NCERT Solutions for Class 12 Physics Chapter 13: Nuclei

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Jasmine Grover

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NCERT Solutions for class 12 physics chapter 13 Nuclei are given in this article. Atomic Nucleus is the small, dense and central part of the Atom consisting of Protons, which are positively charged and Neutrons, which are electrically neutral containing more than 99.9% of the mass of an atom and are ten thousand times smaller than an atom.

Unit 8 Atoms and Nuclei along with Unit 7 Dual Nature of Radiation and Matter has a weightage of 12 marks in the CBSE Board examinations. NCERT Solutions Class 12 Physics Chapter 13 covers concepts of Mass-Energy and Nuclear Binding EnergyRadioactive Decay, and Nuclear Energy.

Download PDF: NCERT Solutions for Class 12 Physics Chapter 13


NCERT Solutions for Class 12 Physics Chapter 13

The NCERT solutions for class 12 physics chapter 13: Nuclei are given below.

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Class 12 Physics Chapter 13 Nuclei – Topics Covered

  • Atomic Mass Unit (amu): The unit used to express atomic masses is known as the atomic mass unit. It is defined as 1/12th of the mass of a Carbon atom (C12).
1 u = 1.660539 x 10-27 kg
  • Atomic number of an element refers to the number of protons that are present inside the nucleus of an atom of an element.
Atomic number = Number of protons = Number of electrons
  • Mass number of an element refers to the total number of protons and neutrons inside the atomic nucleus of the element.
Mass number = Number of protons + Number of neutrons = Number of electrons + Number of neutrons i.e. A = Z + N
  • Size of Nucleus: If R is the radius of the nucleus that has mass number A, then the size of the nucleus can be represented by:
\({4 \over 3} \pi R^3 \propto A => R \propto A^{1/3} => R = R_0A^{1 \over 3}\)
  • Radioactivity Decay Law: According to the Radioactive Decay law, the rate of decay of radioactive atoms at any instant is directly proportional to the number of atoms present at that instant.
\({dN \over dt} \propto N, {dN \over dt}=\ - \lambda N\)

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CBSE CLASS XII Related Questions

  • 1.

    Two cells of emfs 12 V and 6 V are connected in parallel as shown in the figure. Their internal resistances are \(1\,\Omega\) and \(0.5\,\Omega\) respectively. Calculate the emf and internal resistance of the equivalent cell between points A and B.


      • 2.
        An equipotential surface through a point is normal to the electric field at that point.Explain


          • 3.
            When a dielectric is placed in an external electric field, the electric field inside the dielectric is less than that outside it.Explain


              • 4.
                Derive the relation for the refractive index ($\mu$) of a prism in terms of angle of minimum deviation ($\delta_m$) and angle of prism ($A$).

                  • $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
                  • $\mu = \frac{\cos\left(\frac{A + \delta_m}{2}\right)}{\cos\left(\frac{A}{2}\right)}$
                  • $\mu = \frac{\sin\left(\frac{A - \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
                  • $\mu = \frac{\tan\left(\frac{A + \delta_m}{2}\right)}{\tan\left(\frac{A}{2}\right)}$

                • 5.
                  The assertion that V = IR is a statement of Ohm’s law is not true. Why ?


                    • 6.
                      Explain the potential difference between the plates of a charged parallel plate capacitor decreases when its plates are brought closer.

                        CBSE CLASS XII Previous Year Papers

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