CBSE Class 12 Physics Chapter 13: Nuclei Notes

Nuclear Size, Mass, and Constant Density

Unlike an atom, which has a fuzzy outer boundary, experimental evidence from Rutherford's scattering experiments indicates that the nucleus has a well-defined boundary.

The nuclear radius $R$ is found to be directly proportional to the cube root of its mass number $A$. Mathematically, this relation is expressed as $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15} \text{ m}$ (or $1.2 \text{ fm}$) is a constant representing the approximate size of a single nucleon.

Because the volume of a sphere is proportional to $R^3$, the nuclear volume is directly proportional to the mass number $A$ ($V \propto A$). Since nuclear mass is also approximately proportional to $A$, we can derive the nuclear density ($\rho$) as follows:
$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A \cdot m}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{3m}{4\pi R_0^3}$
Where $m$ is the average mass of a nucleon. Notice that the mass number $A$ cancels out completely. This mathematically demonstrates that nuclear density is constant and independent of the size or identity of the nucleus. This density is extraordinarily high, calculated to be approximately $2.3 \times 10^{17} \text{ kg/m}^3$ across all elements.

Must remember

• The atomic nucleus consists of protons and neutrons, collectively termed as nucleons. The atomic number ($Z$) represents protons, while mass number ($A$) represents total nucleons.
• Atomic Mass Unit (u) is defined as exactly 1/12th the mass of a carbon-12 atom. $1 \text{ u} = 1.660539 \times 10^{-27} \text{ kg}$.
• Einstein's Mass-Energy Equivalence: $E = \Delta m \cdot c^2$. A mass defect of $1 \text{ u}$ releases energy equivalent to $931.5 \text{ MeV}$.
• Mass defect ($\Delta m$) is the difference between the mass of individual constituent nucleons and the actual mass of the nucleus: $\Delta m = [Z \cdot m_p + (A - Z) \cdot m_n] - M_{\text{nucleus}}$.
• Binding Energy ($BE$) is the work required to break a nucleus into its constituent individual nucleons: $BE = \Delta m \cdot c^2 = \Delta m \times 931.5 \text{ MeV}$ (when $\Delta m$ is in u).
• Binding Energy per Nucleon ($BE/A$) determines nuclear stability. The higher the $BE/A$, the more stable the nucleus is. The peak stability is at Iron ($^{56}\text{Fe}$) with $BE/A \approx 8.75 \text{ MeV/nucleon}$.
• Nuclear Force is a strong, short-range (up to ~2 fm), attractive force that is non-central and charge-independent (meaning p-p, n-n, and p-n forces are approximately equal).

Glossary of Key Terms

Isotopes

Atoms of the same element having the same atomic number (Z) but different mass numbers (A), such as Protium ($^1_1\text{H}$), Deuterium ($^2_1\text{H}$), and Tritium ($^3_1\text{H}$).

Isobars

Nuclides having the same mass number (A) but different atomic numbers (Z), such as Carbon-14 ($^{14}_{\ 6}\text{C}$) and Nitrogen-14 ($^{14}_{\ 7}\text{N}$).

Isotones

Nuclides containing the exact same number of neutrons ($N = A - Z$), such as Silicon ($^{30}_{14}\text{Si}$) and Phosphorus ($^{31}_{15}\text{P}$).

Mass Defect

The difference between the total mass of the individual nucleons (protons and neutrons) and the rest mass of the bound nucleus.

Binding Energy

The energy equivalent of the mass defect, representing the energy released when a nucleus is assembled from its individual nucleons.

Nuclear Fission

The process of splitting a heavy, unstable nucleus (like Uranium-235) into lighter stable nuclei, accompanied by a massive release of energy.

Nuclear Fusion

The process where two light nuclei combine under extreme temperature and pressure to form a heavier, more stable nucleus (like Deuterium and Tritium combining into Helium).

Nuclear Fission vs Nuclear Fusion

Worked Revision Examples

• {"title":"Example 1: Mass Defect and Binding Energy Calculation","description":"Problem: Calculate the binding energy of an alpha particle ($^4_2\\text{He}$) nucleus. Given:\n- Mass of Helium nucleus = $4.001506 \\text{ u}$\n- Mass of Proton ($m_p$) = $1.007276 \\text{ u}$\n- Mass of Neutron ($m_n$) = $1.008665 \\text{ u}$\n\nSolution:\n1. Number of protons ($Z$) = 2, Number of neutrons ($N$) = $A - Z = 4 - 2 = 2$.\n2. Total mass of individual constituent nucleons:\n $M_{\\text{nucleons}} = 2(1.007276 \\text{ u}) + 2(1.008665 \\text{ u}) = 2.014552 + 2.017330 = 4.031882 \\text{ u}$\n3. Calculate mass defect ($\\Delta m$):\n $\\Delta m = M_{\\text{nucleons}} - M_{\\text{nucleus}} = 4.031882 \\text{ u} - 4.001506 \\text{ u} = 0.030376 \\text{ u}$\n4. Calculate Binding Energy ($BE$):\n $BE = \\Delta m \\times 931.5 \\text{ MeV} = 0.030376 \\times 931.5 \\text{ MeV} \\approx 28.3 \\text{ MeV}$\n\nResult: The Binding Energy of Helium nucleus is $28.3 \\text{ MeV}$."}
• {"title":"Example 2: Nuclear Radius and Volume Comparison","description":"Problem: Two nuclei have mass numbers in the ratio $1:27$. Find the ratio of their nuclear radii and nuclear densities.\n\nSolution:\n1. The nuclear radius is given by $R = R_0 A^{1/3}$.\n2. The ratio of radii is:\n $\\frac{R_1}{R_2} = \\left(\\frac{A_1}{A_2}\\right)^{1/3} = \\left(\\frac{1}{27}\\right)^{1/3} = \\frac{1}{3}$\n3. Since nuclear density is completely independent of mass number $A$, the density of both nuclei is exactly the same.\n\nResult: The ratio of nuclear radii is $1:3$, and the ratio of nuclear densities is $1:1$."}

Exam Traps & Board Preparation Cues

⚠️ The Stability Trap: Students often mistakenly assume that a higher total Binding Energy means greater stability. Always remember: stability depends strictly on Binding Energy per Nucleon ($BE/A$), not total $BE$! For example, Uranium has a very high total $BE$ but lower $BE/A$ than Iron, making Iron much more stable.

⚠️ Unit Pitfalls: If the question provides masses in atomic mass units (u), do not convert them to kilograms to use $E = mc^2$ with $c = 3 \times 10^8 \text{ m/s}$ unless asked. Simply calculate $\Delta m$ in 'u' and multiply directly by $931.5 \text{ MeV}$ to save valuable exam time and avoid calculation mistakes.

⚠️ Graph Accuracy: When sketching the $BE/A$ vs Mass Number graph, ensure the peak clearly points near Iron ($^{56}\text{Fe}$ at $8.75 \text{ MeV}$), shows a drop to $\approx 7.6 \text{ MeV}$ for Uranium ($^{238}\text{U}$), and contains sharp fluctuating peaks for stable light nuclei like $^4\text{He}$, $^{12}\text{C}$, and $^{16}\text{O}$.

Practice Questions with Solutions

• Why does the binding energy per nucleon decrease for very heavy nuclei (A > 170)? For very heavy nuclei, the repulsive electrostatic Coulomb force between the large number of protons starts to dominate over the short-range strong nuclear force, which decreases the overall stability and decreases the BE/A.
• What is the ratio of nuclear radii of two nuclei with mass numbers 8 and 125? Using $R \propto A^{1/3}$, the ratio is $R_1/R_2 = (8/125)^{1/3} = 2:5$.
• How do you explain the release of energy in nuclear fission using the binding energy curve? In the binding energy curve, very heavy nuclei have lower BE/A (~7.6 MeV) than intermediate nuclei (~8.5 MeV). When a heavy nucleus splits into lighter fragments, the products are more tightly bound, resulting in a release of excess binding energy.
• Is the nuclear force charge-dependent or charge-independent? The nuclear force is charge-independent. The attractive force between two protons, two neutrons, or a proton and a neutron is approximately equal, provided they are separated by the same distance.

Frequently Asked Questions