CBSE Class 12 · Physics 
(i) Nuclear Fission: A heavy nucleus in the larger mass region ($A > 200$) breaks into two middle-level nuclei. From the graph, the binding energy per nucleon of the fragments is higher than that of the parent. This increase results in a release of energy.
(ii) Nuclear Fusion: Lighter nuclei in the lower mass region ($A < 20$) fuse to form a nucleus having higher BE/nucleon. Since the final value of BE/nucleon is more than the initial value, energy is released in this process as well. 
Inference (i): The binding energy per nucleon $E_{bn}$ is practically constant and independent of atomic number for nuclei of middle mass number ($30 < A < 170$). The curve has a maximum of about 8.75 MeV for $A = 56$ and has a value of 7.6 MeV for $A = 238$.
Inference (ii): $E_{bn}$ is lower for both light nuclei ($A < 30$) and heavy nuclei ($A > 170$). This implies that fission of heavy nuclei and fusion of light nuclei both release energy, as both processes lead to products with higher BE per nucleon. @import url(‘https://fonts.googleapis.com/css2?family=Montserrat:wght@400;500;600;700&display=swap’);
CBSE Class 12 Physics Nuclei Previous Year Questions
Help your child master CBSE Class 12 Physics Nuclei Previous Year Questions with this curated collection sourced from real board papers. Every question comes with a detailed step-by-step solution, covering nuclear density, binding energy, radioactive decay, and nuclear reactions — topics that consistently carry marks in the board exam.
CBSE Class 12 Physics Nuclei — Questions with Solutions
The ratio of the nuclear densities of two nuclei having mass numbers 64 and 125 is
Solution
Answer: Option (D) is correct.
Explanation: Nuclear density is independent of mass number. Therefore the ratio of nuclear densities of any two nuclei is always 1:1, regardless of their mass numbers.
Explanation: Nuclear density is independent of mass number. Therefore the ratio of nuclear densities of any two nuclei is always 1:1, regardless of their mass numbers.
The curve of binding energy per nucleon as a function of atomic mass number has a sharp peak for helium nucleus. This implies that helium nucleus is
Solution
Answer: Option (D) is correct.
Explanation: A nucleus becomes stable when its binding energy per nucleon is around 8 MeV. The peak indicates the binding energy per nucleon of He is closer to 8 MeV (actually 7.1 MeV). So, it is more stable than its neighbours.
Explanation: A nucleus becomes stable when its binding energy per nucleon is around 8 MeV. The peak indicates the binding energy per nucleon of He is closer to 8 MeV (actually 7.1 MeV). So, it is more stable than its neighbours.
The radius of ${}_{13}^{27}\mathrm{X}$ nucleus is R. The radius of ${}_{53}^{125}\mathrm{Y}$ nucleus will be:
Solution
Answer: Option (A) is correct.
Explanation: Using $\mathrm{R} = \mathrm{R_0} A^{1/3}$:
For X: $R = R_0 (27)^{1/3} = 3R_0$, so $R_0 = \dfrac{R}{3}$
For Y: $R’ = R_0 (125)^{1/3} = 5R_0 = \dfrac{5R}{3}$
Explanation: Using $\mathrm{R} = \mathrm{R_0} A^{1/3}$:
For X: $R = R_0 (27)^{1/3} = 3R_0$, so $R_0 = \dfrac{R}{3}$
For Y: $R’ = R_0 (125)^{1/3} = 5R_0 = \dfrac{5R}{3}$
Which of the following statements about nuclear forces is not true?
Solution
Answer: Option (B) is correct (as the false statement).
Explanation: The nuclear force is much stronger than the Coulomb force acting between charges, not weaker. This is what makes the nucleus stable despite the electrostatic repulsion between protons.
Explanation: The nuclear force is much stronger than the Coulomb force acting between charges, not weaker. This is what makes the nucleus stable despite the electrostatic repulsion between protons.
When two nuclei ($A \leq 10$) fuse together to form a heavier nucleus, the
Solution
Answer: Option (A) is correct.
Explanation: When two light nuclei ($A \leq 10$) fuse, the resulting nucleus has a higher binding energy per nucleon. This makes the product more stable, and the mass defect is converted into energy according to $E = \Delta m \cdot c^2$.
Explanation: When two light nuclei ($A \leq 10$) fuse, the resulting nucleus has a higher binding energy per nucleon. This makes the product more stable, and the mass defect is converted into energy according to $E = \Delta m \cdot c^2$.
In $\beta^{-}$ decay, a
Solution
Answer: Option (A) is correct.
Explanation: In $\beta^{-}$ decay, a neutron inside the nucleus converts into a proton. An electron ($\beta^{-}$ particle) and an antineutrino are emitted in the process.
Explanation: In $\beta^{-}$ decay, a neutron inside the nucleus converts into a proton. An electron ($\beta^{-}$ particle) and an antineutrino are emitted in the process.
Heavy stable nuclei have more neutrons than protons. This is because of the fact that
Solution
Answer: Option (B) is correct.
Explanation: The electrostatic (Coulomb) repulsion between protons causes instability in the nucleus. Extra neutrons provide additional attractive nuclear force without adding to the repulsive electrostatic force, hence heavy stable nuclei have more neutrons than protons.
Explanation: The electrostatic (Coulomb) repulsion between protons causes instability in the nucleus. Extra neutrons provide additional attractive nuclear force without adding to the repulsive electrostatic force, hence heavy stable nuclei have more neutrons than protons.
A certain nucleus M decays into N which further decays to R by undergoing the reactions shown below.
$\mathbf{M} \rightarrow \mathbf{N} + \beta^{-}$
$\mathbf{N} \rightarrow \mathbf{R} + \alpha$
Which of the following options is correct about the above reactions?
$\mathbf{M} \rightarrow \mathbf{N} + \beta^{-}$
$\mathbf{N} \rightarrow \mathbf{R} + \alpha$
Which of the following options is correct about the above reactions?
Solution
Answer: Option (D) is correct.
Explanation: In $\beta^{-}$ decay, atomic number increases by 1 (M → N). In alpha decay, atomic number decreases by 2 (N → R). Net change from M to R: $+1 – 2 = -1$. Therefore, atomic number of R is one less than that of M.
Explanation: In $\beta^{-}$ decay, atomic number increases by 1 (M → N). In alpha decay, atomic number decreases by 2 (N → R). Net change from M to R: $+1 – 2 = -1$. Therefore, atomic number of R is one less than that of M.
Nuclear force is a ________ and ________ force.
Solution
Answer: Option (B) is correct.
Explanation: Nuclear force is the strongest short-range force which binds the neutrons and protons inside a nucleus. It operates only up to a range of a few femtometres.
Explanation: Nuclear force is the strongest short-range force which binds the neutrons and protons inside a nucleus. It operates only up to a range of a few femtometres.
Two nuclei have mass numbers in the ratio $1:2$. The ratio of their nuclear densities is
Solution
Answer: Option (C) is correct.
Explanation: Nuclear density does not depend upon the mass number. It is a universal constant for all nuclei. Hence, the density ratio will be $1:1$.
Explanation: Nuclear density does not depend upon the mass number. It is a universal constant for all nuclei. Hence, the density ratio will be $1:1$.
Consider the following reaction:
${}_{92}^{238}\mathrm{U} \rightarrow {}_{90}^{234}\mathrm{Th} + {}_{2}^{4}\mathrm{He}$
Which of the given options is correct for the above reaction if U was initially at rest?
${}_{92}^{238}\mathrm{U} \rightarrow {}_{90}^{234}\mathrm{Th} + {}_{2}^{4}\mathrm{He}$
Which of the given options is correct for the above reaction if U was initially at rest?
Solution
Answer: Option (B) is correct.
Explanation: By conservation of momentum, both Th and He have equal and opposite momenta. Since kinetic energy $K = \dfrac{p^2}{2m}$ and the mass of He is much less than Th, the kinetic energy of He will be greater. Hence kinetic energy of Th is less than that of He.
Explanation: By conservation of momentum, both Th and He have equal and opposite momenta. Since kinetic energy $K = \dfrac{p^2}{2m}$ and the mass of He is much less than Th, the kinetic energy of He will be greater. Hence kinetic energy of Th is less than that of He.
________ has the mass closest to the mass of a positron.
Solution
Answer: Option (C) is correct.
Explanation: A positron is the antiparticle of an electron. An antiparticle has the same mass as its conjugate particle. Therefore, the positron has the same mass as the electron.
Explanation: A positron is the antiparticle of an electron. An antiparticle has the same mass as its conjugate particle. Therefore, the positron has the same mass as the electron.
$X$ amount of energy is required to remove an electron from its orbit and $Y$ amount of energy is required to remove a nucleon from the nucleus.
Solution
Answer: Option (C) is correct.
Explanation: Nuclear force is much greater than the Coulomb force. Hence, the energy required to remove a nucleon from the nucleus ($Y$) is much greater than the energy required to remove an electron from its orbit ($X$). Therefore $Y > X$.
Explanation: Nuclear force is much greater than the Coulomb force. Hence, the energy required to remove a nucleon from the nucleus ($Y$) is much greater than the energy required to remove an electron from its orbit ($X$). Therefore $Y > X$.
An electron and an $\alpha$-particle have the same de-Broglie wavelength. How are their kinetic energies related?
Solution
Answer: Option (B) is correct.
Explanation: For the same de-Broglie wavelength $\lambda = \dfrac{h}{p}$, both particles have equal momentum $p$. Kinetic energy $K = \dfrac{p^2}{2m}$. Since the electron has much smaller mass than the $\alpha$-particle, its kinetic energy is much larger. Hence $K_e > K_\alpha$.
Explanation: For the same de-Broglie wavelength $\lambda = \dfrac{h}{p}$, both particles have equal momentum $p$. Kinetic energy $K = \dfrac{p^2}{2m}$. Since the electron has much smaller mass than the $\alpha$-particle, its kinetic energy is much larger. Hence $K_e > K_\alpha$.
Which of the following quantities is not conserved in a nuclear reaction?
Solution
Answer: Option (C) is correct.
Explanation: In nuclear reactions, some mass can be converted into energy (mass–energy equivalence, $E = mc^2$), so mass alone is not conserved. However, total mass-energy, charge, momentum, and nucleon number are all conserved.
Explanation: In nuclear reactions, some mass can be converted into energy (mass–energy equivalence, $E = mc^2$), so mass alone is not conserved. However, total mass-energy, charge, momentum, and nucleon number are all conserved.
What do you mean by binding energy per nucleon?
Answer
The binding energy per nucleon is defined as the energy required to break up a nucleus into its constituent protons and neutrons and to separate them to such a large distance that they no longer interact with each other.
It may also be defined as the surplus energy which the nucleons give up by virtue of their attractions when they become bound together to form a nucleus.
The binding energy of a nucleus is given by: $$\Delta m = Zm_p + (A-Z)m_n – M$$ $$E_b = (\Delta m)c^2$$
It may also be defined as the surplus energy which the nucleons give up by virtue of their attractions when they become bound together to form a nucleus.
The binding energy of a nucleus is given by: $$\Delta m = Zm_p + (A-Z)m_n – M$$ $$E_b = (\Delta m)c^2$$
How does the mass density of a nucleus vary with mass number?
Answer
Mass density of a nucleus is independent of mass number. The nuclear radius $R = R_0 A^{1/3}$, so the volume $V \propto A$. Since mass $M \propto A$ as well, the density $\rho = M/V$ remains constant regardless of the mass number $A$.
The binding energies per nucleon for deuteron and an alpha-particle are $x_1$ and $x_2$ respectively. Find the amount of energy released in the following reaction:
$${}_{1}\mathrm{H}^2 + {}_{1}\mathrm{H}^2 \rightarrow {}_{2}\mathrm{He}^4 + Q$$
Answer
Total binding energy of reactants $= 2 \times 2x_1 = 4x_1$
Total binding energy of product $= 4x_2$
Energy released $Q = 4x_2 – 4x_1 = \mathbf{4(x_2 – x_1)}$
Total binding energy of product $= 4x_2$
Energy released $Q = 4x_2 – 4x_1 = \mathbf{4(x_2 – x_1)}$
What do you mean by mass defect of a nucleus?
Answer
The difference between the rest mass of a nucleus and the sum of the rest masses of its constituent nucleons is called its mass defect. It is given by:
$$\Delta m = Zm_p + (A-Z)m_n – M$$
where $Z$ is the atomic number, $A$ is the mass number, $m_p$ is the mass of a proton, $m_n$ is the mass of a neutron, and $M$ is the actual mass of the nucleus.
In the following nuclear reaction, identify the unknown labelled X:
$${}_{11}^{22}\mathrm{Na} + \mathrm{X} \rightarrow {}_{10}^{22}\mathrm{Ne} + v_e$$
Answer
X is an electron (${}_{-1}^{0}e$).
${}_{11}^{22}\mathrm{Na}$ undergoes beta decay (electron capture), emitting a positron and converting into ${}_{10}^{22}\mathrm{Ne}$. Balancing atomic number: $11 + (-1) = 10$ ✓ and mass number: $22 + 0 = 22$ ✓.
${}_{11}^{22}\mathrm{Na}$ undergoes beta decay (electron capture), emitting a positron and converting into ${}_{10}^{22}\mathrm{Ne}$. Balancing atomic number: $11 + (-1) = 10$ ✓ and mass number: $22 + 0 = 22$ ✓.
What is the nuclear radius of ${}^{125}\mathrm{Fe}$, if that of ${}^{27}\mathrm{Al}$ is 3.6 fermi?
Answer
Using $R = R_0 A^{1/3}$:
$$\frac{R_{Fe}}{R_{Al}} = \left(\frac{A_{Fe}}{A_{Al}}\right)^{1/3} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}$$
$$R_{Fe} = \frac{5}{3} \times R_{Al} = \frac{5}{3} \times 3.6 = \mathbf{6 \text{ fermi}}$$
(i) Distinguish between isotopes and isobars.
(ii) Two nuclei have different mass numbers $A_1$ and $A_2$. Are these nuclei necessarily isotopes of the same element? Explain.
(ii) Two nuclei have different mass numbers $A_1$ and $A_2$. Are these nuclei necessarily isotopes of the same element? Explain.
Answer
(i) Isotopes vs Isobars:
Isotopes are atoms of the same element having the same atomic number (Z) but different mass numbers (A). They have the same number of protons but different numbers of neutrons. Example: ${}^{1}_{1}\mathrm{H}$, ${}^{2}_{1}\mathrm{H}$, ${}^{3}_{1}\mathrm{H}$.
Isobars are atoms of different elements having the same mass number (A) but different atomic numbers (Z). Example: ${}^{40}_{18}\mathrm{Ar}$ and ${}^{40}_{20}\mathrm{Ca}$.
(ii) No. Two nuclei with different mass numbers are not necessarily isotopes of the same element. They are isotopes only when they have the same atomic number. If they have different atomic numbers as well, they may be entirely different elements or could be isobars only if their mass numbers happen to be equal.
Isotopes are atoms of the same element having the same atomic number (Z) but different mass numbers (A). They have the same number of protons but different numbers of neutrons. Example: ${}^{1}_{1}\mathrm{H}$, ${}^{2}_{1}\mathrm{H}$, ${}^{3}_{1}\mathrm{H}$.
Isobars are atoms of different elements having the same mass number (A) but different atomic numbers (Z). Example: ${}^{40}_{18}\mathrm{Ar}$ and ${}^{40}_{20}\mathrm{Ca}$.
(ii) No. Two nuclei with different mass numbers are not necessarily isotopes of the same element. They are isotopes only when they have the same atomic number. If they have different atomic numbers as well, they may be entirely different elements or could be isobars only if their mass numbers happen to be equal.
Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.
Answer
(i) Nuclear Fission: A heavy nucleus in the larger mass region ($A > 200$) breaks into two middle-level nuclei. From the graph, the binding energy per nucleon of the fragments is higher than that of the parent. This increase results in a release of energy.
(ii) Nuclear Fusion: Lighter nuclei in the lower mass region ($A < 20$) fuse to form a nucleus having higher BE/nucleon. Since the final value of BE/nucleon is more than the initial value, energy is released in this process as well.
A heavy nucleus P of mass number 240 and binding energy 7.6 MeV per nucleon splits into two nuclei Q and R of mass numbers 110 and 130 and binding energy per nucleon 8.5 MeV and 8.4 MeV, respectively. Calculate the energy released in the fission.
Answer
Total BE of P $= 240 \times 7.6 = 1824$ MeV
BE of Q $= 110 \times 8.5 = 935$ MeV
BE of R $= 130 \times 8.4 = 1092$ MeV
Total BE of Q and R $= 935 + 1092 = 2027$ MeV
Energy released $= 2027 – 1824 = \mathbf{203 \text{ MeV}}$
BE of Q $= 110 \times 8.5 = 935$ MeV
BE of R $= 130 \times 8.4 = 1092$ MeV
Total BE of Q and R $= 935 + 1092 = 2027$ MeV
Energy released $= 2027 – 1824 = \mathbf{203 \text{ MeV}}$
Calculate the energy in the fusion reaction: ${}_{1}^{2}\mathrm{H} + {}_{1}^{2}\mathrm{H} \rightarrow {}_{2}^{3}\mathrm{He} + \mathrm{n}$, where BE of ${}_{1}^{2}\mathrm{H} = 2.23$ MeV and ${}_{2}^{3}\mathrm{He} = 7.73$ MeV.
Answer
Total BE of initial system:
$${}_{1}^{2}\mathrm{H} + {}_{1}^{2}\mathrm{H} = (2.23 + 2.23) \text{ MeV} = 4.46 \text{ MeV}$$
BE of final system (${}_{2}^{3}\mathrm{He}$) $= 7.73$ MeV
Energy released $= 7.73 – 4.46 = \mathbf{3.27 \text{ MeV}}$
Energy released $= 7.73 – 4.46 = \mathbf{3.27 \text{ MeV}}$
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. Assume the coin is entirely made of ${}_{29}^{63}\mathrm{Cu}$ atoms (of mass 62.92960 u). Given $m_p = 1.007825$ u and $m_n = 1.008665$ u.
Answer
Number of atoms in 3 g of Cu coin:
$$N = \frac{6.023 \times 10^{23} \times 3}{63} = 2.86 \times 10^{22}$$
Each atom has 29 protons and 34 neutrons.
Mass defect per atom: $$\Delta M = 29 \times 1.007825 + 34 \times 1.008665 – 62.92960 = 0.59225 \text{ u}$$ Nuclear energy required per atom $= 0.59225 \times 931.5$ MeV
Nuclear energy for 3 g of Cu: $$= 0.59225 \times 931.5 \times 2.86 \times 10^{22} \approx \mathbf{1.58 \times 10^{25} \text{ MeV}}$$
Mass defect per atom: $$\Delta M = 29 \times 1.007825 + 34 \times 1.008665 – 62.92960 = 0.59225 \text{ u}$$ Nuclear energy required per atom $= 0.59225 \times 931.5$ MeV
Nuclear energy for 3 g of Cu: $$= 0.59225 \times 931.5 \times 2.86 \times 10^{22} \approx \mathbf{1.58 \times 10^{25} \text{ MeV}}$$
Draw the graph showing the variation of binding energy per nucleon with mass number. Write two inferences which can be drawn from this graph.
Answer
Inference (i): The binding energy per nucleon $E_{bn}$ is practically constant and independent of atomic number for nuclei of middle mass number ($30 < A < 170$). The curve has a maximum of about 8.75 MeV for $A = 56$ and has a value of 7.6 MeV for $A = 238$.
Inference (ii): $E_{bn}$ is lower for both light nuclei ($A < 30$) and heavy nuclei ($A > 170$). This implies that fission of heavy nuclei and fusion of light nuclei both release energy, as both processes lead to products with higher BE per nucleon.
Identify if the two nuclear reactions mentioned below are endothermic or exothermic. Show your calculations.
${}_{1}^{1}\mathrm{p} + {}_{3}^{7}\mathrm{Li} \rightarrow 2({}_{2}^{4}\mathrm{He})$
${}_{3}^{7}\mathrm{Li} + {}_{2}^{4}\mathrm{He} \rightarrow {}_{0}^{1}\mathrm{n} + {}_{5}^{10}\mathrm{B}$
Given: ${}_{1}^{1}p = 1.00728$ amu, ${}_{3}^{7}\mathrm{Li} = 7.0160$ amu, ${}_{2}^{4}\mathrm{He} = 4.0026$ amu, ${}_{0}^{1}n = 1.0087$ amu, ${}_{5}^{10}\mathrm{B} = 10.01294$ amu
${}_{1}^{1}\mathrm{p} + {}_{3}^{7}\mathrm{Li} \rightarrow 2({}_{2}^{4}\mathrm{He})$
${}_{3}^{7}\mathrm{Li} + {}_{2}^{4}\mathrm{He} \rightarrow {}_{0}^{1}\mathrm{n} + {}_{5}^{10}\mathrm{B}$
Given: ${}_{1}^{1}p = 1.00728$ amu, ${}_{3}^{7}\mathrm{Li} = 7.0160$ amu, ${}_{2}^{4}\mathrm{He} = 4.0026$ amu, ${}_{0}^{1}n = 1.0087$ amu, ${}_{5}^{10}\mathrm{B} = 10.01294$ amu
Answer
Reaction 1:
Mass of reactants $= 1.00728 + 7.0160 = 8.02328$ amu
Mass of products $= 2 \times 4.0026 = 8.0052$ amu
Since mass of reactants > mass of products, energy is released.
Reaction 1 is Exothermic.
Reaction 2:
Mass of reactants $= 7.0160 + 4.0026 = 11.0186$ amu
Mass of products $= 1.0087 + 10.01294 = 11.02164$ amu
Since mass of reactants < mass of products, energy is absorbed.
Reaction 2 is Endothermic.
Mass of reactants $= 1.00728 + 7.0160 = 8.02328$ amu
Mass of products $= 2 \times 4.0026 = 8.0052$ amu
Since mass of reactants > mass of products, energy is released.
Reaction 1 is Exothermic.
Reaction 2:
Mass of reactants $= 7.0160 + 4.0026 = 11.0186$ amu
Mass of products $= 1.0087 + 10.01294 = 11.02164$ amu
Since mass of reactants < mass of products, energy is absorbed.
Reaction 2 is Endothermic.
(i) How is the size of a nucleus found experimentally? Write the relation between the radius and mass number of a nucleus.
(ii) Prove that the density of a nucleus is independent of its mass number.
(ii) Prove that the density of a nucleus is independent of its mass number.
Answer
(i) The size of the nucleus was determined experimentally by Rutherford through his alpha particle scattering experiment. A gold foil was bombarded by energetic $\alpha$-particles. From the analysis of scattering angles and the concept of distance of closest approach, Rutherford concluded that the nuclear size lies between $10^{-15}$ m and $10^{-14}$ m.
Relation between radius and mass number: $$r = R_0 A^{1/3} \quad (R_0 = 1.25 \text{ fm})$$
(ii) Proof of independence of nuclear density from mass number: $$\rho = \frac{M}{V} = \frac{mA}{\frac{4}{3}\pi r^3} = \frac{mA}{\frac{4}{3}\pi R_0^3 A} = \frac{m}{\frac{4}{3}\pi R_0^3}$$ Since $A$ cancels out, $\rho$ is independent of mass number. Hence proved.
Relation between radius and mass number: $$r = R_0 A^{1/3} \quad (R_0 = 1.25 \text{ fm})$$
(ii) Proof of independence of nuclear density from mass number: $$\rho = \frac{M}{V} = \frac{mA}{\frac{4}{3}\pi r^3} = \frac{mA}{\frac{4}{3}\pi R_0^3 A} = \frac{m}{\frac{4}{3}\pi R_0^3}$$ Since $A$ cancels out, $\rho$ is independent of mass number. Hence proved.
(i) Distinguish between nuclear fission and fusion giving an example of each.
(ii) Explain the release of energy in nuclear fission and fusion on the basis of the binding energy per nucleon curve.
(ii) Explain the release of energy in nuclear fission and fusion on the basis of the binding energy per nucleon curve.
Answer
(i)
Nuclear Fission: A heavy nucleus splits into two lighter nuclei of comparable mass. Example: ${}^{235}_{92}\mathrm{U} + \mathrm{n} \rightarrow {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\mathrm{n}$
Nuclear Fusion: Two light nuclei combine to form a heavier nucleus. Example: ${}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \rightarrow {}^{4}_{2}\mathrm{He} + \mathrm{n}$
(ii) In both nuclear fission and nuclear fusion, the binding energy per nucleon of the products is greater than the binding energy per nucleon of the reactants. This difference in binding energy is released in the form of energy.
Nuclear Fission: A heavy nucleus splits into two lighter nuclei of comparable mass. Example: ${}^{235}_{92}\mathrm{U} + \mathrm{n} \rightarrow {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\mathrm{n}$
Nuclear Fusion: Two light nuclei combine to form a heavier nucleus. Example: ${}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \rightarrow {}^{4}_{2}\mathrm{He} + \mathrm{n}$
(ii) In both nuclear fission and nuclear fusion, the binding energy per nucleon of the products is greater than the binding energy per nucleon of the reactants. This difference in binding energy is released in the form of energy.
An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other?
Answer
The kinetic energy $E_K$ of a particle is:
$$E_K = \frac{p^2}{2m}$$
The de-Broglie wavelength is $\lambda = \dfrac{h}{p}$, so:
$$\lambda = \frac{h}{\sqrt{2mE_K}}$$
Since both particles have the same de-Broglie wavelength:
$$\frac{h}{\sqrt{2m_e E_{K_e}}} = \frac{h}{\sqrt{2m_\alpha E_{K_\alpha}}}$$
Simplifying:
$$\frac{m_e}{m_\alpha} = \frac{E_{K_\alpha}}{E_{K_e}}$$
Since $m_\alpha \gg m_e$, it follows that $E_{K_\alpha} \ll E_{K_e}$.
Therefore, the kinetic energy of the electron is much greater than the kinetic energy of the alpha particle: $E_{K_e} > E_{K_\alpha}$.
Therefore, the kinetic energy of the electron is much greater than the kinetic energy of the alpha particle: $E_{K_e} > E_{K_\alpha}$.
Calculate the energy in the fusion reaction:
$${}^2\mathrm{H} + {}^2\mathrm{H} \rightarrow {}^3\mathrm{He} + n$$
where BE of ${}^2\mathrm{H} = 2.23$ MeV and of ${}^3\mathrm{He} = 7.73$ MeV.
Answer
Total binding energy of the initial system $E_i$: $$E_i = {}^2\mathrm{H} + {}^2\mathrm{H} = (2.23 + 2.23) \text{ MeV} = 4.46 \text{ MeV}$$ Binding energy of the final system $E_f$: $$E_f = {}^3\mathrm{He} = 7.73 \text{ MeV}$$ Energy released: $$E_{\text{released}} = E_f – E_i = 7.73 – 4.46 = \mathbf{3.27 \text{ MeV}}$$
Two nuclei have mass numbers in the ratio 1:2. What is the ratio of their nuclear densities?
Answer
Nuclear density $\rho$ is given by:
$$\rho = \frac{\text{Mass of Nucleus}}{\text{Volume of Nucleus}}$$
Since $R = R_0 A^{1/3}$:
$$\rho = \frac{mA}{\frac{4}{3}\pi R_0^3 A} = \frac{m}{\frac{4}{3}\pi R_0^3}$$
The mass number $A$ cancels completely, so nuclear density is independent of $A$.
Therefore, the ratio of nuclear densities is $\mathbf{1:1}$, regardless of the mass number ratio.
Therefore, the ratio of nuclear densities is $\mathbf{1:1}$, regardless of the mass number ratio.
A nucleus with mass number $A = 240$ and $\dfrac{BE}{A} = 7.6$ MeV breaks into two fragments each of $A = 120$ with $\dfrac{BE}{A} = 8.5$ MeV. Calculate the energy released.
Answer
Binding energy of nucleus with $A = 240$:
$$(E_{BN})_1 = 240 \times 7.6 = 1824 \text{ MeV}$$
Binding energy of two fragments ($A = 120$ each):
$$(E_{BN})_2 = 2 \times 120 \times 8.5 = 2040 \text{ MeV}$$
Energy released:
$$\Delta E = (E_{BN})_2 – (E_{BN})_1 = 2040 – 1824 = \mathbf{216 \text{ MeV}}$$
(a) Define radioactive decay constant $\lambda$.
(b) Derive the expression for the number of undecayed nuclei $N(t)$ at time $t$.
(c) A radioactive sample has a half-life of 5 days. If the initial activity is 6400 disintegrations per second, find the activity after 20 days.
(b) Derive the expression for the number of undecayed nuclei $N(t)$ at time $t$.
(c) A radioactive sample has a half-life of 5 days. If the initial activity is 6400 disintegrations per second, find the activity after 20 days.
Answer
(a) The decay constant $\lambda$ is the probability per unit time that a given nucleus will decay. Its unit is s$^{-1}$.
(b) Derivation of $N(t)$:
Rate of decay: $\dfrac{dN}{dt} = -\lambda N$
Separating variables: $\dfrac{dN}{N} = -\lambda \, dt$
Integrating: $\ln N = -\lambda t + C$
At $t = 0$, $N = N_0$, so $C = \ln N_0$. Therefore: $$\boxed{N(t) = N_0 e^{-\lambda t}}$$ This is the radioactive decay law.
(c) Numerical:
$T_{1/2} = 5$ days, $t = 20$ days $= 4 \times T_{1/2}$
Activity after $n$ half-lives: $A = \dfrac{A_0}{2^n}$ $$A = \frac{6400}{2^4} = \frac{6400}{16} = \mathbf{400 \text{ dis/s}}$$
(b) Derivation of $N(t)$:
Rate of decay: $\dfrac{dN}{dt} = -\lambda N$
Separating variables: $\dfrac{dN}{N} = -\lambda \, dt$
Integrating: $\ln N = -\lambda t + C$
At $t = 0$, $N = N_0$, so $C = \ln N_0$. Therefore: $$\boxed{N(t) = N_0 e^{-\lambda t}}$$ This is the radioactive decay law.
(c) Numerical:
$T_{1/2} = 5$ days, $t = 20$ days $= 4 \times T_{1/2}$
Activity after $n$ half-lives: $A = \dfrac{A_0}{2^n}$ $$A = \frac{6400}{2^4} = \frac{6400}{16} = \mathbf{400 \text{ dis/s}}$$
Assertion (A): Two atoms of different elements having the same mass number but different atomic numbers are called isobars.
Reason (R): Atomic number is the number of protons present and atomic mass is the total number of protons and neutrons present in a nucleus.
Reason (R): Atomic number is the number of protons present and atomic mass is the total number of protons and neutrons present in a nucleus.
Solution
Answer: Option (B) is correct.
Explanation: The assertion is true — isobars are atoms of different elements with the same mass number but different atomic numbers. The reason is also true — atomic number is the number of protons and atomic mass number is the total of protons and neutrons. However, the reason does not explain the assertion, so both are true but R is not the correct explanation of A.
Explanation: The assertion is true — isobars are atoms of different elements with the same mass number but different atomic numbers. The reason is also true — atomic number is the number of protons and atomic mass number is the total of protons and neutrons. However, the reason does not explain the assertion, so both are true but R is not the correct explanation of A.
Assertion (A): Nuclear force is same between neutron-neutron, proton-proton and neutron-proton.
Reason (R): Nuclear force is charge independent.
Reason (R): Nuclear force is charge independent.
Solution
Answer: Option (A) is correct.
Explanation: Nuclear force acts between all pairs of nucleons with equal strength — neutron-neutron, proton-proton, and neutron-proton. This is because nuclear force is charge independent. The reason directly and correctly explains the assertion.
Explanation: Nuclear force acts between all pairs of nucleons with equal strength — neutron-neutron, proton-proton, and neutron-proton. This is because nuclear force is charge independent. The reason directly and correctly explains the assertion.
Explore More CBSE Class 12 Physics Chapters
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Frequently Asked Questions
What does the Nuclei chapter cover in CBSE Class 12 Physics? ⌄
The Nuclei chapter in CBSE Class 12 covers the composition and size of nuclei, mass defect and binding energy, the binding energy per nucleon curve, nuclear fission and fusion, and radioactive decay including alpha, beta, and gamma decay. Students also learn about the radioactive decay law, half-life, and mean life — all of which are regularly tested in board exams.
How many marks does the Nuclei chapter carry in the CBSE Class 12 board exam? ⌄
The Nuclei chapter is part of Unit VIII — Atoms and Nuclei, which together carries 12 marks in the CBSE Class 12 Physics board paper. Questions from Nuclei typically appear as 1-mark MCQs, 2-mark short answers, 3-mark calculations, and occasionally 5-mark long-answer or case-study questions, making it a chapter that contributes across all question types.
What are the most important topics students should focus on in Class 12 Nuclei? ⌄
The most frequently tested topics are: nuclear density and its independence from mass number, binding energy per nucleon curve and its inferences, mass defect calculations, energy released in fission and fusion reactions, and the radioactive decay law with numerical problems on half-life and activity. Questions on alpha and beta decay reactions and Assertion-Reason type questions also appear regularly in board papers.
What common mistakes do students make when solving Nuclei questions? ⌄
A very common error is incorrectly applying the nuclear radius formula — students forget that $R = R_0 A^{1/3}$ depends on mass number, not atomic number. Another frequent mistake is confusing mass defect with mass number. Students also sometimes incorrectly conclude that nuclear density varies with mass number, when it is in fact constant. Practising a variety of previous year questions helps your child avoid these errors under exam pressure.
How does Angle Belearn help students score well in Nuclei? ⌄
Angle Belearn’s CBSE specialists carefully curate chapter-wise question banks drawn from real board papers, each paired with clear, step-by-step solutions. Students practising on Angle Belearn develop the habit of showing structured working — something that earns full marks in board exams. Regular practice with these verified questions builds both speed and accuracy so your child walks into the exam confident.
