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13.57, , NUCLEI, , G, , I DELI N ES TO NCERT, , You may find the following, , =, =, =, =, =, , 1/ (4m:o), 1 MeV, 1 year, , mH, , data useful in solving, , the exercises:, , 1.6xlO-19C, , m(~He), , 9x109, , N, , 1.6, , Nm2/C, , x 10-13, , J, , k, , 3.154 x 107 s, , 1amu, , 1.007825 amu, , mn, , respective abundances of 7.5% and 92.5%. These isotopes have, masses 6.01512 amu and 7.01600 amu respectively. Find the, atomic weight of lithium., , (b) Boron has two stable isotopes, 1~ B and 1~ B. Their, respective masses are 10.01294 amu and 11.00931 amu, and the, atomic weight of boron is 10.811 amu. Find the abundances of, 1~ B and 1~ B., Ans. The atomic weight of lithium is, m (Li) = 7.5 x 6.01512 + 92.5 x 7.01600, 100, 45.1134 + 648.98, 694.0934, , mH, , boron, , O, , M, Relative, , abundance, , 1~ B, , isotope, , = 19.9%., , of, , 1~ B isotope, , = 80.1%., , Mass of 7 neutrons, , = 7 x 1.00867 = 7.06069 amu, , Ne21,, , Ne22, , 90.51 x 19.99 + 0.27 x 20.99 + 9.22 x 21.99, m(Ne)=--------1-00--------, , 100, - 1945.31 _ 20 18, -., amu., 100, , and, , = 14.11550 amu, , liN nucleus, , = 14.00307 amu, Sm = 0.11243 amu, , nucleus, , = 0.11243 x 931.5 =104.7 MeV., , 13.4. Obtain the binding energy of the nuclei ~ Fe and, , ~~ Bi in units of MeV from the following data:, mH = 1.007825 amu, mil = 1.008665 amu, , have, , Ans. The, 30 neutrons., , ~Fe, , nucleus, , contains, , 26 protons, , Mass of 26 protons, = 26 x 1.007825 = 26.203450 amu, , Mass of 30 neutrons, , ., , = 30 x 1.008665 = 30.259950 amu, , atomic mass of neon is, , 1736.89 + 5.67 + 202.75, , 7 protons, , Which nucleus has greater binding energy per nucleon ?, , respective abundances of90.51%, 0.27% and 9.22%. The atomic, masses of the three isotopes are 19.99 amu, 20.99 amu and, 21.99 amu respective/yo Obtain the average atomic mass of neon., Ans. The average, , contains, , 1 amu = 931.5 MeV, , of, , 13.2. The three stable isotopes of neon: NiO,, , amu., , m( 2~ Bi) = 208.980388 amu, , = 19.9, , 0.99637, , abundance, , (Ii N), , m( ~Fe) = 55.934939 amu, , 1081.1 = - 0.99637x + 1100.931, , :. Relative, , = 14.00307, , = 7 x 1.00783 = 7.05481 amu, , 1~ B, , x% of, , mN, , Mass of 7 protons, , B.E. of nitrogen, , 100, , = 19.831, , nucleus, , Mass defect,, , 10.811 = ---,---., -----'------'----, , x, , liN, , Ans. The, 7 neutrons., , average of the masses, of two isotopes, x x 10.01294 + (100 - x) x 11.00931, , or, , 1.008665 amu, , Give your answer in MeV., , = Weighted, , or, , 931.5 MeV, , = 1.00783 amu ; mil = 1.00867 amu ;, , Mass of, , H, , mass of natural, , 1.381 x 10-23 JK-1, , Total mass, , 100, , (b) Suppose the natural boron contains, isotope and (100 - x)% of I~B isotope. Then,, Atomic, , 6.023 x 1023 per mole, , from the following data:, , IT, , :::::6.941 amu., , 4.002603 amu, , 13.3. Obtain the binding energy of a nitrogen nucleus, , 13.1. (a) Two stable isotopes of lithium ~ Li and ~ Li have, , 100, , =, =, =, =, =, , SI, R, , e, , EXERCISES, , Total mass, Mass of ~Fe, , = 56.463400 amu, , nucleus, , = 55.934939 amu, Sm = 0.528461amu, , Mass defect,, B.E. of ~Fe nucleus, , = Sm x 931.5 MeV = 0.528461 x 931.5, = 492.26 MeV, , and

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13.58, , PHYSICS-XII, , 56, 492.26, B.E./nucleon of 26Fe = ---s6 = 8.79 MeV., , Ans., (i), , Now, the 2~ Bi nucleus contains 83 protons, 126 neutrons., , and, , (ii), , 32S+e+V, 16, 21Opo+ «: + V, 84, , (iv), , llC~, , (v), , llB+e+, , 6, , +v, , 5, , 97Mo +e+ + v, 42, 1201+v, 53, , 97Tc ~, 43, , (vt), , + e+, (vii) 120Xe, 54, , = 1.760877 arnu, = 1.760877 x 931.5, , B.E. of 2:Bi nucleus, , 32p~, 15, 210B~, 83, , R, , Jim, , 222Rn + 4He, 86, 2, 238U + 4He, 92, 2, , (iii), , Mass of 83 protons, = 83 x 1.007825 = 83.649475 arnu, Mass of 126 neutrons, = 126 x 1.008665 = 127.091790 arnu, = 210.741265 amu, Total mass, Mass of 2: Bi nucleus, = 208.980388 amu, Mass defect,, , 226Ra ~, 88, 242pu ~, 94, , 13.7. A radioactive isotope has a half-life of T years. How, long will it take the activity to reduce to (a) 3.125%, (b) 1% of, its original value?, Ans. (a)~ =.!i = 3.125 = J.., Ro No, 100, 32, , SI, , =1640.3 MeV, , ~, , B.E./nucleon of 2%Bi = 1640.3 = 7.85 MeV, 209, Clearly, ~Fe has a greater B.E. per nucleon. In fact, it, is the maximum value., , ~ =or =or, , or, , 13.5. A given coin has a mass of3.0 g. Calculate the nuclear, , t, , energy that would be required to separate all the neutrons and, protons from each other. For simplicity assume that the coin is, entirely made of ~Cu atoms (of mass 62.92960 amu~ The, masses of proton and neutron are 1.00783amu and, 1.00867 amu, respectively., Ans. The ~Cu nucleus contains 29 protons and, 34 neutrons., Mass of 29 protons = 29 x 1.00783 = 29.22707 arnu, , R, , nT, , N, , =, , 5 T years., , 1, , Ro = No = 100, , IT, , (b), , =, , or n=5, , H, , Mass of 34 protons = 34 x 1.00867 = 34.29478 amu, Total mass, = 63.52185 amu, Mass of ~Cu nucleus, , = 62.92960 amu, , Mass defect,, , Jim, , = 0.59225 arnu, , M, O, , B.E. of ~Cu nucleus, , = 0.59225 x 931.5 MeV = 551.5032 MeV, , Number of atoms in 63 g of Cu, , = Avogadro's number = 6.023 x 1023, , :. Number of atoms in 3 g of Cu, 23, 6.023 x 10 x 3 = 2.868 x 1022, 63, , Energy required to separate all the neutrons, protons from each other of 3 g copper coin, , and, , Required time,, t = 2.303 log No = 2.303 T log 100, A., N, 0.693, 2.303 x 2 x T, --0-.6-9-3 - '" 6.65 T years., , 13.8. The normal activity of living carbon-containing, matter is found to be about 15 decays per minute for every gram, of carbon. This activity arises from the small proportion, radioactive C14 present with the ordinary carbon isotope C1 ., When the organism is dead, its interaction with the atmosphere, (which maintains the above equilibrium activity) ceases and its, activity begins to drop. From the known half life (=5730 years), of C14, and the measured activity, the age of the specimen can be, approximately estimated. This is the principle of ct4 dating, used in archaeology. Suppose a specimen from Mohenjodaro, gives an activity of 9 decays per minute per gram of carbon., Estimate the approximate age of the Indus-Valley civilisation., Ans. Given normal activity,, Ro = 15 decays min-1, , oj, , Present activity,, R = 9 decays min -1,, , = 551.5032 x 2.868 x 1022= 1.582 x 1025 MeV., , 1J.{2 = 5730 years, , 13.6. Write nuclear reaction equations for, (i) a-decay of ~ Ra, (iii), , (v), , rr -decay of:~, , (ii) a-decay of, , P, , (iv), , p+ -decay of 1~ C, , (vii) Electron capture of, , (vi), 1~, , Xe, , p- -decay of, , 2~, 2~, , Pu, , Since activity is proportional, radioactive atoms, therefore,, N0 e -At, R, N, -=-=---=e, , Bi, , p+ -decay of ~ Tc, or, , Ro, , No, , 9, -=, 15, , e, , -At, , -At, , No, or e, , At, , =-, , to the number, , 15, 9, , of

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13.59, , NUCLEI, , 13.11. Obtain approximately the ratio of the nuclear radii of, the gold isotope 1~ Au and the silver isotope 1~ Au. What is the, approximate ratio of their nuclear mass densities ?, Ans. As R = Ro Al/3,, where Ro= 1.1 x 1O-15m, , Taking natural logarithms,, log, , e, , eAt =, , log 15, e 9, , 5, At loge e = 2.303 log 10- = 2.303 x 0.2218, , or, , 3, , or, , =, , _0._51_09_, , [.: loge e = 1], , A, 'T', , As, , 11/2-, , t, , A, , _, -, , 0.5109, , _ 0.5109, , 'T', , -, , 0.693/ 11/2 0.693, 0.5109 x 5730, --0-.-69-3-years, , x, , 11/2, , = 4224, , e, , Since the nuclear mass density is independent, size of the nucleus, so, Pnu(Au) =1, Pnu(Ag), , 0.693, , _, , R( 197Au) (197)1/3_, R 07Ag) = 107, -1.23, , .., , years., , 13.12. Find the Q-value and the kinetic energy of the, emitted a.-particle in the a.-decay of, (a), , 13.9. Obtain the amount of ~Co necessary to provide a, radioactive source of 8.0 mCi strength. The half-life of ~Co is, 5.3 years., Ans. Here R = 8.0 mCi, = 8.0 x 10-3 x 3.7 x io" dis s-l, = 29.6 x 107dis s-l, 11/2 = 5.3 years = 5.3 x 3.16 x 107 s, , 29.6 x 107 x 5.3 x 3.16 x 107, N =-=---------0.693, 0.693, = 7.15 x Uy6 atoms, As 60 g of cobalt contains 6.023 x 1023 atoms, so the, amount necessary to obtain a source of the required, strength, 60 x 7.15 x 1016, -6, =, 23 = 7.123 x10, g., 6.023 x 10, , H, 28, , years. What is the, , O, , 13.10. The half-life of : Sr is, , disintegration rate of 15 mg of this isotope ?, Ans. Here 11/2 = 28 years = 28 x 3.154 x 107 s, m = 15 mg = 0.015 g, M = 90, , .!!!.. x Avogadro's number, , M, , Given, , (b) ~ Rn., , m( Z:Ra) = 226.02540 amu,, , m( ~Rn), , =, , 222.01750 amu,, , m ( ~ Rn), , =, , 220.01137 amu,, , me~, , =, , 216.00189 amu., , Po), , Ans. (a) 22~Ra ~, , ~Rn, , + ~He +Q, , ) - m ( ~He )] c2, , =, , [226.02540 - 222.01750 - 4.00260] x 931.5 MeV, , =, , 0.0053 x 931.5 = 4.937 MeV, A-4, , I\.=j\Q, , 226-4, , = --, , 4.937 = 4.85 MeV., , x, , 226, , 2;!PO + ~He +Q, , (b) 2~Rn ~, , Q = [m ( 22~Rn ) - m ( 2~po ) - m ( ~He )] c2, =, , [220.01137 - 216.00189 - 4.00260] x 931.5 MeV, , =, , 0.00688 x 931.5, , = 6.41, , MeV, , A-4, , Number of atoms in 0.015 g sample of ~Sr,, N=, , Ra, , IT, , 11/2, R11/2, , z:, , Q = [m ( 22~Ra ) - m ( ~Rn, , R = AN = 0.693 . N, , But, , of the, , SI, R, , t, , Ka =j\Q, 220-4, 220, , = --, , M, , x 6.41 = 6.29 MeV., , 0.015 x 6.023 x 1023atoms, , Activity of the sample,, 0.693, R= AN=--., , 11/2, , =, , 13.13. The radionuclide, , 90, , llC ~, , N=, , 0.693 x 0.015 x 6.023 x 1023, 7, , 28 x 3.154 x 10 x 90, , 7.877 x 101°disintegrations/second, , = 7.877 x 1010'Bq, , =, , d0 o, , 3.7 x 1, , ., , = 2.13, , ci,, , decays according to, , llB+e++v:1112=203min., , The maximum energy of the emitted positron is 0.960 MeV. ,, Given the mass values:, me~C) = 11.011434amu, me;B), me, , 7.877 x 101°., , 11 C, , =, , 11.009305 amu, , = 0.000548, , amu, , Calculate Qand compare it with the maximum energy of the, positron emitted.

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13.60, , PHYSICS-XII, , Ans. I~C ~, I~B+ e" + v + Q, where Q is the energy released in the decay process. It is, given by, Q = [mN(I~C) - mN(~B) - me] c2, To express Q-value in terms of atomic masses, we have, to subtract 6me from the atomic mass of carbon and 5me, from the atomic mass of boron to get the corresponding, nuclear masses. So we get, Q = [m (11C)- 6me - m (11B) + 5me - me] ~, = [ me~C) - me~ B) - 2me] c2, , m(~g H) = 19.992439 amu, m(i H) = 4.002603 amu, Ans. (i), , ~H + fH ~, , ~H + ~H, , Q = [m( ~H) + m(fH) - {m( ~H) + m( ~H)}] ~, = [(1.097825+ 3.016049)- 2 x 2.014102] x 931.5 MeV, , x, , R, , MeV, 931.5-amu, = 0.9622 MeV oc 0.96 MeV., , me~ H) = 12.000000 amu, , = (4.023874 - 4.028204) x 931.5 MeV, , = [11.011434 - 11.009305 - 2 x 000548]amu x ~, , = - 0.00433 x 931.5 MeV = - 4.033 MeV, Negative value of Q indicates that the reaction is, endothermic., + 12C ~, (ii) 12C, 6, 6, , 20Ne + 4He, , W, , 2, , SI, , = 0.001033amu, , m(f H) = 3.016049 amu, , Q = [2me~C) - {m(~gNe) + m(~He)}] ~, , t;, , = [2 x 12.000000 - (19.992439 + 4.002603)], x 931.5 MeV, = (24 - 23.995042) x 931.5 MeV, = 0.004958 x 931.5 MeV = 4.618 MeV, , the I3--decay equation and determine the maximum kinetic, energy of the electrons emitted from the following data:, m(liNe) =22.994466 amu, m(1213Na)=22.989770 amu., , Positive value of Q indicates that the reaction is, exothermic., 13.16. Suppose, we think of fission of a ~: Fe nucleus into, , IT, , Q= Ed + Ee +, The daughter nucleus is much heavier than e+ and v,, so its energy Ed = O. When the energy of neutrino is, minimum, the energy of position is maximum and Ee ::::.Q., 13.14. The nucleus Ne23 decays by l3--emission. Write down, , [CBSE D 08], , Ans. The I3--decayof ~Ne may be represented as, 23, 23 '0, Q, 10 Ne ~, 11 Na + _Ie + v +, , =, , H, , Ignoring the rest mass of neutrino, the expression for, the kinetic energy released may be written as, Q = [mN (~Ne) - mN (~fNa) - me]c2, ({mN (~gNe) + lOme}- {mN (~Na) + 11mellc2, , = [m@Ne) - m(~Na)]~, , r. ~, , M, , O, , =931.5 MeV / amu], = [22.994466 - 22.989770], x, 931.5, MeV, ,, = 0.004696 x 931.5 MeV = 4.374 MeV., As 23Na is massive, the kinetic energy released is, mainly shared by electron-positron, pair. When the, neutrino carries no energy, the electron has a maximum, kinetic energy equal to 4.374 MeV., 13.15. The Q value of a nuclear reaction A + b ~ C + d is, defined by, Q = [mA + mb -me - md] ~, , where the masses refer to nuclear rest masses. Determine, from the given data whether the following reactions are, exothermic or endothermic., [CBSE OD 14C], 3H ~, (i) 11 H + 1, , 12 H + 12 H, (ii) 12C+12C ~ 20Ne=t-4He, 6, , 6, , 10, , 2, , Atomic masses are given to be :, m(~ H) = 1.007825 amu, m(~ H) = 2.014102 amu, , two equal fragments, ~: AI. Is the fission energetically possible?, Argue, by working, out Q of the process. Given, m( ~Fe) = 55.93494 amu and m(~: AI) = 27.98191 amu:, Ans. ~Fe ~, , ~:AI + ~:AI +Q, , Q= [m( ~:Fe) - 2m( ~:AI)] c2, , = [55.93494 - 2 x 27.98191] x 931.5 MeV, = - 0.02888 x 931.5 = - 26.90 MeV, As the Q-value is negative, the fission is not possible, energetically., 13.17. The fission properties of ~: Pu are very similar to, those of~ U. The average energy released per fission is 180 MeV., How much energy, in Me V, is released if all the atoms in 1kg of, pure 2~~ Pu undergo fission ?, Ans. Number of atoms present in 239 g of 2~~pu, = 6.023 x 1023, :. Number of atoms present in 1 kg or 1000 g of ~~Pu, 6.023 x 1023 x 1000, 24, ------= 2.52 x 10, 239, Energy released per fission = 180 MeV, Total energy released, = 2.52 x 1024 x 180 MeV = 4.54 x 1026 Me V., 13.18. A 1000 MW fission reactor consumes half of its fuel, in 5.00 y . How much ~~ U did it contain initially? Assume the, reactor operates 80% of the time, that all the energy generated, arises from the fission of 2~~ U and that this nuclide is consumed, only by the fission process.

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13.61, , NUCLEI, , Ans. Power of the reactor, P = 1000 MW = 109 W, Time of power generation,, t = 5y = 5 x 3.154 x 107s, Total energy generated in 5y with 80% on-time, 7, = 80% of Pt = 0.8 x109 x 5 x 3.154 x 10 J, Energy generated in each fission of ~ U, = 200 MeV = 200 x 1.6 x 1O-13J, , Radius of deuteron,, R = 2.0 fm, , Number of atoms in 235 g of ~ U = 6 x 1023, , Z:{U consumed, , R, , in 5y with 80% on-time, , Total energy generated, Energy generated per gram, 0.8 x 109 x5x3.154x107 x 235, 200 x 1.6 x 10-13 x 6 x 1023 g, , This amount is half the fuel taken initially., :. Mass of ~ U taken initially = 3088 kg., , 13.19. How long an electric lamp of 100 W can be kept, glowing by fusion of 2.0 kg of deuterium ? The fusion reaction, can be taken as, , H, , ~He+n+32MeV., , Ans. Number of atoms present in 2 g of deuterium, 23, = 6 x 10, , of atoms present in 2.0 kg or 2000 g of, , O, , Number, deuterium, , Ox 1023 x 2000 = 6 x 1026, 2, Energy released in the fusion of 2 deuterium atoms, , M, , =3.2MeV, Total energy released in the fusion of 2.0 kg of, deuterium atoms, = 3.2 x 6 x 1026 = 9.6 x 1026 MeV, 2, = 9.6 x 1026 x 1.6 x 1O-13J, 13, = 15.34 x 10 J, Energy consumed by the bulb per second, 100 J, Time for which the bulb will ;;low, =, , 15.34 x Hy3, 15.34 x Hy1, 100, s = 3.15 x 107 years, =, , =, , 9, 19, 9 x 10 x{1.6 x 10- )2 keV =360 keY., 15, 4 x 10- x 1.6 x 10-16, , 13.21. From the relation R = RoA1I3, where Ro is a, , constant and A is the mass number of a nucleus, show that the, nuclear matter density is nearly constant (i.e., independent, of A)., Ans. Refer answer to Q. 6 on page 13.4., 13.22. For the p+ (positron) emission from a nucleus, there, is another competing process known as electron capture i.e.,, electron from an inner orbit (say the K - Shell) is captured by the, nucleus and a neutrino is emitted:, , IT, , = 1544 x 106 g = 1544 kg., , iH+iH~, , 9 x 109 x (1.6 x 10-19)2, 2 x 2 x 10-15, J, , 1, , s, , SI, , The amount of, , 2R, , 47t Eo, , J, , 235, , 1, e2, U=--.-, , 235, , Energy generated per gram of ~~ U, 200 x 1.6 x 10-13 x 6 x 1023, , = 2.0 x 10- 15 m, , The Coulomb barrier is given by, , 6 x 1023, Number of atoms in 1 g of ~~ U, , 13.20. Calculate the height of the potential barrier for a head, on collision of two deuterons. (Hint. The height of the potential, barrier is given by the Coulomb repulsion between the two, deuterons when they just touch each other. Assume that they, can be taken as hard spheres of radius 2.0 fm.), Ans. Charge on each deuteron,, e = 1.6 x 1O-19C, , 4.9 x 104 years., , e- + zAX ~, , Ay, Z-1 +v, , Show that ifP+ emission is energetically allowed, electron, capture is necessarily allowed but not vice versa., Ans. Consider the two competing processes:, Positron emission:, , 1X~, , z~y+e++v+Q1, , Electron capture:, e-+, , 1X~, , z_~y+v+Q2, , The energy changes in the two processes are :, , Q1 = [mN, =, , (1X) -, , [m{~X) -2me -m{z_~Y)+, , = [m(1X), Q2 =, , mN (z-1Y), , [mN, , - me] ~, (2 -l)m., , - me] ~, , - m{z _~Y) - 2me] ~, , (1X) + me -, , mN (Z:1 Y)] ~, , =[m{1X)-m(z_~Y)]~, This means Q1 > 0 implies Qz > 0 but Q2 > 0 does not, necessarily imply Q1 > O. Thus if positron emission is, energetically allowed, electron capture is necessarily, allowed but not vice versa., 13.23. In a periodic table, the averageatomic mass of magnesium, is given as 24.312 u. The average value is based on their relative

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PHYSICS-XII, , 13.62, , Neutron separation energy of i~ AI is given by, , natural abundance on Earth. The three isotopes and their, masses are ii Mg (23.98504 u), ~ Mg (24.98564 u), i~ Mg, (25.98259 u). The natural abundance of ii Mg is 78.99"/0by, mass. Calculate the abundances of the other two isotopes., , Sn( i~AI) = [m( i~AI) + mn - m( i~AI)] c2, , = [25.986895 + 1.008665 - 26.981541], x 931 MeV, , Ans. The abundance of iiMg (23.98504 u) is 78.99%., Then abundance, of ~Mg, (24.98564 u), [100 - (x + 78.99 u)flo, where x% is the abundance, i~Mg (25.98259)., , =13.05 MeV., 13.25. A source contains two phosphorous, , radio nuclides, = 14.3d) and ~ P(11/2 = 25.3 d} Initially, 10% of, the decays come from, P. How long one must wait until 90%, do so ?, , i~P(1J./2, , il, , SI, R, , Average atomic mass of Mg, = Weighted average of masses of isotopes, , = 0.014019 x 931 MeV, , is, of, , i~p, , Ans. Clearly, the source has initially 90% of, radionuclides, and 10% of ~p radionuclides. Finally, say, 24.312 =, x_2_4_.9_85_64_+_x--=['--25_.9_8_2_59~], after, t, days,, the, source has 10% of P radionuclides and, 100, 90%, of, ~, P, radionuclides., 2431.2 = 1894.5783 + 2498.564 + x x 0.99615- 1973.6157, :. Initial number of P nuclides = 9x, 4404.8157 - 4393.1423, or, x = -------Initial number of, nuclides = x, 0.99695, 78.99 x 23.98504 + [100-(x, , + 78.99)], , i;, , .., , i~, ilp, of i~pnuclides, , Final number, , = 11.6734 = 11.71, 0.99695, x = 11.71%, , Final number of ~P nuclides, , Abundance of i~Mg = 11.71%, , ~, , IT, , As, , =, , G)" G), =, , =y, , = 9y, , 1, , 1, , li/2 = (2)-li/2, , Abundance of ~Mg = 100-(11.71+ 78.99) = 9.3%., , 13.24. The neutron separation energy is defined to be the, energy required to remove a neutron from a nucleus. Obtain the, neutron separation energies of the nuclei ~~Ca and, Al from, the following data:, mn = 1.008665amu;, , N = No (2), , or, , For first isotope,, , H, , i~, , 9y, , m( ~~Ca) = 40.962278 amu ;, , amu ;, , m(i~ AI), , = 26.981541, , amu., , O, , = 25.986895, , M, , A-iX), , = [{mN(A-iX)+, = [m(, , A-iX), , + »; - mN (~X)], Zme} +, , x (2), , 25.3, , 1, 9 = - (2), , 1(_1 __1 ), 14.3, , 25.3, , 111, , c?, , =; - (mN(~X)+, , Zme}] c?, , Here mN refers to nuclear mass while m refers to, atomic mass., :. Neutron separation energy of ~~Ca is given by, , Sn ( ~~Ca) = [m( ~Ca) + mn - m( ~~Ca)] c?, , = [39.962591+ 1.008665 - 40.962278] x 931 MeV, . [By using c? = 931 MeV / amu], = 8.36 MeV, , =, , On dividing, we get, , 81 = 214.3, , or, , x 25.3, , 11t, , or, , log 81=, , or, , 1.9085 = 11t x 0.3010, 14.3 x 25.3, , + mn - m( ~X)] c?, , = 0.008978 x 931 MeV, , 14.3, , 9, , Ans. Neutron separation energy Sn of a nucleus ~ X is, given by, Sn = [mN(, , 1, , y=9x(2), , For second isotope,, , m( ~Ca) = 39.962591 amu ;, , m(i~ AI), , li/2, , t, , or, , =, , 14.3 x 25.3, , log 2, , 1.9085 x 14.3 x 25.3, 11 x 0.3010, , = 208.5, , d, , ays., , 13.26. Under certain circumstances, a nucleus can decay by, emitting a particle more massive than an a particle. Consider, the following decay process:, Pb +14C, , 223, , Ra ~, , 209, , 223, , Ra ~, , 219Rn +4He, , 88, , 88, , 82, , 86, , 6, , 2, , Calculate the Q-values for these decays and determine that, both are energetically allowed.

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13.63, , NUCLEI, Ans. The Q-value, given by, Q= m(~Ra), , for the first decay process, , - me~Pb), , is, , - me:q, , = [223.01850 - 208.981107 - 14.0003241amu x r?, , (b) Consider the radius of both deuterium and tritium to be, approximately 2.0 fm. What is the kinetic energy needed to, overcome the Coulomb repulsion ? To what temperature must, the gases be heated to initiate the reaction ?, Ans. (a) The net reaction is, , = 0.034109 x 931.5 MeV = 31.85 MeV, , 2H + 13H ~, , Q= m(~Ra), , - m(~Rn), , Q= [mN (iH) + mN (lH) - mN (~He) - mnl r?, , 1~, U by fast, , Adding and subtracting 2 me' we get, , R, , 13.27. Consider the fission of, , Q = [{mN (iH) + me} + {mN (lH + me)}, , - {mN (~He) + 2me} - mnl r?, Q = [m(iH) + maH) - m(~He) - mnl r?, , neutrons. In one, , = [2.014102 + 3.016049, , SI, , fission event, no neutrons are emitted and the final stable end, products, after the beta-decay of the primary fragments, are, Ce and : Ru Calculate Q for this fission process. The, relevant atomic and particle masses are, , 1:~, , m(1~U) = 238.05079 amu, me~ Ce) = 139.90543 amu, , - 4.002603 - 1.00867] x 931.5 MeV, , = 0.018883 x 931 MeV = 17.58 MeV., the nuclei when they almost, touch other is, d = 2r = 2 x 1.5 x 10-15m = 3 x 1O-15m, (b) Distance between, , :. Repulsive potential energy of the two nuclei when, they almost touch other, , m(: Ru) = 98.90594 amu, mn = 1.00867 amu, Ans. The fission may be represented as, 140Ce, 58, , 1, , _1_ M2, 4nso' d, , IT, , + In ~, , +, , 99Ru, 44, , + 10 °e+ Q, , 9 x 109 x (1.6 x 10-19)2, 3 x 10 15, , -1, , The Q-value for the process is, , - 10 mel r?, , H, , - mN (:Ru), , = Average thermal K.E. available with, the interacting particles, , In terms of atomic masses, we can write, , 3, , Q = [( m(~~U) - 92 me} + mn - {m(I~Ce), , - 58 me}, - 44 me} -10 mel c2, , M, O, , - {m(:Ru), , + mn - me:~Ce), , - m(:Ru)l, , 2 H+3H, 112, , ~, , 4 He, , =2x-kT=2kT, 2, , 2kT = 7.68 x 1O-14J, , r?, , = [238.05079 + 1.00867 - 139.90543 - 98.905941, amu xr?, MeV, = (239.05946 - 238.81137) amu x 931.5-amu, = 0.24809 x 931.5 = 231.09 MeV, = 231.1 MeV., , 13.28. Consider the D- T reaction, fusion) given by the equation:, , or, , T=, , 14, , 7.68xlO- J, =1.85x109K., 2x1.38x10 23JK 1, , 13.29. Obtain the maximum kinetic energy of (3-particles, and the radiation frequencies corresponding to y-decays in the, decay scheme shown in Fig. 13.21. You are given that,, m(AJ98) = 197.968233amu, m(HgI98) = 197.966760 amu, , (deuterium-tritium, , +n, , :"'-r-"""t:"'-1.088 MeV, , (a) Calculate the energy released in MeV in this reaction, from the data:, m(i H) = 2.014102 amu, m(l H) = 3.016049 amu, , ~~...,..-"--, , = 1.00867, , amu, , 0.412 MeV, , '---"---0, , m(i He) = 4.002603 amu, m", , J, , = 7.68 x 10-14 J [.: 'It = q2 = 1.6 x 1O-19q, K.E. required for one fusion event, , Q = [mN( ~U) + m" - mNe~Ce), , = [mei~U), , + 10n + Q, , :. Energy released in the reaction is, , - m(~He), , = [223.01850 - 219.00948 - 4.002601amu x r?, = 0.00642 x 931.5 = 5.98 MeV, As the Q-value is positive in both cases, so both decay, processes are energetically possible., , 238U, 92, , 4He, 2, , 1, , For the second process,, , [CBSE 0 03C, 08C], , Fig. 13.21

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13.64, , PHYSICS-XII, , h, , ., (1.088 - 0) x 1.6 x 10-13, 20, •• V ( y1) =, 34, = 2.627 x 10 Hz, 6.63 x 10(0.412 - 0) x 1.6 x 10-13, 19, v(Y2)=, 34, =9.949xl0, Hz, 6.63 x 10, I3, v(Y3)= (1.088-0.412) xl: x 10- =1.632 x 1019 Hz, 6.63 x 10, The P1--decay can be represented as, l':Hg, , + _~e + Q(P1- ) + Q(Y1), , where Q(Y1) = 1.088 MeV., :. Maximum kinetic energy of PI particle is, ~ax, , (~)=[me~AU), , -{m(l':Hg), , + 1.088}] ~, 931.5, , Number of atoms in 1 kg or 1000 g of 235U, 6 x 1023 x 1000, 235, Energy released in per fission of 235U= 200 MeV, Energy released in fission of 1 kg of 235U, 6 x 1026 x 200, 26, -----=5.1xl0, MeV., 235, Thus the energy released in fusion of 1 kg of hydrogen, fusion is about 8 times that of energy released in fission of, 1 kg 235u., 13. 1 Suppose India has a target of producing by 2020, A.D., 200,000 MW of electric power, 10 percent of which is to, be obtained from nuclear power plants. Suppose we are given, that, on average the efficiency of utilisation (i.e., conversion to, electric energy) of thermal energy produced in a reactor is 25%., How much amount of fissionable uranium would our country, need per year at the turn of this century? Take the heat energy, per fission of U235 to be about 200 MeV. Avogadro's Number, N = 6.023 x 1023 mol-I., , SI, , 1~~Au ~, , (b) Number of atoms in 235 g of 235U= 6 x 1023, , R, , Ans. The frequencies of y-radiation will be equal to the, corresponding energy differences divided by Planck's, constant h., v=Ez-E;., , [Neglecting the rest mass ofp-particle], , Target of power by 2020 A.D., = 2 x 105 MW = 2 x 1011W, , = [197.968233 - (197.966760 + 1.088)] x 931.5 MeV, 931.5, , Power required from nuclear power plants, = 10% of 2 x 1011W, , IT, .,' - = Q31.5 MeV / amu], , = (0.001473 x 931.5 - 1.088) MeV, , = (1.372 - 1.088) MeV = 0.284 MeV, The, , P;: decay, , can be represented as, , Energy required from nuclear power plants per, , 19:0Hg + _~e + Q(P;) + Q(Y2), , H, , l~~Au ~, , where Q(Y2) = 0.412 MeV., ., Maximum kinetic energy of P;: particle is, J(, , (P- ) =, , 2, , [m(, , 198Au) _ {, 79, , m (198, ) + 0.412 }] ~, 80Hg, 931.5, , O, , "max, , = [197.068233-(197.966760+, , ~':~~)] x 931.5 MeV, , = [0.001473 x 931.5 - 0.412] MeV, = [1.372 - 0.412] MeV = 0.960 MeV., , M, , 13.30. Calculate and compare the energy released by (a), fusion of 1.0 kg of hydrogen deep within the Sun and (b) the, fission of 1.0 kg of 235U in a fission reactor., Ans. (a) In the sun, 4 hydrogen nuclei combine to form, a helium nucleus with the release of 26 MeV of energy., The net fusion reaction is, 4~H~, ~He + 2e+ + 26 MeV, Number of atoms in 1 g of H = 6 x 1023, Number of atoms in 1 kg or 1000 g of H, = 6 x 1023 x 1000 = 6 x 1026, , i, , =~x2xl(P=2xlOlOW, 100, , i, , The energy released by 1 kg of hydrogen, 26 x 6 x 1026, ---= 39 x 1026 MeV., 4, , year, , = Power x time, = 2 x UyO x 365.25 x 24 x 60 x 60 J, = 6.312 x io" J, Energy released per fission = 200 MeV, Available electrical energy per fission of 235Unucleus, = 25% of 200 MeV, = ~ x 200 MeV = 25 x 2 x 1.6 x 10-13J, 100, = 8 x 1O-12J, [.,' 1 MeV = 1.6 x 10 13 Jl, Number of 235Ufissions required per year, =, , 6.312 x 1017, 28, 12 = 7.89 x 10, 8 x 10-, , Required number of moles of 235U, 7.89 x 1028, , 7.89 x 1028, = Avogadro's number = 6.023 x 1023, , = 13.1 x 104, Mass of 235U required, = Number of moles x mass number, = 13.1 x 104 x 235 g, = 3078.5 x 104 g '" 3.078 x 104 kg.

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