Tutorial 1: SubAtomicPhysics: Nuclear Physics

6) In odd-odd nuclei, an interaction between the last odd neutron and the last odd proton must be takeninto account in order to explain the ground state spins. The coupling favours parallel spins of the oddproton and the odd neutron. On this basis determine the ground state spin and parity for the 14 N(Z=7) nucleus, giving a sketch of the occupation of levels as prescribed by the shell model.7) The rotational model for permanently deformed nuclei predicts excitation energies given byE(J)=J(J+1)ħ 2 / 2I with J=the spin of the nuclear state and I=moment of inertia of the nucleus aboutthe axis of rotation. The energies and spins of the first two excited states of 180 Hf areEnergy (MeV) J0.093 20.309 4Determine whether these values agree with the rotational model and if soa) Predict the energy of the third excited state (J=6) andb) Calculate the moment of inertia for the nucleus (ħ =6.582x10 -22 MeVs)Tutorial 4:SubAtomicPhysics: Nuclear Physics1) Calculate the Q value for 228 Th(Z=90) (mass=228.0287u) to emita) An α particleb) A protonFor the first case calculate the kinetic energies of the alpha particle and the daughter nucleus.224 Ra(Z=88) = 224.0202u; 227 Ac(Z=89) = 227.0278u]α emission Q α = [ M(A,Z) – M(A-4,Z-2) - M( 4 He) ] c 2 = [0.059u]c 2 As 1u = 931.494 MeV/c 2Q α =[0.059u] (931.494 MeV/u = 5.496 MeVQ value is energy released in α decay of 228 Th. This will be split between the daughter and the alpha.