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NCERT Solutions Class 12 Physics Chapter 13

NCERT Solutions Class 12 Physics Chapter 13- Nuclei

The NCERT Solutions Class 12 Physics Chapter 13 provide information about the nucleus and are available on the Extramarks platform. It is a vital study material to make students well equipped for the important CBSE Class 12 Board examinations. The answers given are in a detailed, stepwise, and lucid manner so that students can revise all concepts thoroughly. Every minute detail provided is based on the NCERT books and the latest guidelines issued by the Central Board of Secondary Education (CBSE). 

The NCERT Solutions Class 12 Physics Chapter 13 Nuclei is prepared by subject elites of Extramarks with an aim to get a firm grip on theoretical knowledge and develop problem-solving abilities. 

The Nuclei chapter is very important and carries high weightage. In the chapter, students will learn various properties and fundamentals related to nuclei. Topics like Atomic Masses, Composition Of Nucleus, Mass-energy, Nuclear Binding Energy, and Radioactivity are covered in the NCERT Solutions Class 12 Physics Chapter 13. Experiments based on the scattering of α-particles are also demonstrated using diagrams. Besides derivations and theory, the calculation of numerical problems from various topics is explained in a step-by-step format. 

Extramarks, an online learning platform, assists students in exam preparation by providing extremely reliable, accurate, and simplified notes. Students may refer to the NCERT Solutions Class 11 to brush up on their basic knowledge. In addition to class 11, Extramarks also offers NCERT Solutions for various other classes, from Class 1 to Class 12.

Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 13

The NCERT Solutions Class 12 Physics Chapter 13 helps students in quick revision. It provides many objective and subjective questions from the chapter. The notes contain not only a list of questions that are most likely to be asked in the exam but also previous year-solved questions. These Solutions are prepared by the experienced and expert faculty of Extramarks. 

The NCERT Solutions Class 12 Physics Chapter 13 aims to provide perfect knowledge of all complex topics and an in-depth understanding of all minute details of the chapter. The notes cover the following key topics: 

13.1

13.2

13.3

13.4

13.5

13.6

13.7

Introduction

Atomic Masses And Composition Of Nucleus

Size Of The Nucleus

Mass-energy And Nuclear Binding Energy

Nuclear Force

Radioactivity

Nuclear Energy

13.1 Introduction

Ch 13 physics class 12 is based on the constituents of the nucleus, such as size, mass, density, and stability. Associative nuclear phenomena such as radioactivity, nuclear fission, and nuclear binding energy are also well explained. Many numerical questions are asked from all concepts included in this chapter. 

Students will learn the following topics in the NCERT Solutions Class 12 Physics Chapter 13 Exercise 13.1: 

  • Nuclear binding energy
  • Nuclear fusion
  • Nuclear fission
  • Half-life and Mean life
  • Radioactive decay
  • Soddy and Fajan’s displacement law

13.2 Atomic Masses And Composition Of Nucleus

This section in the NCERT solutions class 12 physics chapter 13 nuclei gives brief information about the atomic mass unit given as 1amu = 931 MeV. Under the composition of the nucleus topic, students learn about the basic fundamentals of the atom. It includes atomic number, atomic mass number, representation of nuclear species, isotopes, isobars, and isotones. A small introduction about the discovery of an atom is also mentioned in this section. 

A firm grip on these concepts is very important to understand complex topics included in the NCERT Solutions Class 12 Physics Chapter 13.

13.3 Size Of The Nucleus

Under this section in the Class 12 physics chapter 13 NCERT solutions, students will learn about the size of the nucleus. These derivations are often asked in exams. This topic is explained in detail in the NCERT Solutions Class 12 Physics Chapter 13. 

The formula to find the size of the nucleus is given as R = RoA1/3, where Ro= 1.2 X 10-15m and A is the atomic mass number. 

13.4 Mass-energy And Nuclear Binding Energy

This section of NCERT solutions class 12 Physics Chapter 13 includes information on Mass-energy given by scientist Einstein. It is based on the study of many discovered particles such as nucleons, nuclei, etc. The nuclear binding energy gives the relation between the mass of the nucleus and the number of protons and neutrons. 

The Mass energy formula is E = mc2

Mass defect= M = {Zmp+ (A-Z) mn]-M

The step-by-step and detailed explanations of the formula in the NCERT solutions class 12 Physics Chapter 13 will help students to perceive the concept better and enhance their learning process significantly.

13.5 Nuclear Force

Students will learn about the forces acting inside the nucleus. Also, some properties by Yukawa are mentioned in this section of NCERT Solutions Class 12 Physics Chapter 13. 

13.6 Radioactivity

The concept of radioactivity and some properties associated with it is explained in detail using the diagram. This section is further divided into four parts in the NCERT Solutions Class 12 Physics Chapter 13. It gives information about the Law of radioactive decay and its types, such as Alpha decay, Beta-decay, and Gamma decay. Students can easily understand these topics with the help of examples. 

Some features of these Becquerel rays are also mentioned in this section. Students are advised to be well-versed with terms like neutrino, antineutrino, and electron capture. The Law of radioactive decay gives the direct relation between radioactive elements and the number of nuclei. The graphical explanation is also given for better understanding. 

Mathematically, it is given as dNdtN. This section includes the SI unit of the radioactive sample, which is given as becquerel. Students also learn about Soddy and Fajan’s displacement law for , , – rays. 

13.7 Nuclear Energy

With the help of this section, students will know what nuclear fission, nuclear fusion and the difference between the two are. The examples for the same are given. Students will be asked to calculate the half-life and mean life of a  radionuclide. Many numerical questions based on these are asked in the exam. 

Students may refer to the NCERT solutions class 12 physics chapter 13 by clicking on the respective chapter.

List of NCERT Solutions Class 12 Physics Chapter 13 Exercise & Answer Solutions

‘Nuclei’ is a crucial chapter. It is important for students to study the entire chapter thoroughly to attain high scores. Students can use the NCERT Solutions Class 10, Class 9, and Class 8 to recall the basic topics studied in previous classes. The study materials are available on the online learning platform Extramarks for free. Using these NCERT Solutions Class 12 Physics Chapter 13, students can solve innumerable questions based on the latest guidelines issued by CBSE (Central board of education). The solution to all questions, based on the following topic, is explained in detail. Students may click on the link below to get access to the NCERT Solutions Class 12 Physics Chapter 13.

Students can also practise Extramarks sample papers to witness notable improvement in their scores. In addition, they may access the NCERT Solutions for other classes by clicking on the respective link below.

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NCERT Exemplar Class 12 Physics

Physics is not an easy subject. It requires practice, consistency, and in-depth knowledge of all concepts. The solutions provided to the NCERT questions help students solve the (HOTS) questions and develop their problem-solving abilities. Apt and minute details of every chapter are available in the Extramarks NCERT Solutions Class 12 Physics Chapter 13. MCQs, Numerical Problems, and CBSE extra questions based on all important concepts are included in these notes. Students can learn various tips and tricks to tackle complex questions in an easier way. 

Key Features of NCERT Solutions Class 12 Physics Chapter 13

The key features of NCERT Solutions Class 12 Physics Chapter 13 provided by Extramarks include

  • The NCERT Solutions can be used for quick revision as it includes all important formulas, equations, and detailed answers to all questions. 
  • The solutions are compiled by some of the best subject matter experts at Extramarks.
  • These notes can be accessed from anywhere on any device, namely, mobile, laptop, tablet, or PC.
  • With the help of detailed solutions, students can clarify their doubts and attain an in-depth understanding of every concept.
  • The class 12 physics chapter 13 NCERT solutions help to improve their problem-solving capabilities, logic, and analytical skills.

Q.1 (a) Two stable isotopes of lithiumLi and36Li have respective37abundance of 7.5 % and 92.5 %. These isotopes have masses6.01512 u and 7.01600 u respectively. Find the atomic weightof lithium.(b) Boron has two stable isotopes B and510B. Their respective511masses are 10.01294 u and 11.00931 u and the atomic mass ofboron is 10.811 u. Find the abundances ofB and510B.511

Ans.

(a) Atomic weight of lithium can be calculated as

Isotope Abundance Mass Isotopic mass
Lithium-6 7.5% 6 6.01512 u
Lithium-7 92.5% 7 7.01600 u

Theatomicweightistheweightedaverageoftheisotopes.atomicweight=6.01512×7.5+7.01600×92.57.5+92.5atomicweight=45.1134+648.98100atomicweight=6.941 u

(b) Let x% and (100-x)% be the abundances of 5B10 and 5B11 respectively

Isotope Abundance Mass Isotopic mass
Boron-10 (x)% 10 10.01294 u
Boron-11 (100-x)% 11 11.00931 u

The atomic weight of Boron can be written as10.811=[x (10.01294)+(100 - x)(11.00931)]10010.811=[10.01294 x+1100.931-11.00931 x]10010.811=[1100.931-0.99637 x]1001081.1=1100.931 - 0.99637 xx=19.91%and (100 - x)=10019.91=80.09%Thus, 19.9 % and 80.09 % are the abudances of B510 and B511.

Q.2 The three stable isotopes of neon:Ne,1020Ne and1021Ne have1022respective abundance of 90.51 %, 0.27 % and 9.22 %. Theatomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u,respectively. Obtain the average atomic mass of neon.

Ans.

Isotope Abundance Mass Isotopic mass
Neon-20 90.51% 20 19.99 u
Neon-21 0.27% 21 20.99 u
Neon-22 9.22% 22 21.99 u

As average atomic mass of neon can be obtained asat. mass=(90.51×19.99)+(0.27×20.99)+(9.22×21.99)100at. mass=1809.29+5.67+202.75100=20.18 u .

Q.3

Obtain the binding energy(in MeV)of a nitrogen nucleus (N147).given m (N147)=14.00307 u.

Ans.

Mass defect of nucleus of Nitrogen will beΔm=[ZmH+(AZ)mn]mNwhereA=14,Z=7 thusΔm=[7(1.00783)+(147)1.00867]14.00307Δm=7.05481+7.0606914.00307=0.11243 uΔm=0.11243 uThus B.E of the nucleus will beB.E=0.11243×931=104.7MeV.

Q.4

Obtain the binding energy of the nuclei Fe and 5626Bi in units of20983MeV from the following data: m(Fe5626)= 55.934939 u            m(Bi20983)= 208.980388 u.

Ans.

For the nucleus ofFe, mass defect can be expressed as5626Δm=[ZmH+(A-Z)mn]–m(Fe5626)Δm=[26(1.007825)+(56-26)  1.008665]–55.934939Δm=0.528461 u .Thus, binding energyB.E=Δm×931 MeV=492 MeVBinding energy per nucleon of Fe nucleus will be=B.EA=49256=8.79 MeV.

Similarly, for the nucleus ofB20983i,mass defect can be expressed asΔm=[ZmH+(A-Z)mn]-m(B20983i)=[83(1.007825)+(209–83)1.008665]–208.980388=1.760877 u.Thus binding energyB.E=Δm×931MeV=1.760877×931=1639.37MeVHence, binding energy per nucleonof Bi nucleus will be=B.EA=1639.37209=7.84MeV.F5626ehas more Binding energy per nucleon compared toB20983i.

Q.5

A given coin has a mass of 3.0 g. Calculate the nuclear energythat would be required to separate all the neutrons and protonsfrom each other. For simplicity assume that the coin is entirelymade ofC6329uatoms(of mass 62.92960 u).

Ans.

As mass defect is expressed asΔm=[Zmp+(AZ)mn]MHere,A=63,Z=29,mp=1.00783 u,mn=1.00867uandM=62.92960uHence, mass defectΔm=[29(1.00783)+34(1.00867)]62.92960Δm=0.59225 u.Thus, binding energy of Cu nucleusisB.E=Δm×931 MeV=0.59225×931B.E=551.38 MeV.Asnumber of atoms in 63g of copper=6.023×1023Number of atoms in 1g of copper=6.023×102363Number of atoms in 3g of copper=6.023×1023×363=2.868×1022If binding energy of one atom=551.38 MeVBinding energy of 2.868×1022atoms=551.38×2.868×1022MeV=1.58×1025MeVThus,This much energy is required to seprate all the neutronsand protons from each other.

Q.6

Write nuclear reaction equations for:(i)  α decay of  Ra22688(ii) α decay of  Pu24294(iii)  βdecay of  P3215(iv)  βdecay ofBi21083(v)  β+decay ofC116(vi) β+decay ofTc9743(vii)  Electron capture ofXe12054

Ans.

In alpha decay, mass number decreases by 4 and charge number decreases by 2.In beta minus decay, mass number remains unaffected and charge number increases byone and there is emission of anti -neutrino from the nucleus.(i)Ra22688Ra22286+He42(ii)Pu24294U23892+He42(iii)P3215S3216+e+ˉv(antineutrino)(iv)Bi21083Po21084+e+ˉv(antineutrino)(v)C116B115+e+(positron)+v(neutrino)(vi)Tc9743Tc9743+e+(positron)+v(neutrino)(vii)Xe12054+e+I12053+v(neutrino)

Q.7 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to
(a) 3.125 % of its original activity (b) 1% of original value ?

Ans.

(a) The rate of disintegration or count rate of a sample of a radioactive material is called activity (A) of the sample, which is directly proportional to the number of atoms left undecayed in the sample.

Thus,AA0=NN0=(12)tT.………………….(1)As, A0=N0λ=initial activity or decay rate at t=0andA=Nλ= activity or decay rate at time t.Here, N0 is original amount of the radioactive isotope andN is the amount of the radioactive isotope after decay.According to question,NN0=3.125%=3.125100=132NN0=132,from eqn(1), we getAA0=132=(12)tTAA0=(12)5=(12)tTtT=n;tT=5t=5THence, the isotope will take about 5 years​ to reduce to 3.125% of its orginal value.(b) According to question, only 1% of N0 remains after decay.Hence,​ NN0=1%=1100NN0=1100,But  NN0=eλt=1100e+λt=100λt=loge100λt=2.3026log100λt=4.6052t=4.6052λSince,​ λ=0.693Tt=4.6052×T0.693t=6.65THence, the isotope will take about 6.65 years​ to reduce to 1% of its orginal value.

Q.8

This activity arises from the small proportion of radioactiveC146present with the stable carbon isotopeC. When the organism is126dead, its interaction with the atmosphere(which maintains the above equilibrium activity)ceases and itsactivity begins to drop. From the known half-life(5730 years)of C, and the measured activity, the age of the specimen can be146approximately estimated. This is the principle of 6C14dating used inarchaeology. Suppose a specimen from Mohenjodaro gives an activityof 9 decays per minute per gram of carbon. Estimate the approximateage of the Indus-Valley civilization.

Ans.

Here normal activity, R0=15 decays/min and activity of a specimen from Mohenjodaro,R=9 decays/min.Asactivityisproportionaltothe​ numberofradioactiveatoms,NN0=RR0=915.…………….(1)From N=N0eλt,NN0=eλtFrom eqn(1)NN0=eλt=915eλt=915eλt=159λt=2.3026log(1.666)λt=0.5109t=0.5109λSince,​ λ=0.693TT=5730year(Given)t=0.5109×T0.693t=0.5109×57300.693=4224.32yearsThus, the approximate age of the Indus ValleyCivilization is 4225 years .

Q.9

Obtain the amount ofCo necessary to provide a radioactive6027source of 8.0 mCi strength. The half-life ofCo is 5.3 years.6027

Ans.

As 1 Curie=3.7 x 1010disintegrations per secondHence8.0mCi=8×103×3.7×1010= 2.96×108disintegrations s-1Also half life,T1/2=5.3 years=5.3×365×24×60×60s=1.67×108sAs per radioactive decay law,dNdt=λNor N=1λ(dNdt)N=(T120.693)(dNdt),  as[T12=0.693λ]Thus, N=(1.67×1080.693)(2.96×108)=7.13×1016As 6.023×1023 atoms of cobalt have mass=60gThus 7.13×1016atoms of cobalt have mass=60×7.13×10166.023×1023=7.1×106 g

Q.10

The half life ofSr is 28 years. What is the disintegration rate9038of 15 mg of this isotope?

Ans.

Given:T=28 years=28×365×24×3600=8.830×108AsNumberof atoms present in 90 g ofSr9038=6.023×1023Numberof atoms present in 1g ofSr9038  =6.023×102390numberof atoms present in 15mg ofSr9038=6.023×1023×1590×1000                                                            =1.0038×1020As disintegration constant λ, is expressed asλ=0.693T12=[0.693(8.830×108)]s1=0.0784×108s1Rate of disintegration(dNdt)=λN(dNdt)=0.0784×108×1.0038×1020(dNdt)=7.87 x 1010Bq.Or 2.13 Ci .

Q.11

Obtain approximately the ratio of the nuclear radii ofthe gold isotopeA19779uand silver isotopeA10747g.Obtainapproximately the ratio of the nuclear radii of the goldisotopeA19779uand silver isotopeA10747g.

Ans.

As nuclear radiusRA13where A is atomic mass of the nuclei.RAuRAg=(AAuAAg)13=(197107)13=(1.84)13=1.23

Q.12

Find the Q-value and the kinetic energy of the emitted particlein the decay of(a)Ra22688  and  (b)Rn22086Given,m(Ra22688)= 226.02540 u;       m(Rn22286)= 222.01750 um(Rn22086)= 220.01137 u;       m(Po21684)= 216.00189 u.

Ans.

(a)Ra22688Rn+22286He42Thus, the value of binding energy or Q-Value=m(Ra22688)[m(Rn22286)+m(He42)]=226.02540–226.0201=0.0053 uThusQ=0.0053×931 MeV(as 1 u = 931 MeV)Q=4.93 MeVThe K.E of an α particle Eα during decay of Ra will beEα=(Mass number after decayMass number before decay)×QEα=(222226)×4.93=4.84 MeV .(b)Rn22086Po21682+He42QValue=m(Rn22086)[m(Po21682)+m(He42)]=220.011373[216.00189+4.00260]=220.011373220.00449=0.006883 u=0.006883×931 MeV=6.41 MeV .K.E.ofαparticle emitted during decay of Rn will beEα=(MassnumberafterdecayMassnumberbeforedecay)×QEα=(216220)×4.93=4.84 MeV

Q.13

The radionuclideC116decays according toC116®B115+ e++ v:half life=20.3 min.The maximum energy of the emitted positron is 0.960 MeV.Given the mass values.m(C116)= 11.011434 u; m(B115)= 11.009305 uCalculate Q and compare it with the maximum energy of the positron emitted.

Ans.

As given the equation of the decay process ofC611isC6115B11+e++v+QThus energyQ=[mN(C611)mN(5B11)me]×931MeVWhere the masses used are of nuclei and not of atoms.In orderto express the Q value in terms of the atomic masses,6 memasshas to be subtracted from atomic mass of carbonand 5 memasshas to be subtracted from atomic mass of boron(because their atoms contain outside electrons).Therefore,in terms of the atomic mass m,the expression for energy will beQ=[m(C611)–6me–m(5B11)+5me–me]×931 MeV=[m(C611)–m(5B11)-2me]×931MeV=[11.01143411.0093052×0.000548]×931=0.001033×931=0.961 MeV.This energy is comparable to actual energy released in the decayprocess.The daughter nucleus5B11is very heavy as compared toe+and neutrino so the daughter nucleus carries minimumenergy.If the kinetic energy carried by the neutrino is minimum,thenthe positron carries maximum energy which is practically same as energy Q.

Q.14

The nucleusNe decays by- emission. Write down the-decay2310equation and determine the maximum kinetic energy of theelectrons emitted. Given that:m(Ne2310)= 22.994466 um(B115)= 22.089770 u

Ans.

The decay process of Ne 2310isNe2310Na2311+e+v+QIf whole of energy is carried byβparticle.ThenQ=[mN(Ne2310)mN(Na2311)me]×931.5 MeVWhere the masses used as mN(Ne2310) and mN(Na2311) are Ne2310andNa respectively2311.If m(Ne2310)and m(Na2311) are atomic masses of Ne2310 andNa2311respectively, thenmN(Ne2310)=m(Ne2310)10me as atom of Ne contains 10electrons.Similarly mN(Na2311)=m(Na2311)11me as atom of Na contains11 electrons.Thus,Q=[m(Ne2310)10mem(Ne2310)+11meme]×931Q=[m(Ne2310)m(Na2311)]x 931 MeVQ=[22.994466  22.989770]×931 MeVQ=0.004689×931=4.37 MeV .

Q.15

The Q value of a nuclear reaction A + b = C + d is defined by Q =[mA+ mb– mC– md]c2where the masses refer to the respective nuclei.Determine from the given data the Q value of the following reactions and state whether thefollowing reactions are exothermic or endothermic.(i)  H +11H31H +21H21(ii) C +126C126Ne +2010He42Atomic masses are given to be ,m(H11)= 1.007825 u, m(H21)= 2.014102 u, m(H31)= 3.016049 u,m(C126)= 12.00000 u, m(Ne2010)= 19.992439 u, m(He42)= 4.002603 u 

Ans.

(i)The given reaction isH11+H31H21+H21Q=[mN(H11)+mN(H31)mN(H21)mN(H21)]×931MeV.(i)where mN refers to nuclear masses,If m refers to atomic masses,thenm(H11)=mN(H11)+memN(H11)=m(H11)meSimilarly,mN(H31)=m(H31)meandmN(H12)=m(H12)meThus, Q=[mN(H11)+mN(H31)mN(H21)mN(H21)]×931MeVBecomes Q=[m(H11)+m(H31)m(H21)m(H21)]×931MeVQ=[1.007825+3.0160492.0141022.014102]×931MeVQ=0.00433×931Q=4.03 MeV .Since Q value is negative,the reaction is endothermic.

(ii) The given reaction isC126+C126Ne2010+He42Q=[mN(C126+mN(C126)mN(Ne2010)mN(He42)]×931 MeV.(2)Where mN refers to the nuclear mass.If m refers to theatomic mass,thenmN(C126)=m(C126)6me,As C126has 6 electrons in its atom.SimilarlymN(Ne2010)=m(Ne2010)10meandmN(He42)=m(He42)2meHence,Q=[mN(C126)+mN(C126)mN(Ne2010)mN(He42)]×931MeVQ=[m(C126)6me+m(C126)6mem(Ne2010)+10mem(He42)+2me]×931MeVQ=[m(C126)+m(C126)m(Ne2010)m(He42)]×931MeVQ=[12.000000+12.00000019.9924394.002603]×931MeVQ=0.004958×931 MeVQ=4.66 MeVSince Q value is positive,the reaction is exothermic .

Q.16

Suppose, we think of fission of aFe nucleus into two equal5626fragments, Al. Is the fission energetically possible? Argue2813by working out Q of the process. Given:m(Fe5626)= 55.93494 u; m(Al2813)= 27.98191 u

Ans.

As suggested in the question,if Fe5626Al2813+Al2813Then Qvalue of process will be=[m(Fe5626)2m(Al2813)]×931MeV=[55.934942(27.98191)]×931MeV=0.02888×931MeV=0.02888×931 MeVHence,Q=26.90MeV .Since, Q value is negative,so the fission is not possible .

Q.17

The fission properties ofPu are very similar to those of 23994U.23592The average energy released per fission is 180 MeV. How muchenergy, in MeV, is released if all atoms in 1 kg of purePu23994undergo fission?The fission properties of94Pu239are very similarto those ofU. The average energy released per fission is 180 MeV.23592How much energy, in MeV, is released if all atoms in 1 kg ofpurePu undergo fission?23994

Ans.

Number of atoms present in 239 g ofPu= 6.023×102323994Number of atoms present in 1 g of  Pu     =239946.023×1023239Number of atoms present in 1000 g of  Pu =239946.023×1023×1000239                                                    = 2.52×1024atomsAs given,the average energy released per fission = 180 MeVThus total energy released , Q=180 x 2.52 x 1024 MeVQ=4.536×1026 MeVHence, 4.536×1026MeV is released if all the atoms in 1 kg of purePu undergo fission.23994

Q.18

A 1000 MW fission reactor consumes half of its fuel in 5.00 y.How muchU did it contain initially? Assume that the reactor23592operates 80 % of the time, that all the energy generated arisesfrom the fission ofU and that this nuclide is consumed by23592the fission process.

Ans.

Power of the fission reactorP=1000 MW=1000×106W=109WAs given the time taken to consume half of the fuel is , t=5 year=5×365×24×60×60=1.577×108sThus, energy consumed during this time of 5 years will beE=P tE=109×1.577×108=1.577×1017 JAs we know energy generated per fission ofU=200 MeV23592=200×1.6×10-13J=3.2×10-11JTherefore number of total fissions that have occurred in five years.=1.577×10173.2×10-11=4.93×1027Now  6.023×1023 atoms(fissions)are produced by 235 g ofU23592Mass of U consumed in five years23592=235×4.93×1027g6.023×1023= 1924 kg=half of the mass of the fuelconsumed during 5 years as given in the question.Hence, initial mass of U=2×192423592=3848 kg

Q.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 asH21+H21H32+ n + 3.27 MeV

Ans.

As given power of lamp, P= 100 W andmass of deuterium= 2 kgFrom the reaction given in the exercise, we find that 2 atoms ofH21combine to give 3.2 MeV of energy.Now 2g ofH21equivalent to 6.023×1023atomsThus 2kg or 2000g ofH21is equivalent to6.023×1023×20002=6023×1023=6.023 ×1026atomsThus the energy released for combination of 6.023×1026atomsshould beE=3.27×6.023×10262= 9.85×1026MeVE=9.85×1026×1.6×10-13JE=1.58×1014JAsP=Etthus t=EPand we gett=1.58×1014100= 1.58×1012s=1.58×1012365×24×60×60years

Q.20 Calculate the height of the potential barrier for the 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.

Before collision,the initial total energy E of the two deuteronsis given by,E=2(Kinetic energy)After collision,when the two deuterons stop,their energy is totallypotential energy U and it is given byU=14πЄ0e.e2R=14πЄ0e22RAs given that radius of each deuteron sphere isR=2 fm=2 x 1015mCharge,e=1.6×1019Cand14πЄ0=9×109Nm2C2ThusP.E=[9×109×(1.6×1019)2](4×1015)jouleP.E=9×1.6×1.6×10144×1.6×1019eV=3.6×105eV=360keVAsper law of conservation of energy,we can say that2K.E=P.EK.E=P.E2=3602=180keV.

Q.21 From the relation R = R0 A1/3, where R0 is a constant and A is the mass number of a nucleus, show that nuclear matter density is nearly constant (i.e independent of A)

Ans

As nuclear density=mass numbervolume=3A(4πR3)=3A4πR03A=34πR03By using relationR=R0A13Hence, nuclear density is independent of A.

Q.22

For the(positron)emission from a nucleus,there is another competingprocess known as electron capture(electron from an inner orbit,say,the Kshell,is captured by the nucleus and a neutrino is emitted).e+XAZYAZ1+vShow that ifemission is energetically allowed,electron capture is necessarily allowed but not viceversa.

Ans.

For the positron emission,we haveXAZYAZ1+e10+v+Q1……(1)And for electron capture,we haveXAZ+e10YAZ1+v+Q2……(2)If mNrepresents the nuclear mass and m represents theatomic mass,then for meas the mass of1e0or+1e0,we have nuclear mass ofXAZ=mN(XAZ)=[m(XAZ)Zme]Similarly nuclear mass ofYAZ1should beYAZ1=mN(YAZ1)=[m(YAZ1)(Z1)me]Hencefromequation(1)Q1=[mN(XAZ)-mN(YAZ1)me]c2=[m(XAZ)Zme-m(YAZ1)+(Z1)meme]c2=[m(XAZ)m(YAZ1)2me]c2……..(3)Similarly from equaton(2)Q2=[mN(XAZ)+memN(YAZ1)]c2=[m(XAZ)– Zme+mem(YAZ1)+(Z1)me]c2=[m(XAZ)m(YAZ1)]c2(4)Thus from eqns.(3)and(4),we find thatif Q1>0,is necessarily greater than zero.However,if Q2>0,it does not always mean that Q1>0.

Q.23

In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative naturalabundance on Earth. The three isotopes and their masses are  Mg2412(23.98504 u),Mg2512(24.98584 u)andMg2612(25.98259 u). The natural abundance ofMg  is 78.99 % by mass. Calculate the abundance of the other two isotopes.2412

Ans.

Let us assume that the natural abundance ofMg be x% then natural abundance of2512Mg  will be2612(100 – x – 78.99)%.Thus,24.312 =[78.99×23.98504 + 24.98584x +(100 – x – 78.99)(25.98259)]10024.312 =2440.474 – 0.99675 x1000.99675 x = 2440.479 – 2431.2x =9.2740.99675= 9.304 %Thus natural abundance of Mg = 9.304 %  and for 2512Mg, (21.01 – x) = 11.71 % .2612

Q.24

The neutron separation energy is defined as the energy requiredto remove a neutron from the nucleus. Obtain the neutron separationenergies of the nuclei Ca and4120Al from the following data:2713m(Ca4020)=39.962591 u,m(Ca4120)=40.962278 um(Al2613)=25.986895 u,m(Al2713)=26.981541 u.

Ans.

The process of neutron separation forCa,4120is given asCa+E=4120Ca+4020n10E=[m(Ca4020)+mn–m(Ca4120)]×931 MeVE=[39.962591+1.008665–40.962278]×931 MeVE=0.008978×931=8.36 MeVThe process of neutron separation for Al is2713Al+E=2713Al+2613n10Thus, neutron separation energy will beE=[m(Al2613)+mn–m(Al2713)]×931 MeVE=[25.986895+1.008665-26.981541]×931 MeVThus, E =0.014019×931=13.05 MeV .

Q.25

A source contains two phosphorus radio nuclidesP3215(T12= 14.3d)andP3315(T12=25.3d). Initially,10 %of the decays come fromP3315.How long one must wait until 90 % do so?

Ans.

Let us assume that R01and R02be the initial activities ofP3315andP3215respectively and R1and R2be their activities at any instant t.As suggested in the question,R01=10[R01+ R02]100as R01=10%of total activity.Hence10R01=R01+ R02OrR02=9R01……….(1)At a later time t ,R1=90[R1+ R2]100asR1= 90%of total activity at time t.Hence10R1=9R1+9R2OrR2=10 R1–9R19=R19……..(2)Now dividing(2)by(1)we get,R2R02=(R19)(19R01)OrR2R02=181R1R01As per decay law,R = R0e-λtR02e-λ2tR02=181R01e-λ1tR01Þ 81=e-λ1te-λ2tOre(λ2– λ1)t =81Taking natural log of both sides,lne(λ2– λ1)t =ln 81Or(λ2– λ1)t=2.303 log 81 …….(3)As ,λ =0.693T12Thusλ1=0.693(T12)1=0.69325.3day-1and λ2=0.693(T12)2=0.69314.3day-1Thus with reference to equation(3), we get(0.69314.30.69325.3)t=2.303 log 81Or0.693×(25.3-14.3)t25.3 ×14.3= 2.303log 810.693×(11)t25.3 ×14.3= 2.303log 81Thust=[(2.303 log 81)(25.3×14.3)11×0.693]t=[2.303×1.9085×25.3×14.3][11×0.693]t =208. 6 days»209 days.

Q.26

Under certain circumstances, a nucleus can decay by emitting aparticlemore massive than an–particle. Consider the following decay process:R22388aP20982b+C146R22388aR21986n+H42eCalculate the Q values for these decays and determine that the both areenergetically allowed.The required data ism(R22388a)=223.01850u,m(P20982b)=208.98107 u,m(R21986n)=219.00948 u,m(C146)=14.00324 u,m(H42e)=4.00260u.

Ans.

The given decay process forR22388aisR22388aP20982b+C146+QHence mass defectΔm=[mN(R22388a)mN(P20982b)mN(C146)]WheremNdenotes the nuclear mass.If m represents the atomic mass,thenΔm=[m(R22388a)88mem(P20982b)+82mem(C146)+me]=[m(R22388a)m(P20982b)m(C146)]As energy Q=Δm×931 MeVQ=[223.01850208.9810714.003241]×931=31.83 MeV.For another decay forR22388a,decay equation isR22388a=R21986n+H42e+QΔm=[mN(R22388a)mN(R21986n)mN(H42e)]=[m(R22388a)m(R21986n)m(H42e)]ThusQ=[223.01850219.009484.00260]×931 MeV=5.98 MeV.

Q.27

ConsiderthefissionofU23892byfastneutrons.Inonefissionevent,noneutronsareemittedandthefinalstableendproducts,afterthebetadecayoftheprimaryfragments,areCe14058andRu9944.CalculateQforthisfissionprocess.Therelevantatomicandparticlemassesare:m(U23892)=238.05079u;m(Ce14058)=139.90543u;m(Ru9944)=98.90594u;mn=1.00867u

Ans.

The fission process may be expressed asU23892+n10Ce14058+Ru9944+QQ=[m(U23892)+mnm(Ce14058)m(Ru9944)]×931 MeV   =[238.05079+1.00867139.9054398.90594]×931   =230.97MeV .

Q.28

(a)Consider the D-T reaction(deuterium-tritium–fusion)given in eqn:H+21H–>31He+42n + Q10Calculate the energy released in MeV in this reaction from the data:m(H21)= 2.014102 u; m(H31)= 3.016049 u;m(He42)= 4.002603 u; mn= 1.00867 u(b)Consider the radius of both deuterium and tritium to be approximately2.0 fm. What is the kinetic energy needed to overcome the Coulombrepulsion between the two nuclei? To what temperature must the gasesbe heated to initiate the reaction?

Ans.

(a) From the equation given in the questionQ=[mN(H21)+mN(H31)–mN(He42)–mn]×931 MeVwhere mN refer to nuclear mass of the element given inthe brackets and mn=mass of neutron.If m represents the atomic mass, thenmN(H21)=m(H21)–memN(H31)=m(H31)–memN(He42)=m(He42)–2eQ=[m(H21)–me+m(H31)–me–m(He42)+2me–mn]×931 MeV=[2.014102+3.016049 – 4.002603–1.00867]×931 MeV=17.575 MeV=17.58 MeV .(b) Repulsive potential energy between two nuclei is=14 π ε0q1 q22r=14 π ε0e22 rAs q1= q2= e and  distance between particle= 2 r .Thus,P.E= [9×109(1.6×10-19)2][2×2×10-15]J=5.76×10-14 JAs kinetic energy required for one fusion event is equal to averagethermal kinetic energy available with the interacting particles, asper the expression K.E = 2(3 kT2)where k is Boltzmann constant.As repulsive P.E = K.EHence,T=K.E3 k=(5.76×10-14)(3×1.38×10-23)T= 1.39×109 K

Q.29

Obtain the maximum kinetic energy ofparticle and the radiationfrequencies of decays in the following decay scheme shown in the fig 13.6.You are given thatm(A19879u)= 197.968233 u,m(H19880g)= 197.966760 u.

Ans.

The total energy released for the transformation ofAu to19879Hg19880can be found by considering the energies of gamma rays.For γ1, the frequencyν1=E1h=1.088×1.6×10-136.626×10-34ν1=2.63×1020 HzFor γ2, the frequency isν2=E2h=0.412×1.6×10-136.626×10-34ν2=9.95×1019 Hz .For γ3, the energyE3=1.088–0.412E3=0.676 Mev=0.676×1.6×10-13 JE3=1.082×10-13 Jν3=E3h=1.082×10-136.626×10-34ν3=1.63×1020 HzNow for β1 decay,Maximum kinetic energy=[m(Au19879)–m(Hg19880)]×931–1.088 MeV=[197.968233–197.966760]×931–1.088 MeV=1.372–1.088=0.283 MeV .Now for β2 decay,Maximum kinetic energy=[m(Au19879)–m(Hg19880)]×931–0.412=(197.968233–1.97.966760)×931–0.412 MeV=1.372–0.412=0.96 MeV .

Q.30

Calculate and compare the energy released by(a)fusion of1.0 kg of hydrogen deep within the sun and(b)the fissionof 1.0 kg ofU in a fission reactor.235

Ans.

(a) In sun fusion takes place according to the equation4H—-11He+ 242e+26 MeV0+1So, 4 hydrogen atoms combine to produce 26 MeV of energy.Now 1 g of hydrogen contains =6.02×1023 atomsHence, 1000 g of hydrogen contains=6.02×1023×1000 atoms                                    ​  =6.02×1026 atomsHence energy releasedE=26 MeV×6.02×10264=39×1026 MeV .(b) Fission of oneU nucleus gives energy=200 MeV23592Now, 235 g of  U has no. of atoms = 6.02×102323592 atoms1 kg(i.e 1000 g)of  U has=235926.02×1023×1000 atoms235                                      = 2.56 x 1024atomsTotal energy releasedE’=200×2.56×1024=5.1×1026 MeVFor comparison of two energies, we getEnergy released by 1 kg of fusion ofH11Energy released by 1 kg of fission of U23592hence, EE’=3.913 x 10275.12 x 1026=7.64

Q.31 Suppose India had a target of producing by 2020 AD, 2 × 105 MW of electric power, ten percent of which was to be obtained from nuclear power plant. 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 did our country need per year by 2020? Take the heat energy per fission of U235 to be about 200 MeV.

Ans.

Total power target=200000 MW=200000×106=2×1011 WNuclear power target is 10%of total power target=10×2×1011 W100=2×1010 WAs efficiency is defined as,η=Total useful powerTotal power generatedHence Total power generated=Total useful powerη=201025=8×1010 WTotal energy required for the year 2020,isE=P×t=8×1010×366×25×60×60as 2020isa leap year .E=2.265×1018 JNow 1 fission ofU23592produces 200MeV of energy=200×1.6×1013 J=3.2×1011JHence, no.of fission required for the generation of energy E=2.265×10183.2×1011=7.90×1028As 6.023 x 1023atomsofU23592havemass=235gThus mass required to produce 7.90x 1028atoms=235×3.95×1028g6.023×1023=3.084×104kg .Hence,massofuraniumneededperyear3.08×104kg .

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