ/2). 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/h̄ = π, where R is [1] the nuclear radius. Therefore for these three nuclei, the radii are given by h̄c = 3.14 × 197/129. = 4.8fm qc h̄c R = π = 3.14 × 197/108. = 5.7fm qc h̄c R = π = 3.14 × 197/87.2 = 7.0fm qc R = π [3] c University of Southampton Copyright 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 [2] information about the distribution of nucleons within the nucleus. TURN OVER c University of Southampton Copyright Page 7 of 13 8 PHYS3002W1 B2 Solution [2] a) Discuss qualitatively what happens to the shape of a nucleus during nuclear fission. b) Hence explain why spontaneous fission is extremely rare, but fission can be induced by bombarding fissile material [4] with neutrons. c) Explain why fission can be induced by low-energy (cold) neutrons for 235 92 U, whereas neutrons must have a kinetic energy of over 1 MeV in order to induce fission in 235 92 U [4] d) Why is nuclear fission usually followed by β-decay? [3] e) Explain how a chain reaction occurs in nuclear fission. [3] f) Why is it necessary for a sample of fissile material to be greater than a critical mass in order for a chain reaction to occur? [2] g) In a nuclear reactor how is the chain reaction controlled in order to produce a steady output of energy? [2] a) The nucleus stretches in one direction. Then it develops a neck and finally [2] splits into two smaller nuclei. b) As the neutron stretches its surface area increases, thereby increasing its surface [1] energy. This is balanced by the Coulomb repulsion between the two positively charged fission products. Between them this produces a high and wide potential barrier. [1] The amplitude for quantum tunnelling though this barrier is highly suppressed. However, when bombarded with neutrons sufficient energy can by imparted to the [2] nucleus to overcome the potential barrier. c) 235 92 U has an odd number of neutrons. When it absorbs an extra neutron the binding energy increases considerably because of the pairing term between the neutrons. This releases enough energy to enable the fission fragments to go past the potential [2] barrier and fission can occur. c University of Southampton Copyright Page 8 of 13 9 238 92 U PHYS3002W1 already has an even number of neutrons so the extra binding energy when a further neutron is absorbed is insufficient to allow the fission fragments to overcome the potential barrier, so it is necessary that the incident neutron have a kinetic energy [2] of at least 1 MeV in order for the fission to occur. d) Although neutrons are emitted along with the primary fission process, the fission products still usually contain more neutrons than the stable isotope. [2] Such nuclides are unstable against β-decay. [1] e) Typically 3 - 4 neutrons are emitted for each fission process. Each of these can induce a fission process when absorbed other parent nuclei, which then undergo fission and each of these produces 3 - 4 neutrons which induce further fissions. [3] This is a chain reaction. f) The sample must be sufficiently large for the neutrons produced during one fission event to be absorbed by another nucleus, thereby inducing a further fission event, [2] rather than passing straight through the sample. g) Rods of high neutron absorbing material (usually cadmium) are inserted into the [2] fissile fuel. These can be moved in and out to control the reaction rate. TURN OVER c University of Southampton Copyright Page 9 of 13 10 PHYS3002W1 C1 Solution (Bookwork + Seen similar on the problem sheets, c) is a new problem) a) Draw the Feynman Diagrams for e+ e− → W+ W− annihilation. [2] b) Find the minimal energy of the e+ (E ) which would be colliding with e− with the energy twice as low as that (E/2) and producing W + W − pair ( MW = 80GeV/c2 ). For simplicity, you may neglect the mass of the electron as compared to its energy. ~ (with |P| ~ = E/c once we The 4-momentum of the electron is Pe− = (E/c, P) ~ . neglect me ), and the 4-momentum of the incoming positron is Pe+ = (E/(2c), −P/2) [2] 2 2 The invariant mass squared of the e+ e− pair is the (Pe− + Pe+ )2 = (2MW )2 c2 = 4MW c [2] for minimal requested E . Evaluating (Pe− + Pe+ )2 we find 1 9 (Pe− + Pe+ )2 = E 2/c2 − E 2 /c2 = 2E 2/c2 4 4 [2] Therefore E = √ 2MW c2 = 113 GeV [2] c) Write down electric charges of all fundamental fermions. fermions e− , µ− , τ− νe , νm u , τ− u, c, t d, s, b charge -1 0 +2/3 -1/3 [1] [1] [1] [1] c University of Southampton Copyright Page 10 of 13 11 PHYS3002W1 d) Calculate R= σ(e+ e− → hadrons) σ(e+ e− → µ+ µ− ) for center-of-mass energy just below the threshold of “charm” production and for center-of-mass energy just below the threshold of b-quark production. Explain your results. The weak diagram can be ignored. e)Explain how the experimentally measured value of R can be used as evidence for the existence of quark colours. In case of hadrons one needs to calculate a cross section for quarks in the final state, i.e. cross section for e+ e− → qq̄ process and therefore take into account a colour factor for quarks which is equal to 3 as well as EM charges Qi of quarks [2] which come squared into formula for the cross section. Therefore, summing over all quarks produced, we have X σ(e+ e− → hadrons) X σ(e+ e− → qi q̄i → hadrons) = = 3 Q2i R= + − + − + − + − σ(e e → µ µ ) σ(e e → µ µ ) i i [1] Just below “charm” threshold u, d and s quarks can be produces, so R=3 X i Q2i = 3 × 4 1 1 + + 9 9 9 = 2. [1] Just below b-quark threshold u, d , s and c quarks can be produces, so R=3 X i Q2i =3× 4 1 1 4 + + + 9 9 9 9 = 10 . 3 [1] The number of quark colours Nc can be directly measured from the R value which is proportional to Nc given that that Qi charges of quarks are known (measured) from [1] different experiments. TURN OVER c University of Southampton Copyright Page 11 of 13 12 PHYS3002W1 C2 Solution(Seen in the problem sheets and in the notes) c University of Southampton Copyright Page 12 of 13 13 PHYS3002W1 END OF PAPER c University of Southampton Copyright Page 13 of 13