2011 Final Exam with Solutions

22.02 Intro to Applied Nuclear Physics Final Exam Tuesday May 17, 2011 Problem 1: Solution Short Questions 20 points These short questions require only short answers (but even for yes/no questions give a brief explanation) 1) True or False: if nuclide A1 X has a larger neutron scattering cross section than nuclide A2 X ′ , moderator than A2 X ′ . (Explain) A1 X is a better Solution: This is true only if A1 ≈ A2 . Otherwise the lighter nuclide might be the best moderator, since the neutron can loose more energy in each single collision. 2) A sheet of paper can stop a beam of alpha particle. What type of collisions are responsible for this? Indicate the type of interaction and target. Solution: The collisions are due to the Coulomb interaction with the electrons in the material. (Interactions with nuclei are rare and will not contribute substantially to the stopping down). 3) Why is the sky blue? Solution: Light from the sun is scattered by the air via Rayleigh scattering. The probability of scattering (cross section) depends on the frequency as σ ∝ ω 4 , thus blue light is scattered more away from the sun than other wavelengths and it can reach our eyes, when we look away from the sun (looking in a direction closer to the sun we see light that has not been scattered, thus it is closer to the original mix of wavelengths: white light). 4) Berillium-8 is an even-even nuclide (N and Z even) in which all the nucleons are paired off, leading to a very energetically favorable configuration. Why then Be-8 is not found in nature? [Hint: what is the decay mode of Be-8?] Solution: Be-8 decays to two alpha particles: since alpha particles are so tightly bound, the decay from Be-8 to He-4 is so fast that Be-8 cannot be found in nature. As hinted in the question, beta decay to either 8 Li or 8 B is not favorable (Q¡0) because of the pairing term. [Note: this was a tricky question, since we usually don’t have alpha decay or f ssion for light isotopes, so I did not detract many points for it] 5) The decay rates for gamma and beta decay can be obtain from the Fermi golden rule. What are the two factors that determine the decay rates? Solution: Besides numerical factors, the Fermi golden rule is given by the product of the matrix element and of the density of states. The f rst one gives the probability of a transition from the initial state to a f nal state. The second one multiplies this probability by the number of states available, to f nally give the total transition rate. 6) Fusion of two deuterium atoms is energetically favorable. Why then it does not happen unless we provide a lot of energy to them? Solution: In order to reach the deep potential well of the bound He-4, the two deuteriums have to overcome the Coulomb potential barrier. 1 7) Consider a free particle in quantum mechanics, initially in a state ψ(x, t = 0) (for example a Gaussian wavepacket). What is the evolution of the expectation value of its position hxi and velocity hv i? (write at least the equations of motion ddhxt i = . . . and ddhvt i = . . . ). 8) Are the equations of motion you found in question 7 different than the classical equations of motion? How would you answer change if the particle was moving in a potential? Solution: d hpi ∂V From Ehrenfest theorem, we have ddhxi t = hpi = m hvi and d t = − ∂x = 0, since a ”free particle” means that is not subjected to any potential or force and V = 0. Thus hvi = v0 (its value at t = 0) and hx(t)i = v0 t + x0 (where x0 = hx(0)i). The equation of motions thus give the classical result. ∂V (hxi) If we have a potential, we might have that ∂V ∂x 6= ∂hxi (or hF (x)i 6= F (hxi)) it the wavefunction is spread out in space, thus the equation of motion for the quantum particle could differ from the classical case. Problem 2: Radioactive Decay I 10 points Molybdenum-99 is produced in f ssion reactors, as one of the possible f ssion fragments. It decays to Technetium-99 with a half-life of 66 hours. a) What is the type of radioactive decay? Write the nuclear reaction describing this decay. Solution: Since the mass number A is left unchanged, but the atomic number changes, the decay must be a beta decay. Specif cally, it is Beta-minus β − decay: 99 99 − 42 Mo →43 Tc + e + ν̄ where the underlying decay converts one neutron into one proton: n → p + e− + ν̄ b) What is the Q-value of this decay? Describe qualitatively how this energy is distributed among the reaction products. Solution: Given the atomic masses of Tc and Mo we can calculate the Q value: 99 2 2 Q = [mA (99 42 Mo) − mA (43 Tc)]c = (98.90771 − 98.90625)amu × 931.46MeV/c /amu = 1.35993MeV Since we’re using the atomic masses, the electron masses are already accounted for (notice this is different for β + decay). The energy is distributed between the electron and anti-neutrino (Tc is very heavy so it does not acquire much kinetic energy due to conservation of momentum). The exact distribution of energy between electron and anti-neutrino can be calculated from the Fermi-Golden rule (in particular the density of state part) which gives the spectra of the beta decay (as calculated in class and plotted in the lecture notes, as well as you calculated in a Pset). Note that because the mass of neutrino is very small we expect it to get more kinetic energy, as we infer from a simple conservation of energy and momentum calculation. c) Bonus question: A metastable level of Technetium-99 (Technetium-99m) is used as a gamma emitter in medical imaging. Its half life is about 6 hours (and it decays to the ground state of Technetium-99 by emitting gamma radiation). Molybdenum-99 produced from fission is transported to hospitals where its radioactive decay product, Technetium99m, is extracted. What is the optimal time at which the largest amount of Technetium-99m can be extracted? Solution: Molybdenum decays as while Technetium-99m dynamics is given by d NM = −λM NM (t) dt d NT = λM NM (t) − λT NT (t) dt 2 with λM = log(2)/t 12 = 2.9 × 10−6 s−1 and λT = log(2)/t 21 = 32 × 10−6 s−1 . The optimum time is obtained when there is a maximum amount of Tc-99m, which can be calculated by putting the time derivative to zero. This impose the condition that the activity of the two isotopes are equal at the optimum time: d NT = λM NM (topt ) − λT NT (topt ) = 0 dt → AM (topt ) = AT (topt ) To f nd the optimal time we still need an explicit expression for the time evolution of the activities. Solving the f rst equation (setting NM (t = 0) = N0 ) we have: NM (t) = N0 e−λM t . The second equation is then d NT = λM N0 e−λM t − λT NT (t) dt with NT (t = 0) = 0. We can guess a solution of the form NT (t) = Ae−λM t + Be−λT t , with A and B two parameters to be determined: d NT = λM N0 e−λM t − λT NT (t) dt → −λM Ae−λM t − λT Be−λT t = λM N0 e−λM t − λT (Ae−λM t + Be−λT t ) The RHS equation is satisf ed for A = N0 and λM λT − λM B = −N0 Thus giving: NT (t) = N0 λM λT − λM λM (e−λM t − e−λT t ) λT − λM The maximum quantity of Technetium-99m is found for d NT dt = 0 or NT = N0 λλM e−λM t T λM λM −λM t (e−λM t − e−λT t ) = e λT − λM λT → λT (e−λM t − e−λT t ) = (λT − λM )e−λM t thus we f nd: topt = log Problem 3:  → e−(λT −λM )t = λM λT  λM /(λT − λM ) ≈ 22 hours λT Shell Model 20 points a) Based on the shell model, what is the spin and parity of the ground states of Molybdenum-99 and of Technetium-99? Solution: + + Molybdenum-99 has 57 neutrons. The most likely spin-parity assignment is 72 , or maybe 52 if the orbit term is + considered. [Note: it turns out to be 12 but we can’t obtain this result from a simple picture of the shell model]. Technetium-99 has 43 protons. In the valence nucleon model, the last unpaired proton determines the spin and parity + of the nuclide. The proton occupies the 1g 29 level, thus Technetium-99 has spin 92 and parity +, 92 b) Consider now Technetium-99. The first two excited states of Technetium-99 have very similar energy. Their spin+ − parity assignments are 72 (E1 = 140.511 keV) and 12 (E2 = 142.63keV). Explain the spin and parity of these two excited states. Draw a scheme of the energy levels involved (including the ground state) and which levels the nucleons occupy (you might want to indicate filled states with filled circles and empty states with empty circles). Solution: The f rst excited state is explained by having the unpaired proton jump to the level 1g 27 , which in the next shell (after the magic number 50). 3 ground, 9/2+ 1st excited, 7/2+ 2nd excited, 1/2- Filled-up Shells Filled-up Shells 2d5/2 1g7/2 1g9/2 2p1/2 Filled-up Shells 3rd excited, 5/2+ Alternatives configurations for 2nd excited wrong ok 2d5/2 1g7/2 1g9/2 2p1/2 Filled-up Shells Filled-up Shells Filled-up Shells Figure 1: Shell-model conf guration for Technetium. Note that students proposed different conf gurations for the second excited state (in the gray box). While the conf guration on the left in the gray box is ok, it most probably would have higher energy than what found experimentally. The conf guration on the right is not acceptable since it leaves three unpaired nucleons. The second excited state is obtained by having one proton in the closed shell 2p 21 being promoted to the ground state − level 1g 29 . The unpaired proton is now in the 2p 21 shell, yielding 12 spin parity. c) The third excited state of Technetium-99 is the level 2d 25 , as we would have expected from a naive picture of the energy arrangement in the nuclear shell model. What characteristic of the nuclear force makes the energy of the level − with spin/parity 12 (the second excited state) smaller than the energy of the level 2d 25 ? Solution: − We expect the level 2d 25 to be higher in energy than the conf guration 2p 21 (with a closed shell1g 29 ) because of a combination of the pairing term and spin-orbit coupling: the angular momentum dependence of the nuclear force makes leaving unpaired nucleons with higher angular momentum less favorable (more energetic). ˆ ˆ V0 ~ d) The spin-orbit potential Vso = ~so s explains the energy difference between the ground state and the first excited 2 l ·~ state of Technetium-99. Given the energy difference between the first excited state and the ground state (E1 = 140.511 keV), what is the value 0 of Vso for Technetium-99? Solution: The difference in energy is given by ∆E = Vso |1g 27 − Vso |1g 92 . For a given value of the total and orbital angular momentum, the spin-orbit coupling is calculated as: Vso = 0 0 Vso ~ˆl · ~sˆ = Vso = Vso ~(I(I + 1) − l(l + 1) − s(s + 1)) ~2 ~2 ˆ ˆ ˆ ˆ since we have the vector operator relationship: I~2 = (l~ + ~sˆ)2 = ~l2 + ~sˆ2 + 2l~ · ~sˆ and given the value of the operators ˆ ˆ I~2 when the system is in an eigenstate with eigenvalue ~2 I(I + 1). Here I~ is the total angular momentum, what we ~ˆ is the last unpaired nucleon spin, which has always eigenvalue ~2 s(s + 1) with s = 1 call the ”spin” of the nucleus. S 2 ~ˆ is the angular momentum (with eigenvalue ~2 l(l + 1) identif ed by the shell spectroscopic notation: here l = 4 and L for a g-shell). Since l (and s) is the same for the two states, we have 9 9 0 0 7 7 0 9 ∆E = Vso [I1 (I1 + 1) − I0 (I0 + 1)] = Vso [ ( + 1) − ( + 1)] = −Vso 2 2 2 2 2 4 Energy Shell order 2d5/2 1g7/2 1g9/2 2p1/2 3rd excited 2d5/2 2p1/2 1g7/2 2nd excited E=142.6keV 1st excited E=140.5keV E3 M1,E2 M4 (E5) 1g9/2 Ground E=0 Figure 2: Gamma decays from the f rst two excited states of Tc-99. Note that although the shell order is as given in the left, the actual order of the energy levels (when ordered by energy) is as in the right side. Three transitions are possible, even when not considering the third excited level. 0 Then we have Vso = −140.511keV × 29 = −31keV . 0 Note that since I = l ± 12 , we had also seen in lecture that we can also always write: ∆E = −Vso (l + 12 ) = 1 0 0 9 −Vso (4 + 2 ) = −Vso 2 . Problem 4: Radioactive Decay II 20 points Consider the gamma decay from the two f rst excited levels of Technetium-99. a) What are the allowed multipole transitions for each one of the three possible decays? Specify if there are possibly competing multipoles. Solution: There are three possible decays: + + 1g 27 → 1g 29 , − + 2p 21 → 1g 29 , − + 2p 21 → 1g 27 , ∆Π = no, 1 ≤ ℓ ≤ 8 ∆Π = yes, 4 ≤ ℓ ≤ 5 ∆Π = yes, 3 ≤ ℓ ≤ 4 M 1, E2, (M 3, E4, M 5, E6, M 7, E8) M 4, (E5) E3, (M 4) The f rst decay allows odd magnetic multipoles and even electric multipoles. Thus we expect to have two competing decays, M1 and E2. The second and third decays allow instead odd electric multipoles and even magnetic multipoles. Thus the second decay will be M4 (with maybe E5 competing) and the third E3. In this last case, we don’t expect to have competing processes, because it would have much smaller decay rate. Note that than we are considering ”decay” processes, in which some energy is released (in the form of a gamma photon) the transition is always from the excited state (which has more energy) to the ground state (see f gure). b) Estimate the decay rates for each transition (use the formulas given at the end of the exam sheet). Solution: We can use the formulas in the appendix to estimate λ for the four multipoles found above, using A=99. Notice that the energy to be considered for the third decay is ∆E3 = E2 − E1 = 142.63 − 140.511keV = 2.12keV We obtain: λ(M 1) = 1.55 × 1011 s−1 λ(E2) = 1.83 × 106 s−1 λ(M 4) = 1.08 × 10−9 s−1 λ(E3) = 6.39 × 10−14 s−1 c) Based on your answer above and the 6 hours lifetime of Technetium-99m can you identify the excited energy level that corresponds to Technetium-99m? If not, what other radioactive process could be happening that could lead to the observed lifetime (by accelerating some of the decays)? Solution: The decay from 1g 27 is too fast, while the decay from 2p 21 is too slow. Thus the level 2p 21 (that we can identify with Tc-99m) must be decaying by internal conversion at a faster rate than gamma decay to explain the observed lifetime. 5 Internal conversion is a decay process that is always present and can have faster rates that gamma decay in some cases, as here. d) Although long-lived, Technetium-99 is not stable. By what radioactive process does its ground state decay? (write the nuclear reaction describing this decay). Solution: We can f rst eliminate alpha decay, since the mass number is too low to allow it. Then we should decide if Tc-99 decays by β − or β + /electron capture processes. Note that the last two processes would create Molybdenum-99 again, but we know this is not possible, since the opposite decay happens. Then, we f nd that Tc-99 decays by β − decay to Ruthenium-99 (99 44 Ru): 99 43 T c − →99 44 Ru + e + ν̄ e) The daughter nuclide of Technetium-99 can be also obtained from the decay of an isotope of Rhodium. What is the decay type? (write the reaction). Use the semi-empirical mass formula to explain this fact. Solution: Ruthenium-99 can be obtained also from Rhodium-99 by changing one proton into a neutron. The two reactions that make this possible are β + decay and electron capture: 99 45 Rh 99 →44 Ru + e+ + ν 99 45 Rh + e− →9449 Ru + ν It turns out that electron capture is the actual decay mode (although you could not have known, but you should at least listed the possibility). Nuclides try to reach an optimal value of their Z/N ratio, as it can be explained by the mass parabola obtained from the semi-empirical mass formula plotted for a f xed A as a function of Z (see lecture notes for the plot of the mass parabola). Problem 5: Neutron Scattering and Capture by Hydrogen 20 points a) Consider the elastic scattering of neutrons of energy E0 by Hydrogen initially at rest. What is the mean energy of the Hydrogen after one collision with a neutron? What is the variance of the Hydrogen energy? [Hint: start from what you know about the energy distribution of neutrons] Solution: In the elastic scattering of a neutron by a proton the two particles share the f nal energy. We know that the probability 2 1 for the neutron to have f nal energy E1 is P (E1 ) = (A+1) 4AE0 = E0 , which is a uniform distribution between αE0 and E0 (here α = 0). See the lecture notes on how we obtained this probability distribution function: qualitatively this is due to the variation in the angle at which the two particle are scattered. Then the probability for the proton to have an energy Ep is P (Ep )dEp = P (E1 )dE1 = E10 for Ep ∈ [0, E0 ]. Then 2 E2 the mean energy of the proton is hEp i = 21 E0 and the variance ∆Ep2 = Ep2 − hEp i = 120 . Note: the variance of a random variable is calculate by calculating the mean (or average) as: hEi = Z ∞ Z ∞ P (E)E dE here : → −∞ E0 Z 0 1 E2 E dE = 0 E0 2E0 and the second moment: E 2 = 2 P (E)E dE → −∞ and then taking the difference: 2 Ep2 − hEp i = here : Z 0 E0 1 2 E3 E dE = 0 E0 3E0 E02 E2 E2 − 0 = 0. 3 4 12 6 You can review probability concepts on the notes (from Griff th) that I had posted some time ago. b) Now consider the synthesis of deuterium D from the collision of a neutron and a proton. The reaction is a neutron capture reaction, n+H→ γ+ D or H(n,γ)D. What is the energy of the gamma photon emitted? Solution: The gamma photon will carry out the Q energy released in the reaction. Thus we need to calculate Q: Q = c2 [m(H) + m(n) − m(D)] = [(938.272 + 0.511 + 939.565) − 2.014 × 931.46]M eV = 2.39M eV Here I approximated the atomic mass of the hydrogen by the sum of the proton and electron mass (neglecting the binding energy). c) In many neutron reaction capture A X(n,γ)A+1X some of the energy released in the reaction is given to the A+1 X nuclide, which is left in an excited state A+1 X∗ . Can this happen in the neutron capture reaction producing deuterium, H(n,γ)D? What does this imply about the relative values of the scattering cross section σsc and the capture cross section σcp of a neutron by Hydrogen? (which one is larger?) Solution: Deuterium does not have any bound excited state, thus the only possible reaction leading to the formation of deuterium is the one in which the gamma possess all the energy Q generated in the reaction. In practice, if some of this energy would to be given to the deuterium, the nucleus would again split apart in a neutron and a proton. This process would be then just a (possibly inelastic) collision. We can deduce that the probability of a collision is much higher than the creation of a deuterium: in other words, σsc ≫ σcp . d) The neutron and proton that collide can have their spins aligned or anti-aligned. Does the probability of a collision leading to neutron capture instead of elastic scattering depend the spins of proton and neutron being aligned or anti-aligned? Explain. [Hint: What is the total spin of deuterium?] Solution: A proton and a neutron are different particles, so we do not have to worry about the Pauli exclusion principle. They both have spin-1/2, so their possible states are (in the coupled and uncoupled representations): s= s= s= s= 1 2 , ms 1 2 , ms 1 2 , ms 1 2 , ms |S |S |S |S = + 12 = + 12 = − 12 = − 12 p p p p s= s= s= s= 1 2 , ms 1 2 , ms 1 2 , ms 1 2 , ms = 0, Ns = 0i = = 1, Ms = −1i = = 1, Ms = 0i = = 1, Ms = 1i = = + 12 = − 12 = + 12 = − 12 n n n n = |↑↑i = |↑↓i = |↓↑i = |↓↓i |↑,↓i−|↓,↑i √ 2 |↓, ↓i |↑,↓i+|↓,↑i √ 2 |↑, ↑i Note that the total spin can only be S = 0, 1 (no negative values!!) while for each value of S we have Ms = −S, −S + 1, . . . , 0, . . . , S. The only bound state of deuterium has total spin S = 1. This spin state is reached if the spins of proton and neutron are aligned but also if they are anti-aligned. In this last case however, the spins can either form the desired triplet (with S = 1) but also a singlet (i.e. the state with S = 0); if this is the case, then neutron capture and deuterium formation cannot happen. Thus we expect that the probability of neutron capture is higher if the spins are aligned (and conversely a lower scattering cross-section). e) Bonus question: Denote by σc↑↑p the cross-section for neutron capture if the neutron and proton spins are aligned and σc↑↓p the cross-section if the two nucleons are anti-aligned. Assuming that the collision and capture does not change the spins of neutron and proton, what is the ratio σc↑↓p /σc↑↑p ? (you do not need to find an explicit expression for these cross-section to answer this question). 7 Solution: If we neglect spin effects, we would have a cross section σcp . If the two nucleons are aligned, the cross section is unchanged, since they can in all the case fusion to form deuterium, σc↑↑p = σcp . However if the nucleons are antialigned, only in half of the cases they are going to be in the triplet state S = 1 that lead to a bound state, thus σc↑↓p = 21 σcp and the ratio of the two cross-sections is σc↑↓p /σc↑↑p = 12 . Problem 6: Time Evolution 10 points Consider again a neutron and a proton (before the collision). Assume that at time t = 0 the proton is in the state |ψ(0)ip = S = 12 , ms = + 21 √ state |ψ(0)in = 21 S = 21 , ms = + 21 n + 23 The states S = 12 , ms = + 21 = |↑i and S energies E+ and E− respectively. S= 1 2 , ms = 12 , ms = = |↑ip , while the neutron is the √ p = − 21 n = 12 |↑in + 23 |↓in . − 21 = |↓i are eigenstates of the spin Hamiltonian, with a) What is the state of the proton at a time t > 0? What is the state of the neutron at the same time? Solution: Since it is in an energy eigenstate, the evolution of the proton is simply |ψ(t)ip = e−iE+ t/~ |↑ip . Notice that this is only an ”apparent” evolution, since it does not change any observable property of the proton. The evolution of the neutron is instead given by: √ 1 −iE+ t/~ 3 −iE− t/~ |ψ(t)in = e |↑in + e |↓in 2 2 which is a different state than the initial one. b) What is the probability for the two spins to be aligned at a time t? (assume that the two particle evolve independently) Solution: Since the state of the proton does not really change with time and it’s always |↑ip = S = 12 , ms = + 21 p , we have to f nd the probability for the neutron to be as well in the state |↑in = S = 12 , ms = + 12 n . This is given by |h↑|ψ(t)in |2 = | 21 e−iE+ t/~ |2 = 41 . c) Bonus question: What is the probability at time t for the proton and neutron to be in a (common) state with total spin S = 1? Solution: The probability of having spin S = 1, P1 is the probability of not having spin S = 0, P1 = 1 − P0 . In turns, =|  h↑↓| − h↓↑| √ 2 Thus, P1 = 1 − 3 8  P0 = |hS = 0, ms = 0|ψp (t)ψn (t)i|2 ! √ √ 1 −iE+ t/~ 3 −iE− t/~ 1 3 −iE− t/~ 3 2 e h↑↓|↑↓i|2 = . e |↑↑i + e |↑↓i | = | √ 8 2 2 2 2 = 58 . 8 Appendix: Useful Formulas and Constants Masses: Proton mass, mp c2 = 938.272 MeV; Neutron mass, mn c2 = 939.565 MeV; Deuteron mass = 2.014amu; 99 Electron mass=0.511 MeV. Atomic mass of 99 42 Mo: 98.90771 amu. Atomic mass of 43 Tc: 98.90625 amu. 2 Constants: ~c = 197MeV fm; ~e c = 1317 ; c = 3 × 108 m/s; R0 = 1.2fm. 1 amu ≈ 931.46 MeV/c2 p Relativistic energy: E = p2 c2 + m2 c4 SEMF coefficients: av ≈ 15.5MeV, as ≈ 16MeV, ac ≈ 0.6MeV, asym ≈ 24MeV, ap ≈ 34MeV. ˆ ˆ V0 ~ Spin-orbit interaction Vso = ~so s 2 l ·~ 2 Fermi’s Golden Rule: W = 2π | hψ f | V |ψi i | ρ(Ef ) ~ Gamma Decay: Estimates for the rates of different electric multipoles (energies in MeV): - λ(E1) = 1.0 × 1014 A2/3 E 3 - λ(M 1) = 5.6 × 1013 E 3 - λ(E2) = 7.3 × 107 A4/3 E 5 - λ(M 2) = 3.5 × 107 A2/3 E 5 - λ(E3) = 34A2 E 7 - λ(M 3) = 16A4/3 E 7 - λ(E4) = 1.1 × 10−5 A8/3 E 9 - λ(M 4) = 4.5 × 10−6 A2 E 9 Cross Sections Differential cross-section: Z2 dσ dΩ Coulomb : σc = 2πre2 βα4 ln Λ = Doubly differential cross section: r(ϑ,ϕ) 4πIa n (with re = d 2σ d ΩdEb 1 e2 4πǫ0 me c2 Stopping power: − dd E x = σc ne ∆E. 4 Rayleigh: σR = 83 πre2 ω ω4 ∼ 2.8f m –or 4πre2 ∼ 1barn– and ln Λ ∼ 10)  2 2
σ Rutherford’s: dd Ω = zZe (4Ta )−2 sin−4 ϑ2 4πǫ0 Thomson: σT = 83 πre2 0 2 Compton: σC = σT m~eωc , with a change in the photon’s wavelength: ∆λ = Number of collisions to reach neutron thermalization 1 H 18 2 H 25 4 He 43 9 12 C 110 238 U 2200 2π~ me c (1 − cos ϑ). 7 Figure 3: Shell Model Figure 4: Periodic Table of the elements 10