{"id":121371,"date":"2022-05-17T18:10:25","date_gmt":"2022-05-17T12:40:25","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=121371"},"modified":"2022-05-17T18:16:59","modified_gmt":"2022-05-17T12:46:59","slug":"chapter-13-nuclei-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-13-nuclei-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 13 \u2013 Nuclei Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"<h2>1. Questions and Answers<br \/>\n(a) Two stable isotopes of lithium 63Li36Liand 73Li37Li have respective abundances of 7.57.5 and 92.592.5These isotopes have masses 6.01512u6.01512uand7.01600u7.01600u, respectively. Find the atomic mass of lithium.<\/h2>\n<h3>Ans: We are given the following information:<br \/>\nMass of 63Li36Li lithium isotope, m1=6.01512um1=6.01512u<br \/>\nMass of 73Li37Li lithium isotope, m2=7.01600um2=7.01600u<br \/>\nAbundance of63Li36Li, n1=7.5n1=7.5<br \/>\nAbundance of 73Li37Li, n2=92.5n2=92.5<br \/>\nThe atomic mass of lithium atom is given by,<br \/>\nm=m1n1+m2n2n1+n2m=m1n1+m2n2n1+n2<br \/>\nSubstituting the given values, we get,<br \/>\nm=6.01512\u00d77.5+7.01600\u00d792.592.5+7.5m=6.01512\u00d77.5+7.01600\u00d792.592.5+7.5<br \/>\n\u2234m=6.940934u\u2234m=6.940934u<br \/>\nTherefore, we found the atomic mass of lithium atoms to be 6.940934u6.940934u.<\/h3>\n<h2>(b) Boron has Two Stable Isotopes 105B510B and115B511B. Their respective masses are 10.01294u10.01294uand11.00931u11.00931u, and the atomic mass of boron is 10.811u10.811u. Find the abundances of 105B510B and 115B511B .<\/h2>\n<h3>Ans: We are given:<br \/>\nMass of 105B510B Boron isotope, m1=10.01294um1=10.01294u<br \/>\nMass of 115B511B lithium isotope, m2=11.00931um2=11.00931u<br \/>\nAbundance of105B510B, n1=xn1=x<br \/>\nAbundance of 115B511B, n2=(100\u2212x)n2=(100\u2212x)<br \/>\nWe know the atomic mass of boron to be, m=10.811um=10.811u<br \/>\nThe atomic mass of lithium atom is given by,<br \/>\nm=m1n1+m2n2n1+n2m=m1n1+m2n2n1+n2<br \/>\nSubstituting the given values, we get,<br \/>\n10.811=10.01294\u00d7x+11.00931\u00d7(100\u2212x)x+(100\u2212x)10.811=10.01294\u00d7x+11.00931\u00d7(100\u2212x)x+(100\u2212x)<br \/>\n\u21d21081.11=10.01294x+1100.931\u221211.00931x\u21d21081.11=10.01294x+1100.931\u221211.00931x<br \/>\n\u2234x=19.8210.99637=19.89\u2234x=19.8210.99637=19.89<br \/>\nAnd, 100\u2212x=80.11100\u2212x=80.11<br \/>\nTherefore, we found the abundance of 105B510Band 115B511Bto be 19.8919.89 and 80.1180.11respectively.<\/h3>\n<h2>2. The three stable isotopes of neon: 2010Ne,2110Neand2210Ne1020Ne,1021Neand1022Ne have respective abundances of 90.5190.51 The atomic masses of the three isotopes are 19.99u,20.99uand21.99u19.99u,20.99uand21.99u , respectively. Obtain the average atomic mass of neon.<\/h2>\n<h3>Ans: We are given that:<br \/>\nAtomic mass of 2010Ne1020Ne, m1=19.99um1=19.99u<br \/>\nAbundance of 2010Ne1020Ne, \u03b71=90.51\u03b71=90.51<br \/>\nAtomic mass of2110Ne1021Ne, m2=20.99um2=20.99u<br \/>\nAbundance of 2110Ne1021Ne, \u03b72=0.27\u03b72=0.27<br \/>\nAtomic mass of 2210Ne1022Ne, m3=21.99um3=21.99u<br \/>\nAbundance of 2210Ne1022Ne, \u03b73=9.22\u03b73=9.22<br \/>\nThe average atomic mass of neon could be given as,<br \/>\nm=m1\u03b71+m2\u03b72+m3\u03b73\u03b71+\u03b72+\u03b73m=m1\u03b71+m2\u03b72+m3\u03b73\u03b71+\u03b72+\u03b73<br \/>\nSubstituting the given values, we get,<br \/>\nm=19.99\u00d790.51+20.99\u00d70.27+21.99\u00d79.2290.51+0.27+9.22m=19.99\u00d790.51+20.99\u00d70.27+21.99\u00d79.2290.51+0.27+9.22<br \/>\n\u2234m=20.1771u\u2234m=20.1771u<br \/>\nThe average atomic mass of neon is thus found to be 20.177u.20.177u.<\/h3>\n<h2>3. Obtain the binding energy (in MeV) of a nitrogen nucleus (147N)(714N), given m(147N)=14.00307um(714N)=14.00307u<\/h2>\n<h3>Ans: We are given:<br \/>\nAtomic mass of nitrogen (7N14)(7N14), m=14.00307um=14.00307u<br \/>\nA nucleus of 7N147N14nitrogen contains 7 protons and 7 neutrons.<br \/>\nHence, the mass defect of this nucleus would be, \u0394m=7mH+7mn\u2212m\u0394m=7mH+7mn\u2212m<br \/>\nWhere, Mass of a proton, mH=1.007825umH=1.007825u<br \/>\nMass of a neutron, mn=1.008665umn=1.008665u<br \/>\nSubstituting these values into the above equation, we get,<br \/>\n\u0394m=7\u00d71.007825+7\u00d71.008665\u221214.00307\u0394m=7\u00d71.007825+7\u00d71.008665\u221214.00307<br \/>\n\u21d2\u0394m=7.054775+7.06055\u221214.00307\u21d2\u0394m=7.054775+7.06055\u221214.00307<br \/>\n\u2234\u0394m=0.11236u\u2234\u0394m=0.11236u<br \/>\nBut we know that, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u21d2\u0394m=0.11236\u00d7931.5MeV\/c2\u21d2\u0394m=0.11236\u00d7931.5MeV\/c2<br \/>\nNow, we could give the binding energy as,<br \/>\nEb=\u0394mc2Eb=\u0394mc2<br \/>\nWhere, c=speed of light =3\u00d7108ms\u22122c=speed of light =3\u00d7108ms\u22122<br \/>\nNow, Eb=0.11236\u00d7931.5(MeVc2)\u00d7c2Eb=0.11236\u00d7931.5(MeVc2)\u00d7c2<br \/>\n\u2234Eb=104.66334MeV\u2234Eb=104.66334MeV<br \/>\nTherefore, we found the binding energy of a Nitrogen nucleus to be 104.66334MeV104.66334MeV.<\/h3>\n<h2>4. Obtain the binding energy of the nuclei 5626Fe2656Fe and 20983Bi83209Bi in units of MeV from the following data: m(5626Fe)=55.934939um(2656Fe)=55.934939u, m(20983Bi)=208.980388um(83209Bi)=208.980388u<\/h2>\n<h3>Ans: We are given the following:<br \/>\nAtomic mass of 5626Fe2656Fe, m1=55.934939um1=55.934939u<br \/>\n5626Fe2656Fe nucleus has 26 protons and 56\u221226=3056\u221226=30neutrons<br \/>\nHence, the mass defect of the nucleus would be, \u0394m=26\u00d7mH+30\u00d7mn\u2212m1\u0394m=26\u00d7mH+30\u00d7mn\u2212m1<br \/>\nWhere, Mass of a proton,mH=1.007825umH=1.007825u<br \/>\nMass of a neutron,mn=1.008665umn=1.008665u<br \/>\nSubstituting these values into the above equation, we get,<br \/>\n\u0394m=26\u00d71.007825+30\u00d71.008665\u221255.934939\u0394m=26\u00d71.007825+30\u00d71.008665\u221255.934939<br \/>\n\u21d2\u0394m=26.20345+30.25995\u221255.934939\u21d2\u0394m=26.20345+30.25995\u221255.934939<br \/>\n\u2234\u0394m=0.528461u\u2234\u0394m=0.528461u<br \/>\nBut we have, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u0394m=0.528461\u00d7931.5MeV\/c2\u0394m=0.528461\u00d7931.5MeV\/c2<br \/>\nThe binding energy of this nucleus could be given as,<br \/>\nEb1=\u0394mc2Eb1=\u0394mc2<br \/>\nWhere, c = Speed of light<br \/>\n\u21d2Eb1=0.528461\u00d7931.5(MeVc2)\u00d7c2\u21d2Eb1=0.528461\u00d7931.5(MeVc2)\u00d7c2<br \/>\n\u2234Eb1=492.26MeV\u2234Eb1=492.26MeV<br \/>\nNow, we have the average binding energy per nucleon to be,<br \/>\nB.E=492.2656=8.79MeVB.E=492.2656=8.79MeV<br \/>\nAlso, atomic mass of 20983Bi83209Bi, m2=208.980388um2=208.980388u<br \/>\nWe know that, 20983Bi83209Bi nucleus has 83 protons and 209\u221283=126neutrons209\u221283=126neutrons<br \/>\nWhere,<br \/>\nMass of a proton, mH=1.007825umH=1.007825u<br \/>\nMass of a neutron, mn=1.008665umn=1.008665u<br \/>\n\u0394m\u2032=83\u00d71.007825+126\u00d71.008665\u2212208.980388\u0394m\u2032=83\u00d71.007825+126\u00d71.008665\u2212208.980388<br \/>\n\u21d2\u0394m\u2032=83.649475+127.091790\u2212208.980388\u21d2\u0394m\u2032=83.649475+127.091790\u2212208.980388<br \/>\n\u2234\u0394m\u2032=1.760877u\u2234\u0394m\u2032=1.760877u<br \/>\nBut we know, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\nHence, the binding energy of this nucleus could be given as,<br \/>\nEb2=\u0394m\u2032c2=1.760877\u00d7931.5(MeVc2)\u00d7c2Eb2=\u0394m\u2032c2=1.760877\u00d7931.5(MeVc2)\u00d7c2<br \/>\n\u2234Eb2=1640.26MeV\u2234Eb2=1640.26MeV<br \/>\nAverage binding energy per nucleon is found to be =1640.26209=7.848MeV=1640.26209=7.848MeV<br \/>\nHence, the average binding energy per nucleon is found to be 7.848MeV7.848MeV.<\/h3>\n<h2>5. A given coin has a mass of 3.0 g. Calculate the nuclear 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 6329Cu2963Cu atoms (of mass 62.92960u62.92960u).<\/h2>\n<h3>Ans: We are given:<br \/>\nMass of a copper coin, m\u2032=3gm\u2032=3g<br \/>\nAtomic mass of 29Cu6329Cu63atom,<br \/>\nm=62.92960um=62.92960u<br \/>\nThe total number of 6329Cu2963Cu atoms in the coin, N=NA\u00d7m\u2032Mass numberN=NA\u00d7m\u2032Mass number<br \/>\nWhere,<br \/>\nNA=Avogadronumber=6.023\u00d71023atoms\/gNA=Avogadronumber=6.023\u00d71023atoms\/g<br \/>\nMass number = 63g<br \/>\n\u21d2N=6.023\u00d71023\u00d7363=2.868\u00d71022\u21d2N=6.023\u00d71023\u00d7363=2.868\u00d71022<br \/>\n29Cu6329Cu63nucleus has 29 protons and<br \/>\n(63\u221229)=34(63\u221229)=34<br \/>\nneutrons<br \/>\nMass defect of this nucleus would be, \u0394m\u2032=29\u00d7mH+34\u00d7mn\u2212m\u0394m\u2032=29\u00d7mH+34\u00d7mn\u2212m<br \/>\nWhere,<br \/>\nMass of a proton, mH=1.007825umH=1.007825u<br \/>\nMass of a neutron, mn=1.008665umn=1.008665u\u0394m\u2032=29\u00d71.007825+34\u00d71.008665\u221262.9296=0.591935u\u0394m\u2032=29\u00d71.007825+34\u00d71.008665\u221262.9296=0.591935u<br \/>\nMass defect of all the atoms present in the coin would be,<br \/>\n\u0394m=0.591935\u00d72.868\u00d71022=1.69766958\u00d71022u\u0394m=0.591935\u00d72.868\u00d71022=1.69766958\u00d71022u<br \/>\nBut we have, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u21d2\u0394m=1.69766958\u00d71022\u00d7931.5MeV\/c2\u21d2\u0394m=1.69766958\u00d71022\u00d7931.5MeV\/c2<br \/>\nHence, the binding energy of the nuclei of the coin could be given as:<br \/>\nEb=\u0394mc2=1.69766958\u00d71022\u00d7931.5(MeVc2)\u00d7c2Eb=\u0394mc2=1.69766958\u00d71022\u00d7931.5(MeVc2)\u00d7c2<br \/>\n\u2234Eb=1.581\u00d71025MeV\u2234Eb=1.581\u00d71025MeV<br \/>\nBut, 1MeV=1.6\u00d710\u221213J1MeV=1.6\u00d710\u221213J<br \/>\n\u21d2Eb=1.581\u00d71025\u00d71.6\u00d710\u221213\u21d2Eb=1.581\u00d71025\u00d71.6\u00d710\u221213<br \/>\n\u2234Eb=2.5296\u00d71012J\u2234Eb=2.5296\u00d71012J<br \/>\nThis much energy is needed to separate all the neutrons and protons from the given coin.<\/h3>\n<h2>6. Write the Nuclear Reactions For:<br \/>\n(a) \u03b1\u2212decay of22688Ra\u03b1\u2212decay of88226Ra<\/h2>\n<h3>Ans: We know that, \u03b1\u03b1is basically a nucleus of Helium (2He4)(2He4)and \u03b2\u03b2is an electron (e\u2212for\u03b2\u2212ande+for\u03b2+)(e\u2212for\u03b2\u2212ande+for\u03b2+). In every \u03b1\u03b1-decay, there is a loss of 2 protons and 2 neutrons. In every \u03b2+\u03b2+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every \u03b2\u2212\u03b2\u2212-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.<br \/>\nFor the given case, the nuclear reaction would be,<br \/>\n88Ra226\u219286Rn222+2He488Ra226\u219286Rn222+2He4<\/h3>\n<h2>(b) \u03b1\u2212decay of24294Pu\u03b1\u2212decay of94242Pu<\/h2>\n<h3>Ans: 24294Pu\u219223892U+42He94242Pu\u219292238U+24He<\/h3>\n<h2>1. \u03b2\u2212decay of3215P\u03b2\u2212decay of1532P<\/h2>\n<h3>Ans: 3215P\u21923216S+e\u2212+\u03bd\u00af1532P\u21921632S+e\u2212+\u03bd\u00af<\/h3>\n<h2>2. \u03b2\u2212decay of21083Bi\u03b2\u2212decay of83210Bi<\/h2>\n<h3>Ans: 21083B\u219221084Po+e\u2212+\u03bd\u00af83210B\u219284210Po+e\u2212+\u03bd\u00af<\/h3>\n<h2>3. \u03b2+\u2212decay of116C\u03b2+\u2212decay of611C<\/h2>\n<h3>Ans: 116C\u2192115B+e++\u03bd611C\u2192511B+e++\u03bd<\/h3>\n<h2>4. \u03b2+\u2212decay of9743Tc\u03b2+\u2212decay of4397Tc<\/h2>\n<h3>Ans: 9743Tc\u21929742Mo+e++\u03bd4397Tc\u21924297Mo+e++\u03bd<\/h3>\n<h2>5. Electron capture of 12054Xe54120Xe<\/h2>\n<h3>Ans: 12054Xe+e+\u219212053I+\u03bd54120Xe+e+\u219253120I+\u03bd<\/h3>\n<h2>7. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to:<br \/>\n1. 3.125% of Its Original Value?<\/h2>\n<h3>Ans: We are said that, Half-life of the radioactive isotope=T years=T years Original amount of the radioactive isotope =N0=N0<br \/>\n(a) After decay, let the amount of the radioactive isotope be N.<br \/>\nIt is given that only 3.1253.125 of N0N0remains after decay. Hence, we could write,<br \/>\nNN0=3.125NN0=3.125<br \/>\nBut we know that, NN0=e\u2212\u03bbtNN0=e\u2212\u03bbt<br \/>\nWhere, \u03bb=\u03bb=Decay constant and t=t=Time<br \/>\n\u21d2\u2212\u03bbt=132\u21d2\u2212\u03bbt=132<br \/>\n\u21d2\u2212\u03bbt=lnl\u2212ln32\u21d2\u2212\u03bbt=ln\u2061l\u2212ln\u206132<br \/>\n\u21d2\u2212\u03bbt=0\u22123.4657\u21d2\u2212\u03bbt=0\u22123.4657<br \/>\n\u21d2t=3.4657\u03bb\u21d2t=3.4657\u03bb<br \/>\nBut, since \u03bb=0.693T\u03bb=0.693T<br \/>\n\u21d2t=3.466(0.693T)\u21d2t=3.466(0.693T)<br \/>\n\u2234t\u22485Tyears\u2234t\u22485Tyears<br \/>\nTherefore, we found that the isotope will take about 5T years in order to reduce to 3.1253.125 of its original value.<\/h3>\n<h2>2. 1% of its original value?<\/h2>\n<h3>Ans: After decay, let the amount of the radioactive isotope be N<br \/>\nIt is given that only 1% of N0N0 remains after decay. Hence, we could write:<br \/>\nNN0=1NN0=1<br \/>\nBut we know, NN0=e\u2212\u03bbtNN0=e\u2212\u03bbt<br \/>\n\u21d2e\u2212\u03bbt=1100\u21d2e\u2212\u03bbt=1100<br \/>\n\u21d2\u2212\u03bbt=ln1\u2212ln100\u21d2\u2212\u03bbt=ln\u20611\u2212ln\u2061100<br \/>\n\u21d2\u2212\u03bbt=0\u22124.602\u21d2\u2212\u03bbt=0\u22124.602<br \/>\n\u21d2t=4.6052\u03bb\u21d2t=4.6052\u03bb<br \/>\nSince we have, \u03bb=0.639T\u03bb=0.639T<br \/>\n\u21d2t=4.6052(0.693T)\u21d2t=4.6052(0.693T)<br \/>\n\u2234t=6.645Tyears\u2234t=6.645Tyears<br \/>\nTherefore, we found that the given isotope would take about 6.645T years to reduce to 1% of its original value.<\/h3>\n<h2>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 of radioactive 146C614C present with the stable carbon isotope 126C612C . 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 146C614C , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 146C614C 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.<\/h2>\n<h3>Ans: We are given that:<br \/>\nDecay rate of living carbon-containing matter,<br \/>\nR=15decay\/minR=15decay\/min<br \/>\nLet N be the number of radioactive atoms present in a normal carbon- containing matter.<br \/>\nHalf life of 146C,T12=5730years614C,T12=5730years<br \/>\nThe decay rate of the specimen obtained from the Mohenjodaro site:<br \/>\nR\u2032=9decays\/minR\u2032=9decays\/min<br \/>\nLet N&#8217; be the number of radioactive atoms present in the specimen during the Mohenjodaro period.<br \/>\nTherefore, we can relate the decay constant, \u03bb\u03bb and time, t as:<br \/>\nN\u2032N=R\u2032R=e\u2212\u03bbtN\u2032N=R\u2032R=e\u2212\u03bbt<br \/>\n\u21d2e\u2212\u03bbt=915=35\u21d2e\u2212\u03bbt=915=35<br \/>\n\u21d2\u2212\u03bbt=loge35=\u22120.5108\u21d2\u2212\u03bbt=loge35=\u22120.5108<br \/>\n\u21d2t=0.5108\u03bb\u21d2t=0.5108\u03bb<br \/>\nBut we know,<br \/>\n\u03bb=0.693T12=0.6935730\u03bb=0.693T12=0.6935730<br \/>\n\u21d2t=0.5108(0.6935730)=4223.5years\u21d2t=0.5108(0.6935730)=4223.5years<br \/>\nTherefore, the approximate age of the Indus-Valley civilization is found to be 4223.5 years.<\/h3>\n<h2>9. Obtain the Amount of 6027Co2760Co necessary to provide a radioactive source of 8.0mCi8.0mCistrength. The half-life of 6027Co2760Co is 5.3 years.<\/h2>\n<h3>Ans: We know that,<br \/>\nThe strength of the radioactive source could be given as,<br \/>\ndNdt=8.0mCidNdt=8.0mCi<br \/>\n\u21d2dNdt=8\u00d710\u22123\u00d73.7\u00d71010=29.6\u00d7107decay\/s\u21d2dNdt=8\u00d710\u22123\u00d73.7\u00d71010=29.6\u00d7107decay\/s<br \/>\nWhere, N is the required number of atoms.<br \/>\nHalf life of 6027Co2760Co, T12=5.3yearsT12=5.3years<br \/>\n\u21d2T12=5.3\u00d7365\u00d724\u00d760\u00d760=1.67\u00d7108s\u21d2T12=5.3\u00d7365\u00d724\u00d760\u00d760=1.67\u00d7108s<br \/>\nFor decay constant \u03bb\u03bb, we could give the rate of decay as,<br \/>\ndNdt=\u03bbNdNdt=\u03bbN<br \/>\nWhere, \u03bb=0.693T12=0.6931.67\u00d7108s\u22121\u03bb=0.693T12=0.6931.67\u00d7108s\u22121<br \/>\n\u21d2N=1\u03bbdNdt=29.6\u00d7107(0.6931.67\u00d7108)=7.133\u00d71016atoms\u21d2N=1\u03bbdNdt=29.6\u00d7107(0.6931.67\u00d7108)=7.133\u00d71016atoms<br \/>\nNow for 27Co6027Co60, Mass of Avogadro number of atoms =60g=60g<br \/>\nThen, mass of 7.133\u00d71016atoms=60\u00d77.133\u00d710166.023\u00d71023=7.106\u00d710\u22126g7.133\u00d71016atoms=60\u00d77.133\u00d710166.023\u00d71023=7.106\u00d710\u22126g<br \/>\nTherefore, the amount of 27Co6027Co60that is required for the purpose is 7.106\u00d710\u22126g7.106\u00d710\u22126g.<\/h3>\n<h2>10. The half life of 9038Sr3890Sr is 28years. What is the disintegration rate of 15mg of this isotope?<\/h2>\n<h3>Ans: We know that,<br \/>\nHalf life of 9038Sr3890Sr, t12=28years=28\u00d7365\u00d724\u00d73600=8.83\u00d7108st12=28years=28\u00d7365\u00d724\u00d73600=8.83\u00d7108s<br \/>\nMass of the isotope, m=15mgm=15mg<br \/>\n90g of 9038Sr3890Sr atom contains Avogadro number of atoms. So, 15mg of 9038Sr3890Sr contains,<br \/>\n6.023\u00d71023\u00d715\u00d710\u2212390=1.0038\u00d71020number of atoms6.023\u00d71023\u00d715\u00d710\u2212390=1.0038\u00d71020number of atoms<br \/>\nRate of disintegration would be, dNdt=\u03bbNdNdt=\u03bbN<br \/>\nWhere, \u03bb\u03bbis the decay constant given by, \u03bb=0.6938.83\u00d7108s\u22121\u03bb=0.6938.83\u00d7108s\u22121<br \/>\n\u2234dNdt=0.693\u00d71.0038\u00d710208.83\u00d7108=7.878\u00d71010atoms\/s\u2234dNdt=0.693\u00d71.0038\u00d710208.83\u00d7108=7.878\u00d71010atoms\/s<br \/>\nTherefore, we found the disintegration rate of 15mg of given isotope to be 7.878\u00d71010atoms\/s7.878\u00d71010atoms\/s.<\/h3>\n<h2>11. Obtain approximately the ratio of the nuclear radii of the gold isotope 19779Au79197Au and the silver isotope 10747Ag47107Ag.<\/h2>\n<h3>Ans: We know that,<br \/>\nNuclear radius of the gold isotope 79Au197=RAu79Au197=RAu<br \/>\nNuclear radius of the silver isotope 47Ag107=RAg47Ag107=RAg<br \/>\nMass number of gold, AAu=197AAu=197<br \/>\nMass number of silver, AAg=107AAg=107<br \/>\nWe also know that the ratio of the radii of the two nuclei is related with their mass numbers as:<br \/>\nRAuRAg=(AAuAAg)13=1.2256RAuRAg=(AAuAAg)13=1.2256<br \/>\nHence, the ratio of the nuclear radii of the gold and silver isotopes is found to be about 1.23.<\/h3>\n<h2>12. Find the Q-value and the kinetic energy of the emitted \u03b1-particle in the \u03b1-decay of: Given:$m(22688Ra)=226.02540u,m(22289Rn)=222.01750u,$$m(22086Rn)=220.01137u,m(21684Po)=216.00189u$Given:$m(88226Ra)=226.02540u,m(89222Rn)=222.01750u,$$m(86220Rn)=220.01137u,m(84216Po)=216.00189u$<br \/>\n1. 22688Ra88226Ra<\/h2>\n<h3>Ans: We know that,<br \/>\nAlpha particle decay of 2688Ra8826Ra emits a helium nucleus. As a result, its mass number reduces to 222=(226\u22124)222=(226\u22124)and its atomic number reduces to86=(88\u22122)86=(88\u22122). This is shown in the following nuclear reaction:<br \/>\n22688Ra\u219222286Ra+42He88226Ra\u219286222Ra+24He<br \/>\nQ\u2212valueofemitted\u03b1\u2212particle=(Sumofinitialmass\u2212Sumoffinalmass)c2Q\u2212valueofemitted\u03b1\u2212particle=(Sumofinitialmass\u2212Sumoffinalmass)c2<br \/>\nWhere, c = Speed of light<br \/>\nIt is also given that:<br \/>\nm(22688Ra)=226.02540um(88226Ra)=226.02540u<br \/>\nm(22086Rn)=220.01137um(86220Rn)=220.01137u<br \/>\nm(42He)=4.002603um(24He)=4.002603u<br \/>\nOn substituting these values into the above equation,<br \/>\nQ value =[226.02540\u2212(222.01750+4.002603)]uc2Q value =[226.02540\u2212(222.01750+4.002603)]uc2<br \/>\nQvalue=0.005297uc2Qvalue=0.005297uc2<br \/>\nBut we know, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u21d2Q=0.005297\u00d7931.5\u22484.94MeV\u21d2Q=0.005297\u00d7931.5\u22484.94MeV<br \/>\nKinetic energy of the \u03b1\u03b1particle=(Mass number after decayMassnumber before decay)\u00d7Q=(Mass number after decayMassnumber before decay)\u00d7Q<br \/>\n\u2234K.E\u03b1=222226\u00d74.94=4.85MeV\u2234K.E\u03b1=222226\u00d74.94=4.85MeV<br \/>\nHence, the Kinetic energy of the alpha particle is found to be 4.85MeV4.85MeV.<\/h3>\n<h2>2. 22086Rn86220Rn<\/h2>\n<h3>Ans: We know that, Alpha particle decay of 22086Rn86220Rn could be given as,<br \/>\n22086Rn\u219221684Po+42He86220Rn\u219284216Po+24He<br \/>\nWe are also given,<br \/>\nMass of 22086Rn=220.01137u86220Rn=220.01137u<br \/>\nMass of 21684Po=216.00189u84216Po=216.00189u<br \/>\nNow, Q value could be given as,<br \/>\nQ\u2212value=[220.01137\u2212(216.00189+4.00260)]\u00d7931.5\u2248641MeVQ\u2212value=[220.01137\u2212(216.00189+4.00260)]\u00d7931.5\u2248641MeV<br \/>\nNow, we have the kinetic energy as,<br \/>\nK.E\u03b1=(220\u22124220)\u00d76.41=6.29MeVK.E\u03b1=(220\u22124220)\u00d76.41=6.29MeV<br \/>\nThe kinetic energy of the alpha particle is found to be 6.29MeV6.29MeV.<\/h3>\n<h2>13. The Radionuclide 116C611C decays according to,<br \/>\n116C\u2192115B+e++\u03bd;T12=20.3min611C\u2192511B+e++\u03bd;T12=20.3min The maximum energy of the emitted positron is 0.960MeV0.960MeV. Given the mass values: m(116C)=11.011434uand m(116B)=11.009305um(611C)=11.011434uand m(611B)=11.009305u Calculate Q and compare it with the maximum energy of the positron emitted.<\/h2>\n<h3>Ans: The given nuclear reaction is,<br \/>\n116C\u2192115B+e++\u03bd611C\u2192511B+e++\u03bd<br \/>\nHalf life of 116C611C nuclei, T12=20.3minT12=20.3min<br \/>\nAtomic masses are given to be:<br \/>\nm(116C)=11.011434um(611C)=11.011434u<br \/>\nm(116B)=11.009305um(611B)=11.009305u<br \/>\nMaximum energy that is possessed by the emitted positron would be<br \/>\n0.960MeV0.960MeV<br \/>\n. The change in the Q &#8211; value<br \/>\n(\u0394Q)(\u0394Q)<br \/>\nof the nuclear masses of the 116C611C<br \/>\n\u0394Q=[m\u2032(6C11)\u2212[m\u2032(115B)+me]]c2\u0394Q=[m\u2032(6C11)\u2212[m\u2032(511B)+me]]c2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (1)<br \/>\nWhere,me=me=Mass of an electron or positron =<br \/>\n0.000548u0.000548u<br \/>\nc=c= Speed of light<br \/>\nm\u2032=m\u2032=Respective nuclear masses<br \/>\nIf atomic masses are used instead of nuclear masses, then we will have to add 6me6me in the case of 11C11C and 5me5mein case of 11B11B.<br \/>\nHence, equation (1) would now reduce to,<br \/>\n\u0394Q=[m(6C11)\u2212m(115B)\u22122me]c2\u0394Q=[m(6C11)\u2212m(511B)\u22122me]c2<br \/>\nWhere, m(6C11)and m(115B)m(6C11)and m(511B)are the atomic masses.<br \/>\nNow, we have the change in Q value as,<br \/>\n\u0394Q=[11.011434\u221211.009305\u22122\u00d70.000548]c2=(0.001033c2)u\u0394Q=[11.011434\u221211.009305\u22122\u00d70.000548]c2=(0.001033c2)u<br \/>\nBut we know, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u2234\u0394Q=0.001033\u00d7931.5\u22480.962MeV\u2234\u0394Q=0.001033\u00d7931.5\u22480.962MeV<br \/>\nWe see that the Q value is almost comparable to the maximum energy of the emitted positron.<\/h3>\n<h2>14. The nucleus 2310Ne1023Ne decays by \u03b2\u2212\u03b2\u2212emission. Write down the \u03b2\u03b2decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:<br \/>\nm(2310Ne)=22.994466um(1023Ne)=22.994466u<br \/>\nm(2311Na)=22.989770um(1123Na)=22.989770u<\/h2>\n<h3>Ans: We know that: In \u03b2\u2212\u03b2\u2212emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus. \u03b2\u2212\u03b2\u2212emission of the nucleus could be given by,<br \/>\n2310Ne\u21922311Na+e\u2212+\u03bd\u00af+Q1023Ne\u21921123Na+e\u2212+\u03bd\u00af+Q<br \/>\nIt is also given that:<br \/>\nAtomic mass of 2310Ne=22.994466u1023Ne=22.994466u<br \/>\nAtomic mass of 2311Na=22.989770u1123Na=22.989770u<br \/>\nMass of an electron, me=0.000548ume=0.000548u<br \/>\nQ value of the given reaction could be given as:<br \/>\nQ=[m(2310Ne)\u2212[m(2311Na)+me]]c2Q=[m(1023Ne)\u2212[m(1123Na)+me]]c2<br \/>\nThere are 10 electrons in 10Ne2310Ne23and 11 electrons in 2311Na1123Na. Hence, the mass of the electron is cancelled in the Q-value equation.<br \/>\nQ=[22.994466\u221222.9897770]c2=(0.004696c2)uQ=[22.994466\u221222.9897770]c2=(0.004696c2)u<br \/>\nBut we have, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u21d2Q=0.004696\u00d7931.5=4.374MeV\u21d2Q=0.004696\u00d7931.5=4.374MeV<br \/>\nThe daughter nucleus is too heavy as compared to that of e\u2212and\u03bd\u00afe\u2212and\u03bd\u00af. Hence, it carries negligible energy. The kinetic energy of the antineutrino is found to be nearly zero.<br \/>\nHence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e.,<br \/>\n4.374MeV4.374MeV.<\/h3>\n<h2>15. The Q-value of a nuclear reaction A+b\u2192C+dA+b\u2192C+d is defined by Q=[mA+mb\u2212mC\u2212md]c2Q=[mA+mb\u2212mC\u2212md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.<br \/>\nAtomic masses are given to be: m(21H)=2.014102u,m(31H)=3.016049u,m(12H)=2.014102u,m(13H)=3.016049u,<br \/>\nm(126C)=12.000000u,m(2010Ne)=19.992439um(612C)=12.000000u,m(1020Ne)=19.992439u<\/h2>\n<h2>1. 11H+31H\u219221H+21H11H+13H\u219212H+12H<\/h2>\n<h3>Ans:The given nuclear reaction is:<br \/>\n11H+31H\u219221H+21H11H+13H\u219212H+12H<br \/>\nAtomic mass of 11H=1.007825u11H=1.007825u<br \/>\nAtomic mass of 31H=3.0164049u13H=3.0164049u<br \/>\nAtomic mass of 21H=2.014102u12H=2.014102u<br \/>\nAccording to the question, the Q-value of the reaction could be written as:<br \/>\nQ=[m(11H)+m(31H)\u22122m(21H)]c2Q=[m(11H)+m(13H)\u22122m(12H)]c2<br \/>\n\u21d2Q=[1.007825+3.016049\u22122\u00d72.014102]c2=(\u22120.00433c2)u\u21d2Q=[1.007825+3.016049\u22122\u00d72.014102]c2=(\u22120.00433c2)u<br \/>\nBut we know, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u2234Q=\u22120.00433\u00d7931.5=\u22124.0334MeV\u2234Q=\u22120.00433\u00d7931.5=\u22124.0334MeV<br \/>\nThe negative Q-value of this reaction shows that the given reaction is endothermic.<\/h3>\n<h2>2. 126C+126C\u21922010Ne+42He612C+612C\u21921020Ne+24He<\/h2>\n<h3>Ans: We are given that,<br \/>\nAtomic mass of 126C=12.0u612C=12.0u<br \/>\nAtomic mass of 1210Ne=19.992439u1012Ne=19.992439u<br \/>\nAtomic mass of 42He=4.002603u24He=4.002603u<br \/>\nThe Q-value here could be given as,<br \/>\nQ=[2m(126C)\u2212m(2010Ne)\u2212m(42He)]c2Q=[2m(612C)\u2212m(1020Ne)\u2212m(24He)]c2<br \/>\n\u21d2Q=[2\u00d712.0\u221219.992439\u22124.002603]c2=(0.004958c2)u=0.004958\u00d7931.5\u21d2Q=[2\u00d712.0\u221219.992439\u22124.002603]c2=(0.004958c2)u=0.004958\u00d7931.5<br \/>\n\u2234Q=4.618377MeV\u2234Q=4.618377MeV<br \/>\nSince the Q-value is found to be positive, the reaction could be considered exothermic.<\/h3>\n<h2>16. Suppose, we think of fission of a 5626Fe2656Fe nucleus into two equal fragments of 13Al2813Al28. Is fission energetically possible? Argue by working out Q of the process. Given: m(5626Fe)=55.93494uand m(2813Al)=27.98191um(2656Fe)=55.93494uand m(1328Al)=27.98191u<\/h2>\n<h3>Ans: We know that the fission of 5626Fe2656Fe could be given as,<br \/>\n5626Fe\u219222813Al2656Fe\u219221328Al<br \/>\nWe are also given, atomic masses of 5626Feand2813Al2656Feand1328Al as 55.93494uand 27.98191u55.93494uand 27.98191u respectively.<br \/>\nThe Q-value here would be given as,<br \/>\nQ=[m(5626Fe)\u22122m(2813Al)]c2Q=[m(2656Fe)\u22122m(1328Al)]c2<br \/>\n\u21d2Q=[55.93494\u22122\u00d727.98191]c2=(\u22120.02888c2)u\u21d2Q=[55.93494\u22122\u00d727.98191]c2=(\u22120.02888c2)u<br \/>\nBut, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u2234Q=\u22120.02888\u00d7931.5=\u221226.902MeV\u2234Q=\u22120.02888\u00d7931.5=\u221226.902MeV<br \/>\nThe Q value is found to be negative and hence we could say that the fission is not possible energetically. In order for a reaction to be energetically possible, the Q-value must be positive.<\/h3>\n<h2>17. The fission properties of 23994Pu94239Pu are very similar to those of<br \/>\n23592U92235U.The average energy released per fission is 180MeV180MeV. How much energy, in MeV, is released if all the atoms in 1kg1kg of pure 23994Pu94239Pu undergo fission?<\/h2>\n<h3>Ans: We are given that the average energy released per fission of 23994Pu94239Pu, Eav=180MeVEav=180MeV<br \/>\nThe amount of pure 94Pu23994Pu239, m=1kg=1000gm=1kg=1000g<br \/>\nAvogadro number, NA=6.023\u00d71023NA=6.023\u00d71023<br \/>\nMass number of 23994Pu=239g94239Pu=239g<br \/>\n1 mole of 94Pu23994Pu239contains Avogadro number of atoms.<br \/>\n1gof94Pu239contains(NAmassnumber\u00d7m)atoms1gof94Pu239contains(NAmassnumber\u00d7m)atoms<br \/>\n\u21d2(6.023\u00d71023239\u00d71000)=2.52\u00d71024atoms\u21d2(6.023\u00d71023239\u00d71000)=2.52\u00d71024atoms<br \/>\nTotal energy released during the fission of 1kg of 23994Pu94239Pucould be calculated as:<br \/>\nE=Eav\u00d72.52\u00d71024=180\u00d72.52\u00d71024=4.536\u00d71026MeVE=Eav\u00d72.52\u00d71024=180\u00d72.52\u00d71024=4.536\u00d71026MeV<br \/>\nTherefore, 4.536\u00d71026MeV4.536\u00d71026MeV is released if all the atoms in 1kg of pure 94Pu23994Pu239undergo fission.<\/h3>\n<h2>18. A 1000MW fission reactor consumes half of its fuel in 5.00 y. How much<br \/>\n23592U92235U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 23592U92235U and that this nuclide is consumed only by the fission process.<\/h2>\n<h3>Ans: We are said that the half life of the fuel of the fission reactor, t12=5yearst12=5years<br \/>\n\u21d2t12=5\u00d7365\u00d724\u00d760\u00d760s\u21d2t12=5\u00d7365\u00d724\u00d760\u00d760s<br \/>\nWe know that in the fission of 1g of 23592U92235U nucleus, the energy released is equal to 200MeV.<br \/>\n1 mole, i.e., 235g of 23592U92235U contains<br \/>\n6.023\u00d710236.023\u00d71023<br \/>\natoms.<br \/>\n1gof23592Ucontains6.023\u00d71023234atoms1gof92235Ucontains6.023\u00d71023234atoms<br \/>\nThe total energy generated per gram of 23592U92235U is calculated as:<br \/>\nE=6.023\u00d71023235\u00d7200MeV\/g=200\u00d76.023\u00d71023\u00d71.6\u00d710\u221219\u00d7106235=8.20\u00d71010J\/gE=6.023\u00d71023235\u00d7200MeV\/g=200\u00d76.023\u00d71023\u00d71.6\u00d710\u221219\u00d7106235=8.20\u00d71010J\/g<br \/>\nThe reactor operator operates only 80% of the time. Therefore, the amount of 23592U92235U consumed in 5years by the 1000MW fission reactor could be calculated as,<br \/>\n5\u00d780\u00d760\u00d760\u00d7365\u00d724\u00d71000\u00d7106100\u00d78.20\u00d71010g\u22481538kg5\u00d780\u00d760\u00d760\u00d7365\u00d724\u00d71000\u00d7106100\u00d78.20\u00d71010g\u22481538kg<br \/>\nSo, the initial amount of 23592U=2\u00d71538=3076kg92235U=2\u00d71538=3076kg<br \/>\nHence, we found the initial amount of uranium to be 3076kg.3076kg.<\/h3>\n<h2>19. How long can an electric lamp of 100W be kept glowing by fusion of 2.0kg of deuterium? Take the fusion reaction as 21H+21H\u219232He+n+3.27MeV12H+12H\u219223He+n+3.27MeV<\/h2>\n<h3>Ans: The fusion reaction is given to be:<br \/>\n21H+21H\u219232He+n+3.27MeV12H+12H\u219223He+n+3.27MeV<br \/>\nAmount of deuterium,<br \/>\nm=2kgm=2kg<br \/>\n1 mole, i.e., 2 g of deuterium contains<br \/>\n6.023\u00d71023atoms6.023\u00d71023atoms<br \/>\n.<br \/>\n2.0 kg of deuterium contains 6.023\u00d710232\u00d72000=6.023\u00d71026atoms6.023\u00d710232\u00d72000=6.023\u00d71026atoms atoms<br \/>\nIt could be inferred from the given reaction that when two atoms of deuterium fuse,<br \/>\n3.27MeV3.27MeV<br \/>\nenergy is released.<br \/>\nTherefore, the total energy per nucleus released in the fusion reaction would be:<br \/>\nE=3.272\u00d76.023\u00d71026MeV=3.272\u00d76.023\u00d71026\u00d71.6\u00d710\u221219\u00d7106E=3.272\u00d76.023\u00d71026MeV=3.272\u00d76.023\u00d71026\u00d71.6\u00d710\u221219\u00d7106<br \/>\n\u2234E=1.576\u00d71014J\u2234E=1.576\u00d71014J<br \/>\nPower of the electric lamp is given to be, P=100W=100J\/sP=100W=100J\/s, that is, the energy consumed by the lamp per second is 100J.<br \/>\nNow, the total time for which the electric lamp glows could be calculated as,<br \/>\nt=1.576\u00d71014100=1.576\u00d71014100\u00d760\u00d760\u00d724\u00d7365t=1.576\u00d71014100=1.576\u00d71014100\u00d760\u00d760\u00d724\u00d7365<br \/>\n\u2234t\u22484.9\u00d7104years\u2234t\u22484.9\u00d7104years<br \/>\nHence, the total time for which the electric lamp glows is found to be 4.9\u00d7104years.4.9\u00d7104years.<\/h3>\n<h2>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 2.0fm.)<\/h2>\n<h3>Ans: When two deuterons collide head-on, the distance between their centres, d could be given as:<br \/>\nRadius of 1st deuteron ++Radius of 2nd deuteron<br \/>\nRadius of a deuteron nucleus=2fm=2\u00d710\u221215m=2fm=2\u00d710\u221215m<br \/>\n\u21d2d=2\u00d710\u221215+2\u00d710\u221215=4\u00d710\u221215m\u21d2d=2\u00d710\u221215+2\u00d710\u221215=4\u00d710\u221215m<br \/>\nAlso, charge on a deuteron == Charge on an electron =e=1.6\u00d710\u221219C=e=1.6\u00d710\u221219C<br \/>\nPotential energy of the two-deuteron system could be given by,<br \/>\nV=e24\u03c0\u03b50dV=e24\u03c0\u03b50d<br \/>\nWhere, \u03b50\u03b50is the permittivity of free space.<br \/>\nAlso, 14\u03c0\u03b50=9\u00d7109Nm2C\u2212214\u03c0\u03b50=9\u00d7109Nm2C\u22122<br \/>\n\u21d2V=9\u00d7109\u00d7(1.6\u00d71019)24\u00d71015J=9\u00d7109\u00d7(1.6\u00d710\u221219)24\u00d710\u221215\u00d7(1.6\u00d710\u221219)eV\u21d2V=9\u00d7109\u00d7(1.6\u00d71019)24\u00d71015J=9\u00d7109\u00d7(1.6\u00d710\u221219)24\u00d710\u221215\u00d7(1.6\u00d710\u221219)eV<br \/>\n\u2234V=360keV\u2234V=360keV<br \/>\nTherefore, we found the height of the potential barrier of the two-deuteron system to be 360keV.<\/h3>\n<h2>21. From the relation R=R0A13R=R0A13, where R0R0 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).<\/h2>\n<h3>Ans: We know the expression for nuclear radius to be:<br \/>\nR=R0A13R=R0A13<br \/>\nWhere, R0R0 is a Constant and AA is the mass number of the nucleus<br \/>\nNuclear matter density would be,<br \/>\n\u03c1=Mass of the nucleusVolume of the nucleus\u03c1=Mass of the nucleusVolume of the nucleus<br \/>\nNow, let m be the average mass of the nucleus, then, mass of the nucleus =mA=mA<br \/>\nNuclear density,<br \/>\n\u03c1=mA43\u03c0R3=3mA4\u03c0(R0A13)3=3mA4\u03c0R03A\u03c1=mA43\u03c0R3=3mA4\u03c0(R0A13)3=3mA4\u03c0R03A<br \/>\n\u2234\u03c1=3m4\u03c0R03\u2234\u03c1=3m4\u03c0R03<br \/>\nTherefore, we found the nuclear matter density to be independent of A and it is found to be nearly constant.<\/h3>\n<h2>22. For the \u03b2+\u03b2+(positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K\u2212 shell, is captured by the nucleus and a neutrino is emitted).<br \/>\ne++AZX\u2192AZ\u22121Y+\u03bde++ZAX\u2192Z\u22121AY+\u03bd<br \/>\nShow that if \u03b2+\u03b2+emission is Energetically Allowed, Electron Capture is Necessarily Allowed but Not Vice\u2212versa.<\/h2>\n<h3>Ans: Let the amount of energy released during the electron capture process be Q1Q1 . The nuclear reaction could be written as:<br \/>\ne++AZX\u2192AZ\u22121Y+\u03bd+Q1e++ZAX\u2192Z\u22121AY+\u03bd+Q1\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (1)<br \/>\nLet the amount of energy released during the positron capture process be Q2Q2. The nuclear reaction could be written as:<br \/>\nAZX\u2192AZ\u22121Y+e++\u03bd+Q2ZAX\u2192Z\u22121AY+e++\u03bd+Q2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (2)<br \/>\nLet, mN(AZX)be the nuclear mass ofAZXmN(ZAX)be the nuclear mass ofZAX, mN(AZ\u22121Y)be the nuclear mass ofAZ\u22121YmN(Z\u22121AY)be the nuclear mass ofZ\u22121AY<br \/>\nm(AZX)be the atomic mass ofAZXm(ZAX)be the atomic mass ofZAX<br \/>\nm(AZ\u22121Y)be the nuclear mass ofAZ\u22121Ym(Z\u22121AY)be the nuclear mass ofZ\u22121AY<br \/>\nmebe the mass of an electronmebe the mass of an electron, cbe the speed of lightcbe the speed of light, then, the Q-value of the electron capture reaction could be given as,<br \/>\nQ1=[mN(AZX)+me\u2212mN(AZ\u22121Y)]c2Q1=[mN(ZAX)+me\u2212mN(Z\u22121AY)]c2<br \/>\n\u21d2Q1=[m(AZX)\u2212Zme+me\u2212m(AZ\u22121Y)+(Z\u22121)me]c2\u21d2Q1=[m(ZAX)\u2212Zme+me\u2212m(Z\u22121AY)+(Z\u22121)me]c2<br \/>\n\u21d2Q1=[m(AZX)\u2212m(AZ\u22121Y)]c2\u21d2Q1=[m(ZAX)\u2212m(Z\u22121AY)]c2\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (3)<br \/>\nThe Q-value of the positron capture reaction could be given as,<br \/>\nQ2=[mN(AZX)\u2212mN(AZ\u22121Y)\u2212me]c2Q2=[mN(ZAX)\u2212mN(Z\u22121AY)\u2212me]c2<br \/>\n\u21d2Q1=[m(AZX)\u2212Zme\u2212m(AZ\u22121Y)+(Z\u22121)me\u2212me]c2\u21d2Q1=[m(ZAX)\u2212Zme\u2212m(Z\u22121AY)+(Z\u22121)me\u2212me]c2<br \/>\n\u21d2Q1=[m(AZX)\u2212m(AZ\u22121Y)\u22122me]c2\u21d2Q1=[m(ZAX)\u2212m(Z\u22121AY)\u22122me]c2\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (4)<br \/>\nIt can be inferred that if<br \/>\nQ2&gt;0Q2&gt;0<br \/>\n, then; Also, if<br \/>\nQ1&gt;0Q1&gt;0<br \/>\n, it does not necessarily mean that<br \/>\nQ2&gt;0Q2&gt;0<br \/>\n. In other words, we could say that if \u03b2+\u03b2+emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is so because the Q-value must be positive for an energetically-allowed nuclear reaction.<\/h3>\n<h2>23. In a periodic table the average atomic mass of magnesium is given as<br \/>\n24.312u24.312u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are:<br \/>\n2412Mg(23.98504u),2512Mg(24.98584u)and2612Mg(25.98259u)1224Mg(23.98504u),1225Mg(24.98584u)and1226Mg(25.98259u)The natural abundance of 12Mg2412Mg24is 78.99% by mass. Calculate the abundances of the other two isotopes.<\/h2>\n<h3>Ans: We are given:<br \/>\nAverage atomic mass of magnesium,<br \/>\nm=24.312um=24.312u<br \/>\nMass of magnesium 2412Mg1224Mg isotope,<br \/>\nm1=23.98504um1=23.98504u<br \/>\nMass of magnesium 2512Mg1225Mg isotope,<br \/>\nm2=24.98584um2=24.98584u<br \/>\nMass of magnesium 2612Mg1226Mg isotope,<br \/>\nm3=25.98259um3=25.98259u<br \/>\nAbundance of 2412Mg,\u03b71=78.991224Mg,\u03b71=78.99<br \/>\nAbundance of 2512Mg,\u03b72=x1225Mg,\u03b72=x<br \/>\nNow, the abundance of 2612Mg1226Mg,<br \/>\n\u03b73=100\u2212x\u221278.99\u03b73=100\u2212x\u221278.99<br \/>\nAlso, we have the relation for the average atomic mass as:<br \/>\nm=m1\u03b71+m2\u03b72+m3\u03b73\u03b71+\u03b72+\u03b73m=m1\u03b71+m2\u03b72+m3\u03b73\u03b71+\u03b72+\u03b73<br \/>\n\u21d224.312=23.98504\u00d778.99+24.98584\u00d7x+25.98259\u00d7(21.01\u2212x)100\u21d224.312=23.98504\u00d778.99+24.98584\u00d7x+25.98259\u00d7(21.01\u2212x)100<br \/>\n\u21d20.99675x=9.2725255\u21d20.99675x=9.2725255<br \/>\n\u2234x\u22489.3\u2234x\u22489.3<br \/>\nAnd, 21.01\u2212x=11.7121.01\u2212x=11.71<br \/>\nTherefore, we found the abundance of 2512Mg1225Mg to be 9.3% and that of 2612Mg1226Mg to be 11.71%.<\/h3>\n<h2>24. The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei 4120Ca2041Ca and 2713Al1327Al from the following data:<br \/>\nm(4020Ca)=39.962591u,m(4120Ca)=40.962278u,m(2613Al)=25.986995u,m(2040Ca)=39.962591u,m(2041Ca)=40.962278u,m(1326Al)=25.986995u,<br \/>\nm(2713Al)=26.981541um(1327Al)=26.981541u<\/h2>\n<h3>Ans: For a neutron removal from 20Ca4120Ca41nucleus, the corresponding nuclear reaction could be written as,<br \/>\n4120Ca\u21924020Ca+10n2041Ca\u21922040Ca+01n<br \/>\nWe are given:<br \/>\nm(4020Ca)=39.962591um(2040Ca)=39.962591u<br \/>\nm(4120Ca)=40.962278um(2041Ca)=40.962278u<br \/>\nm(0n1)=1.008665um(0n1)=1.008665u<br \/>\nNow, the mass defect for this reaction could be given by,<br \/>\n\u0394m=m(4020Ca)+(10n)\u2212m(4120Ca)\u0394m=m(2040Ca)+(01n)\u2212m(2041Ca)<br \/>\n\u21d2\u0394m=39.962591+1.008665\u221240.962278=0.008978u\u21d2\u0394m=39.962591+1.008665\u221240.962278=0.008978u<br \/>\nBut we know, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u21d2\u0394m=0.008978\u00d7931.5MeV\/c2\u21d2\u0394m=0.008978\u00d7931.5MeV\/c2<br \/>\nNow, we could calculate the energy required for the neutron removal by,<br \/>\nE=\u0394mc2E=\u0394mc2<br \/>\n\u21d2E=0.008978\u00d7931.5=8.363007MeV\u21d2E=0.008978\u00d7931.5=8.363007MeV<br \/>\nFor the case of 2713Al1327Al, the neutron removal reaction could be written as,<br \/>\n2713Al\u21922613Al+10n1327Al\u21921326Al+01n<br \/>\nWe are given,<br \/>\nm(2613Al)=25.986995um(1326Al)=25.986995u<br \/>\nm(2713Al)=26.981541um(1327Al)=26.981541u<br \/>\nNow, the mass defect here could be given by,<br \/>\n\u0394m=m(2613Al)+m(10n)\u2212m(2713Al)\u0394m=m(1326Al)+m(01n)\u2212m(1327Al)<br \/>\n\u21d2\u0394m=25.986895+1.008665\u221226.981541=0.014019u\u21d2\u0394m=25.986895+1.008665\u221226.981541=0.014019u<br \/>\n\u21d2\u0394m=0.014019\u00d7931.5MeV\/c2\u21d2\u0394m=0.014019\u00d7931.5MeV\/c2<br \/>\nTherefore, the energy that is required for the removal of neutron would be,<br \/>\nE=\u0394mc2=0.014019\u00d7931.5E=\u0394mc2=0.014019\u00d7931.5<br \/>\n\u2234E=13.059MeV\u2234E=13.059MeV<\/h3>\n<h2>25. A source contains two phosphorous radio nuclides 3215P(T12=14.3d)1532P(T12=14.3d) and 3315P(T12=25.3d)1533P(T12=25.3d)Initially, 10% of the decays come from 3315P1533P. How long must one wait until 90% do so?<\/h2>\n<h3>Ans: We are given:<br \/>\nHalf life of 3215P(T12=14.3d)1532P(T12=14.3d)<br \/>\nHalf life of 3315P(T12=25.3d)1533P(T12=25.3d)<br \/>\nNow, we know that nucleus decay is 10% of the total amount of decay.<br \/>\nAlso, the source has initially 10% of 3215P1532P nucleus and 90% of 3215P1532P nucleus.<br \/>\nSuppose after t days, the source has 10% of 3215P1532P nucleus and 90% of 3315P1533P<br \/>\nnucleus.<br \/>\nInitially we have:<br \/>\nNumber of 3315P1533P nucleus =N=N<br \/>\nNumber of 3215P1532P nucleus =9N=9N<br \/>\nFinally:<br \/>\nNumber of 3315P1533P nucleus =9N\u2032=9N\u2032<br \/>\nNumber of 3215P1532P nucleus=N\u2032=N\u2032<br \/>\nFor 3215P1532P nucleus, we could write the number ratio as:<br \/>\nN\u20329N=(12)tT12N\u20329N=(12)tT12<br \/>\n\u21d2N\u2032=9N(2)\u2212t14.3\u21d2N\u2032=9N(2)\u2212t14.3\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (1)<br \/>\nNow, for 3315P1533P, we could write the number ratio as,<br \/>\n9N\u2032N=(12)1T\u2032129N\u2032N=(12)1T\u203212<br \/>\n\u21d29N\u2032=N(2)\u2212t25.3\u21d29N\u2032=N(2)\u2212t25.3\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (2)<br \/>\nWe could now divide equation (1) by equation (2) to get,<br \/>\n19=9\u00d72(t25.3\u2212t14.3)19=9\u00d72(t25.3\u2212t14.3)<br \/>\n\u21d2181=2(\u221211t25.3\u00d714.3)\u21d2181=2(\u221211t25.3\u00d714.3)<br \/>\n\u21d2log1\u2212log81=\u221211t25.3\u00d714.3log1\u21d2log\u20611\u2212log\u206181=\u221211t25.3\u00d714.3log\u20611<br \/>\n\u21d2\u221211t25.3\u00d714.3=0\u22121.9080.301\u21d2\u221211t25.3\u00d714.3=0\u22121.9080.301<br \/>\n\u2234t=25.3\u00d714.3\u00d71.90811\u00d70.301\u2248208.5days\u2234t=25.3\u00d714.3\u00d71.90811\u00d70.301\u2248208.5days<br \/>\nTherefore, we found that it would take about 208.5days for 90% decay of 3315P1533P.<\/h3>\n<h2>26. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an \u03b1\u2212\u03b1\u2212particle. Consider the following decay processes:<br \/>\n22388Ra\u219220982Pb+146C88223Ra\u219282209Pb+614C<br \/>\n22388Ra\u219221986Rn+42He88223Ra\u219286219Rn+24He<br \/>\nCalculate the Q-values for these decays and determine that both are energetically allowed.<\/h2>\n<h3>Ans: Consider a 146C614C emission nuclear reaction,<br \/>\n22388Ra\u219220982Pb+146C88223Ra\u219282209Pb+614C<br \/>\nWe know that:<br \/>\nMass of 22388Ra,m1=223.01850u88223Ra,m1=223.01850u<br \/>\nMass of 146C,m3=14.00324u614C,m3=14.00324u<br \/>\nNow, the Q-value of the reaction could be given as:<br \/>\nQ=(m1\u2212m2\u2212m3)c2Q=(m1\u2212m2\u2212m3)c2<br \/>\n\u21d2Q=(223.01850\u2212208.98107\u221214.00324)c2=(0.03419c2)u\u21d2Q=(223.01850\u2212208.98107\u221214.00324)c2=(0.03419c2)u<br \/>\nBut we have, 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u21d2Q=0.03419\u00d7931.5\u21d2Q=0.03419\u00d7931.5<br \/>\n\u2234Q=31.848MeV\u2234Q=31.848MeV<br \/>\nHence, the Q-value of the nuclear reaction is found to be 31.848 MeV. Since the value is positive, the reaction is energetically allowed.<br \/>\nNow consider a 42He24He emission nuclear reaction:<br \/>\n22388Ra\u219222986Rn+42He88223Ra\u219286229Rn+24He<br \/>\nWe know that:<br \/>\nMass of 22388Ra,m1=223.0185088223Ra,m1=223.01850<br \/>\nMass of 21982Rn,m2=219.0094882219Rn,m2=219.00948<br \/>\nMass of 42He,m3=4.0026024He,m3=4.00260<br \/>\nQ-value of this nuclear reaction could be given as:<br \/>\nQ=(m1\u2212m2\u2212m3)c2Q=(m1\u2212m2\u2212m3)c2<br \/>\n\u21d2Q=(223.01850\u2212219.00948\u22124.00260)c2\u21d2Q=(223.01850\u2212219.00948\u22124.00260)c2<br \/>\n\u21d2Q=(0.00642c2)u\u21d2Q=(0.00642c2)u<br \/>\n\u2234Q=0.00642\u00d7931.5=5.98MeV\u2234Q=0.00642\u00d7931.5=5.98MeV<br \/>\nTherefore, the Q-value of the second nuclear reaction is found to be 5.98MeV. Since the value is positive, we could say that the reaction is energetically allowed.<\/h3>\n<h2>27. Consider the fission of 23892U92238U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are 14058Ceand9944Ru58140Ceand4499Ru. Calculate Q for this fission process. The relevant atomic and particle masses are:<br \/>\nm(23892U)=238.05079um(92238U)=238.05079u<br \/>\nm(14058Ce)=139.90543um(58140Ce)=139.90543u<br \/>\nm(9944Ru)=98.90594um(4499Ru)=98.90594u<\/h2>\n<h3>Ans: We are given:<br \/>\nIn the fission of 23892U92238U, 10 \u03b2\u2212\u03b2\u2212particles decay from the parent nucleus. The nuclear reaction can be written as:<br \/>\n23892U+10n\u219214058Ce+9944Ru+100\u22121e92238U+01n\u219258140Ce+4499Ru+10\u221210e<br \/>\nIt is also given that:<br \/>\nMass of a nucleus of 23892U,m1=238.05079u92238U,m1=238.05079u<br \/>\nMass of a nucleus of 14058Ce,m2=139.90543u58140Ce,m2=139.90543u<br \/>\nMass of nucleus of 9944Ru,m3=98.90594u4499Ru,m3=98.90594u,<br \/>\nMass of a neutron 10n,m4=1.008665u01n,m4=1.008665u<br \/>\nQ-value of the above equation would be,<br \/>\nQ=[m\u2032(23892U)+m(10n)\u2212m\u2032(14058Ce)\u2212m\u2032(9944Ru)\u221210me]c2Q=[m\u2032(92238U)+m(01n)\u2212m\u2032(58140Ce)\u2212m\u2032(4499Ru)\u221210me]c2<br \/>\nWhere, m\u2032=m\u2032=Represents the corresponding atomic masses of the nuclei<br \/>\nm\u2032(23892U)=m1\u221292mem\u2032(92238U)=m1\u221292me<br \/>\nm\u2032(14058Ce)=m2\u221258mem\u2032(58140Ce)=m2\u221258me<br \/>\nm\u2032(9944Ru)=m3\u221244mem\u2032(4499Ru)=m3\u221244me<br \/>\nm(10n)=m4m(01n)=m4<br \/>\nQ=[m1\u221292me+m4\u2212m2+58me\u2212m3+44me\u221210me]c2Q=[m1\u221292me+m4\u2212m2+58me\u2212m3+44me\u221210me]c2<br \/>\n\u21d2Q=[m1+m4\u2212m2\u2212m3]c2=[238.0507+1.008665\u2212139.90543\u221298.90594]c2\u21d2Q=[m1+m4\u2212m2\u2212m3]c2=[238.0507+1.008665\u2212139.90543\u221298.90594]c2<br \/>\n\u21d2Q=[0.247995c2]u\u21d2Q=[0.247995c2]u<br \/>\nBut 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u2234Q=0.247995\u00d7931.5=231.007MeV\u2234Q=0.247995\u00d7931.5=231.007MeV<br \/>\nTherefore, the Q-value of the fission process is found to be 231.007MeV.<\/h3>\n<h2>28. Consider the D\u2212T reaction (deuterium\u2212tritium fusion)<br \/>\n21H+31H\u219242He+n12H+13H\u219224He+n<br \/>\n1. Calculate the energy released in MeV in this reaction from the data: m(21H)=2.014102um(12H)=2.014102u , m(31H)=3.016049um(13H)=3.016049u<\/h2>\n<h3>Ans: Consider the D-T nuclear reaction,<br \/>\n21H+31H\u219242He+n12H+13H\u219224He+n<br \/>\nWe are also given that:<br \/>\nMass of<br \/>\n21H,m1=2.014102u12H,m1=2.014102u<br \/>\nMass of<br \/>\n31H,m2=3.016049u13H,m2=3.016049u<br \/>\nMass of 42He,m3=4.002603u24He,m3=4.002603u<br \/>\nMass of 10n,m4=1.008665u01n,m4=1.008665u<br \/>\nNow, the Q-value of the given D-T reaction would be:<br \/>\nQ=[m1+m2\u2212m3\u2212m4]c2Q=[m1+m2\u2212m3\u2212m4]c2<br \/>\n\u21d2Q=[2.014102+3.016049\u22124.002603\u22121.008665]c2\u21d2Q=[2.014102+3.016049\u22124.002603\u22121.008665]c2<br \/>\n\u21d2Q=[0.018883c2]u\u21d2Q=[0.018883c2]u<br \/>\nBut 1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\n\u2234Q=0.018883\u00d7931.5=17.59MeV\u2234Q=0.018883\u00d7931.5=17.59MeV<\/h3>\n<h2>2. Consider the radius of both deuterium and tritium to be approximately 2.0fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles 2(3kT2)2(3kT2); k = Boltzmann\u2019s constant, T = absolute temperature.)<\/h2>\n<h3>Ans: We are given:<br \/>\nRadius of deuterium and tritium, r\u22482.0fm=2\u00d710\u221215mr\u22482.0fm=2\u00d710\u221215m<br \/>\nDistance between the two nuclei at the moment when they touch each other,<br \/>\nd=r+r=4\u00d710\u221215md=r+r=4\u00d710\u221215m<br \/>\nCharge on the deuterium nucleus =e=e<br \/>\nCharge on the tritium nucleus =e=e<br \/>\nHence, the repulsive potential energy between the two nuclei could be given as:<br \/>\nV=e24\u03c0\u03b50dV=e24\u03c0\u03b50d<br \/>\nWhere, \u03b50=\u03b50=Permittivity of free space<br \/>\n14\u03c0\u03b50=9\u00d7109Nm2c\u2212214\u03c0\u03b50=9\u00d7109Nm2c\u22122<br \/>\n\u21d2V=9\u00d7109\u00d7(1.6\u00d710\u221219)24\u00d710\u221215=5.76\u00d710\u221214J=5.76\u00d710\u2212141,6\u00d710\u221219\u21d2V=9\u00d7109\u00d7(1.6\u00d710\u221219)24\u00d710\u221215=5.76\u00d710\u221214J=5.76\u00d710\u2212141,6\u00d710\u221219<br \/>\n\u2234V=3.6\u00d7105eV=360keV\u2234V=3.6\u00d7105eV=360keV<br \/>\nTherefore, 5.76\u00d710\u221214J5.76\u00d710\u221214J or 360keV360keVof kinetic energy (KE) is needed to overcome the coulomb repulsion between the two nuclei.<br \/>\nHowever, we are also given that:<br \/>\nKE=2\u00d73kT2KE=2\u00d73kT2<br \/>\nWhere, k=k=Boltzmann constant<br \/>\nT=T=Temperature required for triggering the reaction<br \/>\n\u2234T=KE3k=5.76\u00d710\u2212143\u00d71.38\u00d710\u221223=1.39\u00d7109K\u2234T=KE3k=5.76\u00d710\u2212143\u00d71.38\u00d710\u221223=1.39\u00d7109K<br \/>\nTherefore, we found that the gas must be heated to a temperature of 1.39\u00d7109K1.39\u00d7109K to initiate the reaction.<\/h3>\n<h2>29. Obtain the maximum kinetic energy of \u03b2\u2212\u03b2\u2212particles, and the radiation frequencies of \u03b3\u03b3 decays in the decay scheme shown in figure. You are given that:<br \/>\nm(198Au)=197.968233um(198Au)=197.968233u<br \/>\nm(198Hg)=197.966760um(198Hg)=197.966760u<\/h2>\n<h3>Ans: It can be observed from the given \u03b3\u03b3-decay diagram that \u03b31\u03b31decays from the 1.088MeV energy level to the 0MeV energy level. Hence, the energy corresponding to \u03b31\u03b31-decay is given as:<br \/>\nE1=1.088\u22128=1.088MeVE1=1.088\u22128=1.088MeV<br \/>\n\u21d2h\u03bd1=1.088\u00d71.6\u00d710\u221219\u00d7106J\u21d2h\u03bd1=1.088\u00d71.6\u00d710\u221219\u00d7106J<br \/>\nWhere, Planck&#8217;s constant h=6.6\u00d710\u221234Jsh=6.6\u00d710\u221234Js<br \/>\n\u03bd1=\u03bd1=Frequency of radiation radiated by \u03b31\u2212\u03b31\u2212decay<br \/>\n\u03bd1=E1h\u03bd1=E1h<br \/>\n\u21d2\u03bd1=1.088\u00d71.6\u00d710\u221219\u00d71066.6\u00d710\u221234=2.637\u00d71020Hz\u21d2\u03bd1=1.088\u00d71.6\u00d710\u221219\u00d71066.6\u00d710\u221234=2.637\u00d71020Hz<br \/>\nIt can be observed from the given \u03b3\u2212\u03b3\u2212decay diagram that \u03b32\u03b32decays from the<br \/>\n0.412MeV0.412MeV<br \/>\nenergy level to the<br \/>\n0MeV0MeV<br \/>\nenergy level.<br \/>\nNow, the energy corresponding to \u03b32\u03b32-decay could be given as:<br \/>\nE2=0.412\u22120=0.412MeVE2=0.412\u22120=0.412MeV<br \/>\n\u21d2h\u03bd2=0.412\u00d71.6\u00d710\u221219\u00d7106J\u21d2h\u03bd2=0.412\u00d71.6\u00d710\u221219\u00d7106J<br \/>\nWhere, \u03bd2=\u03bd2=Frequency of radiation radiated by \u03b32\u2212\u03b32\u2212decay<br \/>\n\u03bd2=E2h=0.412\u00d71.6\u00d710\u221219\u00d71066.6\u00d710\u221234=9.988\u00d71019Hz\u03bd2=E2h=0.412\u00d71.6\u00d710\u221219\u00d71066.6\u00d710\u221234=9.988\u00d71019Hz<br \/>\nIt can be observed from the given \u03b3\u03b3-decay diagram that \u03b33\u03b33-decays from the<br \/>\n1.088MeV1.088MeV<br \/>\nenergy level to the<br \/>\n0.412MeV0.412MeV<br \/>\nenergy level.<br \/>\nNow, the energy corresponding to \u03b33\u03b33 -decay is given as:<br \/>\nE3=1.088\u22120.412=0.676MeVE3=1.088\u22120.412=0.676MeV<br \/>\n\u21d2h\u03bd3=0.676\u00d710\u221219\u00d7106\u21d2h\u03bd3=0.676\u00d710\u221219\u00d7106<br \/>\nWhere, \u03bd3=\u03bd3=Frequency of radiation radiated by \u03b33\u2212\u03b33\u2212decay<br \/>\n\u03bd3=E3h=0.676\u00d71.6\u00d710\u221219\u00d71066.6\u00d710\u221234=1.639\u00d71020Hz\u03bd3=E3h=0.676\u00d71.6\u00d710\u221219\u00d71066.6\u00d710\u221234=1.639\u00d71020Hz<br \/>\nMass of<br \/>\nm(19878Au)=197.968233um(78198Au)=197.968233u<br \/>\nMass of m(19880Hg)=197.966760um(80198Hg)=197.966760u<br \/>\n1u=931.5MeV\/c21u=931.5MeV\/c2<br \/>\nEnergy of the highest level could be given as:<br \/>\nE=[m(19878Au)\u2212m(19080Hg)]=197.968233\u2212197.966760=0.001473uE=[m(78198Au)\u2212m(80190Hg)]=197.968233\u2212197.966760=0.001473u<br \/>\n\u21d2E=0.001473\u00d7931.5=1.3720995MeV\u21d2E=0.001473\u00d7931.5=1.3720995MeV<br \/>\n\u03b21\u03b21decays from the<br \/>\n1.3720995MeV1.3720995MeV<br \/>\nlevel to the<br \/>\n1.088MeV1.088MeV<br \/>\nlevel<br \/>\nMaximum kinetic energy of the \u03b21\u03b21particle =1.3720995\u22121.088=1.3720995\u22121.088<br \/>\n\u21d2K.E=0.2840995MeV\u21d2K.E=0.2840995MeV<br \/>\n\u03b22\u03b22 decays from the 1.3720995MeV1.3720995MeVlevel to that of the 0.412MeV0.412MeVlevel. Now, we find the maximum kinetic energy of the \u03b22\u03b22particle to be,<br \/>\nK.Emax=1.3720995\u22120.412=0.9600995MeVK.Emax=1.3720995\u22120.412=0.9600995MeV<br \/>\nTherefore, we found the maximum kinetic energy of the \u03b22\u03b22particle to be 0.9600995MeV0.9600995MeV.<\/h3>\n<h2>30. Calculate and Compare the Energy Released By<br \/>\n1. fusion of 1.0kg of hydrogen deep within Sun and<\/h2>\n<h3>Ans: We are given:<br \/>\nAmount of hydrogen,<br \/>\nm=1kg=1000gm=1kg=1000g<br \/>\n1 mole, i.e., 1g of hydrogen (11H)(11H) contains<br \/>\n6.023\u00d71023atoms6.023\u00d71023atoms<br \/>\n.<br \/>\nThat is, 1000g of 11H11H contains<br \/>\n6.023\u00d71023atoms6.023\u00d71023atoms<br \/>\n.<br \/>\nWithin the sun, four 11H11H nuclei combine and form one 42He24He nucleus. In this process<br \/>\n26MeV26MeV<br \/>\nof energy is released.<br \/>\nHence, the energy released from the fusion of 1 kg 11H11H is:<br \/>\nE1=6.023\u00d71023\u00d726\u00d71034=39.1495\u00d71026MeVE1=6.023\u00d71023\u00d726\u00d71034=39.1495\u00d71026MeV<br \/>\nTherefore, we found the energy released during the fusion of 1kg 11H11H is:<br \/>\nE1=6.023\u00d71023\u00d726\u00d71034=39.1495\u00d71026MeVE1=6.023\u00d71023\u00d726\u00d71034=39.1495\u00d71026MeV<br \/>\nHence, the energy released during the fusion of 1kg of 11H11H to be 39.1495\u00d71026MeV39.1495\u00d71026MeV.<\/h3>\n<h2>2. The Fission of 1.0kg of 235U235U in a fission reactor.<\/h2>\n<h3>Ans: We are given:<br \/>\nAmount of 92U235=1000gm92U235=1000gm<br \/>\n1 mole, i.e., 235g of 23592U92235U contains<br \/>\n6.023\u00d71023atoms6.023\u00d71023atoms<br \/>\n.<br \/>\n1000g of 23592U92235U contains 6.023\u00d71023\u00d71000235atoms6.023\u00d71023\u00d71000235atoms<br \/>\nWe know that the amount of energy released in the fission of one atom of 23592U92235U is<br \/>\n200MeV200MeV<br \/>\n. Therefore, energy released from the fission of 1kg of 23592U92235U is:<br \/>\nE2=6\u00d71023\u00d71000\u00d7200235=5.106\u00d71026MeVE2=6\u00d71023\u00d71000\u00d7200235=5.106\u00d71026MeV<br \/>\nE1E2=39.1495\u00d710265.106\u00d71026=7.67\u22488E1E2=39.1495\u00d710265.106\u00d71026=7.67\u22488<br \/>\nHence, we found the energy released during the fusion of 1kg of hydrogen is nearly 8 times the energy released during the fusion of 1kg of uranium.<\/h3>\n<h2>31. Suppose India Had a Target of Producing, by 2020 AD, 200,000 MW of Electric Power, Ten Percent of Which Was to be Obtained from Nuclear Power Plants. Suppose We are Given That, on an Average, the Efficiency of Utilization (i.e., Conversion to Electric Energy) of Thermal Energy Produced in a Reactor Was 25%. How Much Amount of Fissionable Uranium Would Our Country Need Per Year by 2020? Take the Heat Energy Per Fission of 235U235U to be about<br \/>\n200MeV200MeV<br \/>\n.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nAmount of electric power to be generated,<br \/>\nP=2\u00d7105MWP=2\u00d7105MW<br \/>\n, 10% of this amount has to be obtained from nuclear power plants.<br \/>\nAmount of nuclear power, P1=10100\u00d72\u00d7105=2\u00d7104MWP1=10100\u00d72\u00d7105=2\u00d7104MW<br \/>\n\u21d2P1=2\u00d7104\u00d7106J\/s=2\u00d71010\u00d73600\u00d724\u00d7365J\/y\u21d2P1=2\u00d7104\u00d7106J\/s=2\u00d71010\u00d73600\u00d724\u00d7365J\/y<br \/>\nHeat energy released per fission of a 235U235U nucleus,<br \/>\nE=200MeVE=200MeV<br \/>\nEfficiency of a reactor<br \/>\n=25=25<br \/>\nHence, the amount of energy converted into the electrical energy per fission is calculated as:<br \/>\n25100\u00d7200=50MeV=50\u00d71.6\u00d710\u221219\u00d7106=8\u00d710\u221212J25100\u00d7200=50MeV=50\u00d71.6\u00d710\u221219\u00d7106=8\u00d710\u221212J<br \/>\nThe number of atoms required for fission per year would be:<br \/>\n2\u00d71010\u00d760\u00d760\u00d724\u00d73658\u00d710\u221212=78840\u00d71024atoms2\u00d71010\u00d760\u00d760\u00d724\u00d73658\u00d710\u221212=78840\u00d71024atoms<br \/>\n1 mole, i.e., 235g of U235U235contains<br \/>\n6.023\u00d71023atoms6.023\u00d71023atoms<br \/>\nThat is, the mass of<br \/>\n6.023\u00d71023atoms6.023\u00d71023atoms<br \/>\nof U235=235g=235\u00d710\u22123kgU235=235g=235\u00d710\u22123kg<br \/>\nAlso, the mass of:78840\u00d71024atomsofU235=235\u00d710\u221236.023\u00d71023\u00d778840\u00d71024=3.076\u00d7104kg78840\u00d71024atomsofU235=235\u00d710\u221236.023\u00d71023\u00d778840\u00d71024=3.076\u00d7104kg<br \/>\nHence, the mass of uranium needed per year is found to be,<br \/>\n3.076\u00d7104kg3.076\u00d7104kg.<\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Class 12 Physics NCERT book solutions for Chapter 13 &#8211; Nuclei Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":118347,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-121371","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121371","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=121371"}],"version-history":[{"count":3,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121371\/revisions"}],"predecessor-version":[{"id":121498,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121371\/revisions\/121498"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/118347"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=121371"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=121371"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=121371"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}