NCERT Solutions for Class 12 Physics Chapter 13 Nuclei, PDF Download

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei: Get here the NCERT Class 12 Nuclei Solutions in PDF format. The solutions follow the revised NCERT Class 12 Physics textbook.

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei
NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

NCERT Class 12 Physics Solutions: Students of Class 12 can refer to this article to get NCERT Solutions for Class 12 Physics Chapter 13. These solutions will be helpful in understanding Chapter 13 Nuclei. The NCERT Solutions for Class 12 Chapter 13 Nuclei will include all unsolved theories and numerical problems. Read to get NCERT solutions for Class 12 Physics, Chapter 13 all exercises.

Read: NCERT Class 12 Physics Solutions

Read: CBSE Class 12 Physics Syllabus 2023-24

Read: RBSE Class 12 Physics Syllabus 2023-24

Read: Punjab Board Class 12 Physics Syllabus 2023-24

Get, detailed solutions to the questions of the chapter Nuclei from NCERT textbook Class 12 Physics. The objective is to helping students regarding the pattern of answering the question as per the cbse latest marking scheme. Check the NCERT Class 12 Physics Nuclei solutions below. 

Shiv Khera

Nuclei Solutions

Some questions of this chapter are given here.

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


Binding energy of 7N14 = ΔMc2
Mass defect = [mass of 7 protons + mass of 7 neutrons – actual mass of N]
                    = [14.11543 – 14.00307]
                    = 0.11236 u
Binding energy = 0.11236 u x 931.5 MeV / u
                         = 104.66 MeV

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Q. 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 63 29Cu atoms (of mass 62.92960 u).


Mass of coin = 0.003 kg
Mass of 29Cu63 = 62.9296 u
                            = 1.045261 x 10-25 kg

No. of Cu atoms in given sample
                         D = 0.003 / 1.045261 x 10-25
                             = 2.8701 x 1022 atoms

Mass defect of 1 Cu atom = 0.591935

BE of 1 Cu atom = 0.591935 x 931.5 = 551.387 MeV
   BE of  D atoms = 2.8701 x 1022 x 551.387
                             = 1.583 x 1025 MeV
                             = 2.53 x 1012 J

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NCERT Rationalised Content Chapter: Nuclei

Chapter Name

Page No.

Deleted Topics

Chapter 13: Nuclei


13.6.1 Law of Radioactive Decay

13.6.2 Alpha Decay

13.6.3 Beta Decay

13.6.4 Gamma Decay



13.7.2 Nuclear Reactor


Exercises 13.1, 13.2, 13.6–13.10, 13.12–13.14, 13.18, 13.22–13.31


CBSE Class 12 Physics Updated Question Paper Design 2023-24

Typology of Questions

Total Marks

Approximate % 

Remembering: Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers. 

Understanding: Demonstrate understanding of facts and ideas by organising, comparing, translating, interpreting, giving descriptions, and stating main ideas.



Applying: Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.



Analysing: Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalisations. 

Evaluating: Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. 

Creating: Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions.









Gross Total



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