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: 









Atomic Masses And Composition Of Nucleus

Size Of The Nucleus

Mass-energy And Nuclear Binding Energy

Nuclear Force


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.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT Solutions Class 10
  • NCERT Solutions Class 11
  • NCERT Solutions Class 12

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


(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+7.01600×92.5 7.5+92.5 atomicweight= 45.1134+648.98 100 atomicweight =6.941 u MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabsfacaqGObGaaeyzaiaaysW7caqGHbGaaeiDaiaab+gacaqGTbGaaeyAaiaabogacaaMe8Uaae4DaiaabwgacaqGPbGaae4zaiaabIgacaqG0bGaaGjbVlaabMgacaqGZbGaaGjbVlaabshacaqGObGaaeyzaiaaysW7caqG3bGaaeyzaiaabMgacaqGNbGaaeiAaiaabshacaqGLbGaaeizaiaaysW7caqGHbGaaeODaiaabwgacaqGYbGaaeyyaiaabEgacaqGLbGaaGjbVlaab+gacaqGMbGaaGjbVlaabshacaqGObGaaeyzaiaaysW7caqGPbGaae4Caiaab+gacaqG0bGaae4BaiaabchacaqGLbGaae4Caiaab6caaeaacaqGHbGaaeiDaiaab+gacaqGTbGaaeyAaiaabogacaaMe8Uaae4DaiaabwgacaqGPbGaae4zaiaabIgacaqG0bGaeyypa0ZaaSaaaeaacaqG2aGaaiOlaiaaicdacaqGXaGaaeynaiaabgdacaqGYaGaey41aqRaae4naiaac6cacaqG1aGaey4kaSIaae4naiaac6cacaaIWaGaaeymaiaabAdacaaIWaGaaGimaiabgEna0kaabMdacaqGYaGaaiOlaiaabwdaaeaacaGGGcGaae4naiaac6cacaqG1aGaey4kaSIaaeyoaiaabkdacaGGUaGaaeynaiaacckacaGGGcGaaiiOaiaacckacaGGGcaaaaqaaiaabggacaqG0bGaae4Baiaab2gacaqGPbGaae4yaiaaysW7caqG3bGaaeyzaiaabMgacaqGNbGaaeiAaiaabshacqGH9aqpcaGGGcWaaSaaaeaacaqG0aGaaeynaiaac6cacaqGXaGaaeymaiaabodacaqG0aGaey4kaSIaaeOnaiaabsdacaqG4aGaaiOlaiaabMdacaqG4aaabaGaaiiOaiaabgdacaaIWaGaaGimaiaacckacaGGGcaaaiaacckacaGGGcGaaiiOaaqaaiaabggacaqG0bGaae4Baiaab2gacaqGPbGaae4yaiaaysW7caqG3bGaaeyzaiaabMgacaqGNbGaaeiAaiaabshacaqGGcGaaiiOaiabg2da9iaabAdacaGGUaGaaeyoaiaabsdacaqGXaGaaeiiaiaabwhaaaaa@DA27@

(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 x x=19.91%and (100 - x)=10019.91=80.09 %Thus, 19.9 % and 80.09 % are the abudances of B105 and B115.

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.


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)100 at. mass =1809.29+5.67+202.75 100 =20.18 u .


Obtain the binding energy in MeV of a nitrogen nucleus ( N 7 14 ). given m ( N 7 14 )=14.00307 u.


Mass defect of nucleus of Nitrogen will be Δm=[ Zm H +( A Z ) m n ] m N where A=14, Z=7 thus Δm=[ 7( 1.00783 )+( 147 )1.00867 ]14.00307 Δm=7.05481+7.06069 14.00307 =0.11243 u Δm= 0.11243 u Thus B.E of the nucleus will be B.E =0.11243×931 =104.7MeV. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@084A@


Obtain the binding energy of the nuclei Fe and 2656Bi in units of83209MeV from the following data: m Fe2656 = 55.934939 u            m Bi83209 = 208.980388 u.


For the nucleus of Fe, mass defect can be expressed as2656Δm=ZmH+A-Zmn–mFe2656 Δm=261.007825+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.E A =49256 =8.79 MeV.

Similarly, for the nucleus of B 83 209 i, mass defect can be expressed as Δm= Zm H + A-Z m n -m B 83 209 i = 83 1.007825 + 209–83 1.008665 –208.980388 =1.760877 u. Thus binding energy B.E=Δm×931MeV =1.760877×931 =1639.37MeV Hence, binding energy per nucleon of Bi nucleus will be= B.E A = 1639.37 209 =7.84MeV. F 26 56 ehas more Binding energy per nucleon compared to B 83 209 i.


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 C 29 63 uatoms of mass 62.92960 u .


As mass defect is expressed as Δm=[ Zm p +( AZ ) m n ]M Here, A=63, Z=29, m p =1.00783 u, m n =1.00867uand M=62.92960 u Hence, mass defect Δm=[ 29( 1.00783 )+34( 1.00867 ) ]62.92960 Δm=0.59225 u. Thus, binding energy of Cu nucleusis B.E=Δm×931 MeV =0.59225×931 B.E =551.38 MeV. Asnumber of atoms in 63g of copper= 6.023× 10 23 Number of atoms in 1g of copper= 6.023× 10 23 63 Number of atoms in 3g of copper= 6.023× 10 23 ×3 63 =2.868× 10 22 If binding energy of one atom=551.38 MeV Binding energy of 2.868×1 0 22 atoms =551.38×2.868×1 0 22 MeV =1.58×1 0 25 MeV Thus,This much energy is required to seprate all the neutrons and protons from each other. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabgeacaqGZbGaaeiiaiaab2gacaqGHbGaae4CaiaabohacaqGGaGaaeizaiaabwgacaqGMbGaaeyzaiaabogacaqG0bGaaeiiaiaabMgacaqGZbGaaeiiaiaabwgacaqG4bGaaeiCaiaabkhacaqGLbGaae4CaiaabohacaqGLbGaaeizaiaabccacaqGHbGaae4Caaqaaiabfs5aejaab2gacqGH9aqpdaWadaqaaiaabQfacaqGTbWaaSbaaSqaaiaabchaaeqaaOGaey4kaSYaaeWaaeaacaqGbbGaeyOeI0IaaeOwaaGaayjkaiaawMcaaiaab2gadaWgaaWcbaGaaeOBaaqabaaakiaawUfacaGLDbaacaGGtaIaaeytaaqaaiaabIeacaqGLbGaaeOCaiaabwgacaqGSaaabaGaaeyqaiabg2da9iaabAdacaqGZaGaaiilaiaabccacaqGAbGaeyypa0JaaeOmaiaabMdacaGGSaGaaGjbVlaab2gadaWgaaWcbaGaaeiCaaqabaGccqGH9aqpcaqGXaGaaiOlaiaaicdacaaIWaGaae4naiaabIdacaqGZaGaaeiiaiaabwhacaGGSaGaaeiiaiaab2gadaWgaaWcbaGaaeOBaaqabaGccqGH9aqpcaqGXaGaaiOlaiaaicdacaaIWaGaaeioaiaabAdacaqG3aGaamyDaiaaysW7caWGHbGaamOBaiaadsgaaeaacaqGnbGaeyypa0JaaeOnaiaabkdacaGGUaGaaeyoaiaabkdacaqG5aGaaeOnaiaaicdacaqGGaGaaeyDaaqaaiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaeilaiaabccacaqGTbGaaeyyaiaabohacaqGZbGaaeiiaiaabsgacaqGLbGaaeOzaiaabwgacaqGJbGaaeiDaaqaaiabfs5aejaab2gacqGH9aqpdaWadaqaaiaabkdacaqG5aWaaeWaaeaacaqGXaGaaiOlaiaaicdacaaIWaGaae4naiaabIdacaqGZaaacaGLOaGaayzkaaGaey4kaSIaae4maiaabsdadaqadaqaaiaabgdacaGGUaGaaGimaiaaicdacaqG4aGaaeOnaiaabEdaaiaawIcacaGLPaaaaiaawUfacaGLDbaacaGGtaIaaeOnaiaabkdacaGGUaGaaeyoaiaabkdacaqG5aGaaeOnaiaaicdacaqGGaaabaGaeuiLdqKaaeyBaiabg2da9iaaicdacaGGUaGaaeynaiaabMdacaqGYaGaaeOmaiaabwdacaqGGaGaaeyDaiaaykW7caqGUaaabaGaaeivaiaabIgacaqG1bGaae4CaiaabYcacaqGGaGaaeOyaiaabMgacaqGUbGaaeizaiaabMgacaqGUbGaae4zaiaabccacaqGLbGaaeOBaiaabwgacaqGYbGaae4zaiaabMhacaqGGaGaae4BaiaabAgacaqGGaGaae4qaiaabwhacaqGGaGaaeOBaiaabwhacaqGJbGaaeiBaiaabwgacaqG1bGaae4CaiaaysW7caqGPbGaae4CaaqaaiaadkeacaGGUaGaamyraiabg2da9iabfs5aejaab2gacqGHxdaTcaqG5aGaae4maiaabgdacaqGGaGaaeytaiaabwgacaqGwbGaaiiOaaqaaiabg2da9iaaicdacaGGUaGaaeynaiaabMdacaqGYaGaaeOmaiaabwdacqGHxdaTcaqG5aGaae4maiaabgdacaGGGcGaaiiOaiaacckacaGGGcaabaGaamOqaiaac6cacaWGfbGaaiiOaiabg2da9iaabwdacaqG1aGaaeymaiaac6cacaqGZaGaaeioaiaabccacaqGnbGaaeyzaiaabAfacaaMc8UaaeOlaaqaaiaabgeacaqGZbGaaGjbVlaad6gacaqG1bGaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeiDaiaab+gacaqGTbGaae4CaiaabccacaqGPbGaaeOBaiaabccacaqG2aGaae4maiaabEgacaqGGaGaae4BaiaabAgacaqGGaGaae4yaiaab+gacaqGWbGaaeiCaiaabwgacaqGYbGaeyypa0JaaiiOaiaaysW7caaI2aGaaiOlaiaaicdacaaIYaGaaG4maiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaIYaGaaG4maaaaaOqaaiaad6eacaqG1bGaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeiDaiaab+gacaqGTbGaae4CaiaabccacaqGPbGaaeOBaiaabccacaqGXaGaae4zaiaabccacaqGVbGaaeOzaiaabccacaqGJbGaae4BaiaabchacaqGWbGaaeyzaiaabkhacqGH9aqpcaaMe8UaaiiOamaalaaabaGaaGOnaiaac6cacaaIWaGaaGOmaiaaiodacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaaGOmaiaaiodaaaaakeaacaaI2aGaaG4maaaaaeaacaWGobGaaeyDaiaab2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGaaeyyaiaabshacaqGVbGaaeyBaiaabohacaqGGaGaaeyAaiaab6gacaqGGaGaae4maiaabEgacaqGGaGaae4BaiaabAgacaqGGaGaae4yaiaab+gacaqGWbGaaeiCaiaabwgacaqGYbGaeyypa0JaaiiOaiaaysW7daWcaaqaaiaaiAdacaGGUaGaaGimaiaaikdacaaIZaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiaaikdacaaIZaaaaOGaey41aqRaaG4maaqaaiaaiAdacaaIZaaaaaqaaiabg2da9iaaysW7caaIYaGaaiOlaiaaiIdacaaI2aGaaGioaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaIYaGaaGOmaaaaaOqaaiaabMeacaqGMbGaaeiiaiaabkgacaqGPbGaaeOBaiaabsgacaqGPbGaaeOBaiaabEgacaqGGaGaaeyzaiaab6gacaqGLbGaaeOCaiaabEgacaqG5bGaaeiiaiaab+gacaqGMbGaaeiiaiaab+gacaqGUbGaaeyzaiaabccacaqGHbGaaeiDaiaab+gacaqGTbGaaGPaVlabg2da9iaabwdacaqG1aGaaeymaiaac6cacaqGZaGaaeioaiaabccacaqGnbGaaeyzaiaabAfaaeaacaqGcbGaaeyAaiaab6gacaqGKbGaaeyAaiaab6gacaqGNbGaaeiiaiaabwgacaqGUbGaaeyzaiaabkhacaqGNbGaaeyEaiaabccacaqGVbGaaeOzaiaabccacaqGYaGaaiOlaiaabIdacaqG2aGaaeioaiabgEna0kaabgdacaaIWaWaaWbaaSqabeaacaqGYaGaaeOmaaaakiaabggacaqG0bGaae4Baiaab2gacaqGZbGaaeiiaiaaykW7cqGH9aqpcaqG1aGaaeynaiaabgdacaGGUaGaae4maiaabIdacqGHxdaTcaqGYaGaaiOlaiaabIdacaqG2aGaaeioaiabgEna0kaabgdacaaIWaWaaWbaaSqabeaacaqGYaGaaeOmaaaakiaaykW7caWGnbGaamyzaiaadAfaaeaacqGH9aqpcaqGXaGaaiOlaiaabwdacaqG4aGaey41aqRaaeymaiaaicdadaahaaWcbeqaaiaabkdacaqG1aaaaOGaaeytaiaabwgacaqGwbaabaGaaeivaiaabIgacaqG1bGaae4CaiaacYcacaqGubGaaeiAaiaabMgacaqGZbGaaeiiaiaab2gacaqG1bGaae4yaiaabIgacaqGGaGaaeyzaiaab6gacaqGLbGaaeOCaiaabEgacaqG5bGaaeiiaiaabMgacaqGZbGaaeiiaiaabkhacaqGLbGaaeyCaiaabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabshacaqGVbGaaeiiaiaabohacaqGLbGaaeiCaiaabkhacaqGHbGaaeiDaiaabwgacaqGGaGaaeyyaiaabYgacaqGSbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGUbGaaeyzaiaabwhacaqG0bGaaeOCaiaab+gacaqGUbGaae4CaaqaaiaabggacaqGUbGaaeizaiaabccacaqGWbGaaeOCaiaab+gacaqG0bGaae4Baiaab6gacaqGZbGaaeiiaiaabAgacaqGYbGaae4Baiaab2gacaqGGaGaaeyzaiaabggacaqGJbGaaeiAaiaabccacaqGVbGaaeiDaiaabIgacaqGLbGaaeOCaiaab6caaaaa@56DD@


Write nuclear reaction equations for:(i)  α decay of  Ra88226(ii)  α decay of  Pu94242(iii)  β decay of  P1532(iv)  β decay of Bi83210(v)  β+ decay of C611(vi)  β+ decay of Tc4397(vii)  Electron capture of Xe54120


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)Ra 88226 Ra 86222+ He24 (ii)Pu94242 U 92238+ He24 (iii)P1532 S1632 + e + v¯ (antineutrino)(iv)Bi 83210 Po84210 + e+ v¯ (antineutrino) (v)C611 B 511+ e+ (positron) + v (neutrino)(vi)Tc 4397 Tc 4397+ e+(positron ) + v (neutrino)(vii)Xe54120 + e+ I 53120+ 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 ?


(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.


This activity arises from the small proportion of radioactive C614present with the stable carbon isotope C. When the organism is612dead, its interaction with the atmospherewhich maintains the above equilibrium activity ceases and its activity begins to drop. From the known half-life 5730 yearsof C, and the measured activity, the age of the specimen can be614approximately 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.


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 .


Obtain the amount of Co necessary to provide a radioactive2760source of 8.0 mCi strength. The half-life of Co is 5.3 years.2760


As 1 Curie = 3.7 x 1010disintegrations per secondHence8.0 mCi =8×103×3.7×1010= 2.96×108 disintegrations s-1Also half life,T1/2=5.3 years=5.3 ×365×24×60×60 s=1.67×108sAs per radioactive decay law, dNdt=λN or 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 = 60 gThus 7.13×1016 atoms of cobalt have mass=60×7.13×1016 6.023×1023 = 7.1×106 g


The half life of Sr is 28 years. What is the disintegration rate3890of 15 mg of this isotope?


Given:T=28 years = 28×365×24×3600= 8.830×108AsNumberof atoms present in 90 g of Sr3890=6.023×1023Numberof atoms present in 1g of Sr3890  =6.023×102390numberof atoms present in 15mg of Sr3890=6.023×1023×15 90×1000                                                             =1.0038×1020As disintegration constant λ, is expressed asλ=0.693T12= 0.693 8.830×108 s1=0.0784×108s1Rate of disintegrationdNdt=λN dNdt= 0.0784×108×1.0038×1020dNdt =7.87 x 1010Bq. Or 2.13 Ci .


Obtain approximately the ratio of the nuclear radii of the gold isotope A 79 197 uand silver isotope A 47 107 g.Obtain approximately the ratio of the nuclear radii of the gold isotope A 79 197 uand silver isotope A 47 107 g.


As nuclear radius R A 1 3 where A is atomic mass of the nuclei. R Au R Ag = ( A Au A Ag ) 1 3 = ( 197 107 ) 1 3 = ( 1.84 ) 1 3 =1.23 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9597@


Find the Q-value and the kinetic energy of the emitted particlein the decay ofaRa88226  and  bRn86220Given, mRa88226 = 226.02540 u;       mRn86222 = 222.01750 umRn86220 = 220.01137 u;       mPo84216 = 216.00189 u.


aRa88226Rn+86222He24Thus, the value of binding energy or Q-Value=mRa88226mRn86222+mHe24 =226.02540–226.0201=0.0053 uThusQ=0.0053×931 MeVas 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×Q Eα=222226×4.93 =4.84 MeV .(b)Rn86220Po82216+He24QValue=mRn86220mPo82216+mHe24 =220.011373216.00189+4.00260 =220.011373220.00449=0.006883 u =0.006883×931 MeV= 6.41 MeV .K.E. of αparticle emitted during decay of Rn will be Eα=MassnumberafterdecayMassnumberbeforedecay×Q Eα=216220×4.93 =4.84 MeV


The radionuclide C 6 11 decays according to C 6 11 ® B 5 11 + e + + v: half life=20.3 min.The maximum energy of the emitted positron is 0.960 MeV. Given the mass values. m C 6 11 = 11.011434 u; m B 5 11 = 11.009305 u Calculate Q and compare it with the maximum energy of the positron emitted.


As given the equation of the decay process of C 6 11 is C 6 11 5 B 11 + e + + v+Q Thus energy Q= m N C 6 11 m N 5 B 11 m e ×931MeV Where the masses used are of nuclei and not of atoms. In order to express the Q value in terms of the atomic masses, 6 m e mass has to be subtracted from atomic mass of carbon and 5 m e mass has 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 be Q= m C 6 11 –6m e –m 5 B 11 +5m e –m e ×931 MeV = m C 6 11 –m 5 B 11 -2m e ×931MeV = 11.01143411.0093052×0.000548 ×931 =0.001033×931 =0.961 MeV. This energy is comparable to actual energy released in the decay process. The daughter nucleus 5 B 11 is very heavy as compared to e + and neutrino so the daughter nucleus carries minimumenergy. If the kinetic energy carried by the neutrino is minimum, then the positron carries maximum energy which is practically same as energy Q.


The nucleus Ne decays by- emission. Write down the-decay1023equation and determine the maximum kinetic energy of theelectrons emitted. Given that:m Ne1023 = 22.994466 um B511 = 22.089770 u


The decay process of Ne 1023 isNe1023Na1123+e+v+QIf whole of energy is carried by β particle.ThenQ= mNNe1023 mNNa1123 me×931.5 MeVWhere the masses used as  mNNe1023 and  mNNa1123 are Ne1023 and Na respectively1123.If mNe1023 and  mNa1123 are atomic masses of Ne1023  and Na1123respectively, thenmNNe1023=mNe102310me  as atom of Ne contains 10 electrons.Similarly mNNa1123=mNa112311me as atom of Na contains11 electrons. Thus,Q=mNe102310memNe1023+11meme×931 Q=mNe1023mNa1123 x 931 MeV Q= 22.994466  22.989770×931 MeV Q=0.004689×931=4.37 MeV .


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 the following reactions are exothermic or endothermic.(i)   H + 11H13H + 12H12(ii)  C +612C612Ne + 1020He24Atomic masses are given to be ,m(H11) = 1.007825 u, m(H12) = 2.014102 u, m(H13) = 3.016049 u,m( C612) = 12.00000 u, m(Ne1020) = 19.992439 u, m(He24) = 4.002603 u 


i The given reaction isH11+H13H12+H12 Q=mNH11+mNH13mNH12mNH12 ×931MeV .iwhere mN refers to nuclear masses,If m refers to atomic masses, thenmH11=mNH11+memNH11=mH11meSimilarly,mNH13=mH13me and mNH21=mH21meThus, Q=mNH11+mNH13 mNH12mNH12×931MeVBecomes Q=mH11+mH13mH12m H12×931MeV Q= 1.007825+3.0160492.0141022.014102 ×931MeV Q= 0.00433×931 Q= 4.03 MeV .Since Q value is negative, the reaction is endothermic.

ii The given reaction isC612+C612Ne1020+He24 Q=[mNC612+mNC612mNNe1020mNHe24×931 MeV.2Where  mN refers to the nuclear mass. If m refers to theatomic mass, thenmNC612=mC6126me, As C612 has 6 electrons in its atom.SimilarlymNNe1020=mNe102010me and mNHe24=mHe242me Hence,Q= mNC612+mNC612mNNe1020mNHe24 ×931MeVQ=mC6126me+mC6126memNe1020+10memHe24+2me ×931MeV Q=mC612+mC612mNe1020 mHe24 ×931MeV Q=12.000000+12.00000019.9924394.002603 ×931MeV Q=0.004958×931 MeVQ=4.66 MeVSince Q value is positive, the reaction is exothermic .


Suppose, we think of fission of a Fe nucleus into two equal2656fragments, Al. Is the fission energetically possible? Argue1328by working out Q of the process. Given:m Fe2656 = 55.93494 u; m Al1328 = 27.98191 u


As suggested in the question, if Fe2656Al1328+Al1328Then Qvalue of process will be=mFe26562mAl1328×931MeV =55.93494227.98191×931MeV = 0.02888 ×931MeV = 0.02888×931 MeV Hence,Q= 26.90 MeV .Since, Q value is negative, so the fission is not possible .


The fission properties of Pu are very similar to those of 94239U.92235The average energy released per fission is 180 MeV. How muchenergy, in MeV, is released if all atoms in 1 kg of pure Pu94239undergo fission?The fission properties of 94Pu239are very similar to those of U. The average energy released per fission is 180 MeV.92235How much energy, in MeV, is released if all atoms in 1 kg of pure Pu undergo fission?94239


Number of atoms present in 239 g ofPu= 6.023×102394239Number of atoms present in 1 g of  Pu     = 942396.023×1023239Number of atoms present in 1000 g of  Pu = 942396.023×1023×1000239                                                     = 2.52×1024 atomsAs given,the average energy released per fission = 180 MeVThus total energy released , Q=180 x 2.52 x 1024 MeV Q=4.536×1026 MeVHence, 4.536×1026MeV is released if all the atoms in 1 kg of purePu undergo fission.94239


A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much U did it contain initially? Assume that the reactor92235operates 80 % of the time, that all the energy generated arisesfrom the fission of U and that this nuclide is consumed by92235the fission process.


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 t E=109×1.577×108=1.577×1017 J As we know energy generated per fission of U=200 MeV 92235=200×1.6×10-13J=3.2×10-11J Therefore 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 of U92235 Mass of U consumed in five years 92235= 235×4.93×1027g 6.023×1023 = 1924 kg=half of the mass of the fuelconsumed during 5 years as given in the question.Hence, initial mass of U=2×192492235=3848 kg


How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as H 1 2 + H 1 2 H 2 3 + n + 3.27 MeV


As given power of lamp, P= 100 W andmass of deuterium= 2 kg From the reaction given in the exercise, we find that 2 atoms of H 1 2 combine to give 3.2 MeV of energy. Now 2g of H 1 2 equivalent to 6 .023×10 23 atoms Thus 2kg or 2000g of H 1 2 is equivalent to 6 .023×10 23 ×2000 2 =6023×10 23 =6 .023 ×10 26 atoms Thus the energy released for combination of 6 .023×10 26 atoms should be E= 3.27×6 .023×10 26 2 = 9 .85×10 26 MeV E=9 .85×10 26 ×1 .6×10 -13 J E=1 .58×10 14 J As P= E t thus t= E P and we get t= 1 .58×10 14 100 = 1 .58×10 12 s = 1 .58×10 12 365×24×60×60 years

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.)


Before collision, the initial total energy E of the two deuterons is given by, E=2( Kinetic energy ) After collision, when the two deuterons stop,their energy is totally potential energy U and it is given by U= 1 4π Є 0 e.e 2R = 1 4π Є 0 e 2 2R As given that radius of each deuteron sphere is R =2 fm = 2 x 1 0 15 m Charge,e= 1.6×1 0 19 Cand 1 4π Є 0 = 9×1 0 9 Nm 2 C 2 Thus P.E = [ 9×1 0 9 × ( 1.6×1 0 19 ) 2 ] ( 4×1 0 15 ) joule P.E= 9×1.6×1.6×1 0 14 4×1.6×1 0 19 eV =3.6×1 0 5 eV =360keV As per law of conservation of energy, we can say that 2K.E=P.E K.E= P.E 2 = 360 2 =180keV. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@33DD@

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)


As nuclear density= mass number volume = 3A ( 4π R 3 ) = 3A 4πR 0 3 A = 3 4π R 0 3 By using relation R= R 0 A 1 3 Hence, nuclear density is independent of A. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C482@


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 + X Z A Y Z1 A +v Show that ifemission is energetically allowed, electron capture is necessarily allowed but not viceversa.


For the positron emission, we have X Z A Y Z1 A + e 1 0 +v + Q 1 …… 1 And for electron capture, we have X Z A + e 1 0 Y Z1 A +v+ Q 2 …… 2 If m N represents the nuclear mass and m represents the atomic mass, then for m e as the mass of 1 e 0 or +1 e 0 , we have nuclear mass of X Z A = m N X Z A = m X Z A Zm e Similarly nuclear mass of Y Z1 A should be Y Z1 A = m N Y Z1 A = m Y Z1 A Z1 m e Hencefromequation(1) Q 1 = m N X Z A -m N Y Z1 A m e c 2 = m X Z A Z m e -m Y Z1 A + Z1 m e m e c 2 = m X Z A m Y Z1 A 2m e c 2 …….. 3 Similarly from equaton 2 Q 2 = m N X Z A + m e m N Y Z1 A c 2 = m X Z A – Zm e + m e m Y Z1 A + Z1 m e c 2 = m X Z A m Y Z1 A c 2 4 Thus from eqns. 3 and 4 , we find that if Q 1 > 0,is necessarily greater than zero. However, if Q 2 > 0, it does not always mean that Q 1 >0.


In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on Earth. The three isotopes and their masses are  Mg1224(23.98504 u), Mg1225(24.98584 u) and Mg1226(25.98259 u). The natural abundance of Mg  is 78.99 % by mass. Calculate the abundance of the other two isotopes.1224


Let us assume that the natural abundance of Mg be x% then natural abundance of 1225Mg  will be 1226(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 1225Mg, (21.01 – x) = 11.71 % .1226


The neutron separation energy is defined as the energy requiredto remove a neutron from the nucleus. Obtain the neutron separationenergies of the nuclei Ca and 2041Al from the following data:1327mCa2040=39.962591 u, mCa2041=40.962278 u mAl1326=25.986895 u, mAl1327=26.981541 u.


The process of neutron separation for Ca,2041is given asCa+E=2041Ca+2040n 01E= mCa2040+mn–mCa2041×931 MeV E=39.962591+1.008665–40.962278×931 MeV E=0.008978×931=8.36 MeVThe process of neutron separation for Al is1327Al+E=1327Al+1326n01Thus, neutron separation energy will be E=mAl1326+mn–mAl1327×931 MeVE=25.986895+1.008665-26.981541×931 MeVThus, E =0.014019×931=13.05 MeV .


A source contains two phosphorus radio nuclides P 15 32 T 1 2 = 14.3d and P 15 33 T 1 2 =25.3d . Initially,10 %of the decays come from P 15 33 . How long one must wait until 90 % do so?


Let us assume that R 01 and R 02 be the initial activities of P 15 33 and P 15 32 respectively and R 1 and R 2 be their activities at any instant t. As suggested in the question, R 01 = 10 R 01 + R 02 100 as R 01 =10%of total activity. Hence 10R 01 =R 01 + R 02 Or R 02 =9R 01 ………. 1 At a later time t , R 1 = 90 R 1 + R 2 100 as R 1 = 90%of total activity at time t. Hence 10R 1 =9R 1 +9R 2 Or R 2 = 10 R 1 –9R 1 9 = R 1 9 …….. 2 Now dividing 2 by 1 we get, R 2 R 02 = R 1 9 1 9R 01 Or R 2 R 02 = 1 81 R 1 R 01 As per decay law, R = R 0 e -λt R 02 e -λ2t R 02 = 1 81 R 01 e -λ1t R 01 Þ 81= e -λ1t e -λ2t Or e λ 2 – λ 1 t =81 Taking natural log of both sides, lne λ 2 – λ 1 t =ln 81 Or λ 2 – λ 1 t=2.303 log 81 ……. 3 As ,λ = 0.693 T 1 2 Thus λ 1 = 0.693 T 1 2 1 = 0.693 25.3 day -1 and λ 2 = 0.693 T 1 2 2 = 0.693 14.3 day -1 Thus with reference to equation 3 , we get 0.693 14.3 0.693 25.3 t=2.303 log 81 Or 0.693× 25.3-14.3 t 25.3 ×14.3 = 2.303 log 81 0.693× 11 t 25.3 ×14.3 = 2.303 log 81 Thus t= 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.


Under certain circumstances, a nucleus can decay by emitting aparticle more massive than an–particle. Consider the following decay process: R 88 223 a P 82 209 b + C 6 14 R 88 223 a R 86 219 n+ H 2 4 e Calculate the Q values for these decays and determine that the both are energetically allowed.The required data is m R 88 223 a =223.01850 u, m P 82 209 b =208.98107 u, m R 86 219 n =219.00948 u, m C 6 14 =14.00324 u, m H 2 4 e =4.00260 u.


The given decay process for R 88 223 ais R 88 223 a P 82 209 b+ C 6 14 +Q Hence mass defect Δ m= m N R 88 223 a m N P 82 209 b m N C 6 14 Where m N denotes the nuclear mass. If m represents the atomic mass, then Δm= m R 88 223 a 88m e m P 82 209 b + 82m e m C 6 14 + m e = m R 88 223 a m P 82 209 b m C 6 14 As energy Q=Δm×931 MeV Q= 223.01850208.9810714.003241 ×931 =31.83 MeV. For another decay for R 88 223 a, decay equation is R 88 223 a= R 86 219 n+ H 2 4 e+Q Δm= m N R 88 223 a m N R 86 219 n m N H 2 4 e = m R 88 223 a m R 86 219 n m H 2 4 e Thus Q= 223.01850219.009484.00260 ×931 MeV =5.98 MeV.


Consider the fission of U92238 by fast neutrons. In one fission event,no neutrons are emitted and the final stable end products, afterthe betadecay of the primary fragments, areCe58140 and Ru4499.Calculate Q for this fission process. The relevant atomic andparticle masses are:mU92238=238.05079 u; mCe58140=139.90543 u;mRu4499=98.90594 u; mn=1.00867 u


The fission process may be expressed asU92238+n01Ce58140+Ru4499+Q Q= mU92238+mnmCe58140mRu4499×931 MeV    = 238.05079+1.00867139.9054398.90594×931    = 230 .97 MeV .


a Consider the D-T reaction deuterium-tritium–fusion given in eqn:H+12H–>13He+24n + Q01Calculate the energy released in MeV in this reaction from the data:m H12 = 2.014102 u; m H13 = 3.016049 u; m He24 = 4.002603 u; mn= 1.00867 ub 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?


(a) From the equation given in the questionQ=mNH12+mNH13–mNHe24–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, thenmNH12=mH12–me mNH13=mH13–me mNHe24=mHe24–2e Q=mH12–me+mH13–me–mHe24+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= 1 4 π ε0 q1 q22r = 14 π ε0 e2 2 rAs q1= q2= e and  distance between particle= 2 r .Thus, P.E= 9×1091.6×10-192 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 = 23 kT2where 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


Obtain the maximum kinetic energy ofparticle and the radiation frequencies of decays in the following decay scheme shown in the fig 13.6. You are given that m A 79 198 u = 197.968233 u, m H 80 198 g = 197.966760 u.


The total energy released for the transformation ofAu to79198Hg80198can be found by considering the energies of gamma rays.For γ1, the frequencyν1=E1h = 1.088×1.6×10-136.626×10-34 ν1=2.63×1020 Hz For γ2, the frequency isν2 =E2h =0.412×1.6×10-136.626×10-34 ν2 =9.95×1019 Hz .For γ3, the energyE3=1.088–0.412 E3=0.676 Mev=0.676×1.6×10-13 J E3=1.082×10-13 J ν3=E3h= 1.082×10-136.626×10-34 ν3=1.63×1020 HzNow for β1 decay,Maximum kinetic energy=mAu79198–mHg80198×931–1.088 MeV =197.968233–197.966760 ×931–1.088 MeV =1.372–1.088=0.283 MeV .Now for β2 decay, Maximum kinetic energy=mAu79198–mHg80198×931–0.412 =197.968233–1.97.966760 ×931–0.412 MeV =1.372–0.412=0.96 MeV .


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


a In sun fusion takes place according to the equation4H—-11He+ 224e+26 MeV+10 So, 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×6.02×1026 4 =39×1026 MeV .b Fission of one U nucleus gives energy=200 MeV92235Now, 235 g of  U has no. of atoms = 6.02×102392235 atoms1 kg i.e 1000 g of  U has= 92235 6.02×1023×1000 atoms 235                                       = 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 of H11Energy released by 1 kg of fission of U92235hence, EE’= 3.913 x 1027 5.12 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.


Total power target=200000 MW=200000×106=2×1011 W Nuclear power target is 10% of total power target= 10×2×1011 W 100 =2×1010 W As 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=8×1010×366×25×60×60 as 2020 is a leap year .E=2.265×1018 JNow 1 fission ofU92235 produces 200 MeV 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 1023atomsofU92235havemass=235gThus mass required to produce 7.90 x 1028atoms= 235×3.95×1028g 6.023×1023 =3.084×104kg .Hence,massofuraniumneededperyear3.08×104kg .

For viewing question paper please click here

FAQs (Frequently Asked Questions)

1. Which essential concepts are in NCERT Solutions Class 12 Physics Chapter 13?

The NCERT Solutions cover all questions from the NCERT books based on topics like Nuclear Force, Mass-energy, and Nuclear Binding Energy. Students also gain detailed information about the Law of radioactive decay and its types, such as Alpha decay, Beta-decay, and Gamma decay. These NCERT solutions give clear and precise information about these topics.

2. Which chapters are included in the NCERT Exemplar Class 12 Physics?

Extramarks has provided well-structured answers to all the following chapters:

  1. Electric Charges and Fields
  2. Electrostatic Potential and Capacitance
  3. Current Electricity
  4. Moving Charges and Magnetism
  5. Magnetism and Matter
  6. Electromagnetic Induction
  7. Alternating Current
  8. Electromagnetic Waves
  9. Ray Optics and Optical Instruments
  10. Wave Optics
  11. Dual Nature of Radiation and Matter
  12. Atoms
  13. Nuclei
  14. Semiconductor Electronics: Materials, Devices, and Simple Circuits
  15. Communication Systems