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Instructions Answer THREE questions out of the SIX in this quiz. You have 80 mins to do all THREE questions. Read the paper through carefully , the questions are structured to be increasingly difficult, and so the start to any one question will in general be easier than the end. Make sure you can do all of a question or pay attention to the point distribution. Plan your time carefully, and then begin. ANSWER IN THE SPACES PROVIDED. Ensure that you show your working, so that points can be given even if the final answer is incorrect.
(i) Name 3 types of ionizing radiation 3 points (ii) List 4 possible sources of background radiation, natural or man-made List 2 sources of intense radiation Any 3 of Alpha, Beta, Gamma, Neutron, X-ray etc. (1 each) ! point per background radiation (K-40, Cosmic rays, Nuclear waste, Medical procedures etc.) 1 point per intense radiation (accelerators, reactor core etc.)
(iii) Describe or define (a single sentence is sufficient) the following issues in radiation safety and exposure: a. Radiation dose, equivalent dose and effective dose b. Deterministic or stochastic exposures c. Internal and external exposure d. Low dose effects and extrapolation e. ALARA 5 points (iv) Shielding is one way of reducing radiation exposure, name two other ways that it can be reduced: a) Dose is the radiation energy absorbed per unit mass, Equivalent dose is the dose weighted to account for the energy deposition (LET), Effective dose is Equivalent dose weighted to account for absorption in different organs b) Deterministic " has a certain outcome (e.g. death in the case of extreme radiation dose), Stochastic " outcome determined by probability c) Exposure to either the internal organs via ingestion or inhalation, or skin d) Low dose effects are hard to measure due to low incidence of events (e.g. cancer), approach is to extrapolate effect from high doses e) As Low As Reasonably Achievable (1 point each) 2 from: Increasing distance, decreasing time of exposure, decreasing amount of radioactive substance used 1 point each
(i) A nuclear reactor distributes power to consumers 305 days in a given year. What is its capacity factor for that year? 3 points (ii) Another low carbon alternative to nuclear power is solar energy. Given that solar panels can produce 40 watts of electricity per square meter, how many square meters would be needed to produce the electricity equivalent of one nuclear reactor (assume a nuclear reactor produces 1200 megawatts). How many square miles is this? Assume 1 foot =0.3048 m and 1 mile = 5280 ft 4 points (iii) A reactor outputs 30 MW of heat and has a thermal efficiency of 30%. How much electrical energy in megawatt-days does it produce in one day? How much electrical energy in kilowatt-hours? C = 305 / 365 = .8356 = 83.56% Capacity Factor !"## $ !#%^ &' (# ' ) *
- ,-. ---. --- /^0 1#.###.###&) ^ $ 2 !^ &34 * 5 1#(6 *^ ) *^ 7 2 ! ^ &)89:; * <"6#^ &34 *^ 7 + ==5 >?&/@ABC^0 1#&D'4E $ #51 + F&D': F&D' $ 2 !# %^ &' D' 7 $ G !&H !&' I ; J $ G"(&EKLM; $ 1N##&; EM J $ G !&O' I EM 1.N##.###&H J
- 05 =P $ =->^ QR I STC The amount of energy from a 9 MWe reactor released in 1 day is 9 MW-days
(iv) A state-of-the-art 1500 megawatt-electric nuclear reactor costs $25 Billion for its construction and lifetime operation costs. Given that the plant has an 89% lifetime capacity factor and all of the produced electricity is sold to the consumer at 12 ¢/kilowatt-hour, how many years will it be until the plant reaches financial breakeven (i.e. produces $25 billion dollars of income). Assume all of the $25 Billion dollars are paid upfront. 6 points (v) Convert 15 tons of TNT equivalent energy per century to gigawatt-hours per minute. !"# $ %&'(^ )* %"++# $ %&,^
- ) . #"/%!0! $ %&^1 )-23456- 7 89" :);<=>? %#&&)@A $ &"B0. %"++# $ %&,^ A. %"++# $ %&,^ C D = power production rate E FG$'HI^ )J "'F (^) KLMNOJ P. Q"&B++ $ %&''^ )RA S TU of energy must be sold Q"&B++ $ %&''^ )RA S TU $ V WXYHHXHHH)C ')Z[]^ _. !"# $ %&'(^ )* of energy must be sold %)
A S TU $ E %& ,^ )AA P $ a %)* %)A S - b $ a +/&&)- %)TU b. +/&& $ %&,^ )* %0"0!!)
$ %)`A S TU +/&& $ %&,^ * $ /& - % cd . e" efgh $ 8i\8i^ jk S l> mno 'G)pqrD sts ')uvrpw^x $^ ')uvrpw^x 'HH)xvy^D $^ z"F$'HIC ' pqr sts $^ ' xvy^ WYG {yxD $^ ' {yx Fz ]^D $^ ' ]^ WYHH D . %0"0!!)A. %0"0!!) C D [take this power and convert to desired unit]
How is the number density of Uranium ions in the solution calculated from the mass of Uranium added to the volume of water? What is the value of %? What mass of Uranium would result in a critical reaction if L=50 cm? (Hint: Be careful with units. Use the cross section from part ii and the method for obtaining number-density from part iii. The mean-free-path is related to these, so using the expression for the criticality factor in part iii, work out what mass yields k=1). rho = M/V (mass density) n = rho/molar mass x NA " Avogadros number (3 points) !
0.188 $ % (^) f ( U & 235 ) 0.188 $ % (^) a ( U & 235 ) + ( 1 & 0.188) $ % (^) a ( U & 238 ) !
PL = 1 - 6 #mfp/L k = PL% for k= #mfp = L(1-PL)/6= L(1-1/%)/6=4.26 cm also #mfp = 1/!n (use cross section for fission weighted by 18.8%, convert barns to 1/cm^2 ) n = 1/!#mfp =1/(583x10-^24 x4.26x0.188) = 2.14x10^21 cm-^3 from part iii mass = Volume (cm 3 ) x molarmass (g/mol) x n / NA =(50)^3 x(238x0.812+235x0.188)x2.14x10^21 /6.023x10^23 =100 kg (4 points) (3 points)
(iv) The workers that died received radiation equivalent doses of around 10 Sv each. If this equivalent dose is split equally between gamma-rays and 1 MeV neutrons, what is the absorbed dose, in Gy, equivalent to the 10 Sv equivalent dose in this case? 3 points
(i) Define “Q value” as it pertains to nuclear reactions. 3 points (ii) The fundamental charge of an electron is 1.602 x 10-^19 Coulomb [C]. Define the electron-volt [eV] and explain why it is equal to 1.602 x 10-^19 Joules [J]. 5 Sv 1MeV neutrons = 0.25 Gy, 5 Sv gammas = 5 Gy, Absorbed dose = 5.25 Gy The amount of energy required or yielded from a nuclear reaction. Defined as the energy equivalent of mass defect between initial and final nuclides. An electron volt is the amount of energy which is imparted upon a single electron of charge 1.602 x 10-^19 Coulomb as it is accelerated through a 1 volt potential. 1 Volt is equal to 1 Joule per coulomb of charge. Therefore, when an electron of charge of 1.602 x 10
- 19 coulomb is accelerated over 1 volt, it gains 1.602 x 10-^19 J of kinetic energy.
(iv) The above figure shows the binding energy per nucleon for different elements and isotopes. a. Identify the most tightly bound nuclide visible on the chart. b. What nucleus has a greater absolute binding energy? Circle one.
75 As 235 U
c. Approximate the binding energy of a 98 Mo NUCLEUS using the graph? (^56) Fe BE/A = 8.6 MeV/A A = 98 BE = 8.6 * 98 = 842.8 MeV
d. A nuclear reaction occurs which results in a net average decrease in binding energy per nucleon. Did this reaction require energy or release energy? 6 points (v) Which fission is more likely to occur? Why? (^235) U + n 97 Rb + (^137) Cs + 2n+Q [Q=176.88] (^235) U + n 117 Rh + (^116) Ag + 3n+Q [Q=177.76] 2 points
(i) What is a cross section? An interaction probability expressed as an area 3 points This nuclear reaction required energy. In general, moving from more tightly bound nuclei to less tightly bound nuclei is an “endothermic” process. The first fission is more likely occur because the fission fragments are asymmetric. We know from the fission product distribution chart that fission fragments of similar or matching mass number are highly improbable compared to fission fragments with a large mass difference.
(iii) The equations for energy and momentum conservation state that the total momentum and energy in a closed system is constant. If a neutron (A=1) with velocity v 0 elastically scatters from a nucleus of atomic mass number A so that its new direction is at an angle & with respect to its original propagation direction, as in the figure. The nucleus recoils at an angle & also. What is the velocity v of the neutron as a function of A after scattering? (Hint: Use energy and momentum conservation and the fact the angles are equal.) What is the velocity of the neutron after scattering if the nucleus is C- 12 and the initial velocity is v 0 = 22000 m/s? v = 22000/(1+1/12) 1/ = 21200 m/s Energy + momentum conservation: v 02 =Au^2 +v^2 , v 0 =(v+Au) cos&, vsin&=uAsin& ' v=uA v 0 2 =v 2 /A+v 2 " v=v 0 /(1+1/A) 1/ 4 points
(iv) Now the neutron scatters directly back along its original direction of motion (reflected). Given that the motion is one-dimensional, and using the equations of momentum and energy conservation, what is the final velocity of the neutron now? 4 points In the inelastic scattering of a neutron from a nucleus of Pu- 239 , the incident neutron loses 500 keV of energy, is this enough to cause fission and why? 2 points
(i) What was the purpose of the very first nuclear reactors to be built (in addition to research). Plutonium production
v 02 =Au^2 +v^2 , v 0 =v+Au " u 2 =(v 0 - v) 2 /A 2 = (v 0 2
- 2vv 0 +v 2 )/A 2 " v 02 =(v 02 – 2vv 0 +v^2 )/A+v^2 " (A-1)v 02 =– 2vv 0 +(1+A)v^2 v= 2v 0 !"(4v 0 2
- 4(1+A)(1-A)v 0 2 )/2(1+A) = v 0! v 0 "(1 - 1+A^2 )/(1+A) = v 0 (1!A)/(1+A) " two solutions, v=v 0 (forward scatter) and v=v 0 (1-A)/(1+A) v = 22000 x (1-12)/(1+12) = - 18600 m/s No, fission generally requires ~5 MeV of energy (photofission)
(iv) Suppose that we fuel a thermal reactor with 30 00 kg of natural Uranium enriched to a 3 % level (i.e. 3 % U-235). Calculate the average neutron flux if the reactor is operating at a power of 1000 MW. The number of fissions per unit time (rate) is given by the product of the neutron flux with the fission cross-section and number of target nuclei. You may use 200 MeV as the energy released in a single typical fission. (U- 235 fission cross-section !f = 5 83 b) 6 points (v) At what rate is U- 235 being consumed? Express your answer in units of mass/time. Power = E/fission x rate (fissions) Rate = flux x sigmaf x n Flux = Power/(E/fission x sigmaf x n) n = mass / molar mass x NA = 3000e30.03 /2356.023e = 2.31e Flux = 1000x10^6 /(200 x 1.6e-13 x 583x10-^24 x 2.31e26) = 2.32x 14 s
- 1 cm - 2 Rate (fissions) = P/(E/fission) = 1000e6/(200 x 1.6e-13) = 3.19x10^19 s-^1 dM/dt = MU235 x rate = 235 x 1.67x10-^27 kg x 3.19x10^19 s-^1 = 1.24e-05 gs
- 1 = 12.4 #gs
- 1
