Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Nuclei for CBSE Class 12 Physics, unlocking the massive energies hidden within the atom.
Key Concepts
Nuclear Physics
The mechanics of protons, neutrons, and unimaginable energy
1
Nuclear Size & Radius
Formula
The volume of a nucleus is directly proportional to its mass number ($A$). Therefore, the radius ($R$) scales with the cube root of $A$.
$$R = R_0 A^{1/3} \quad (R_0 \approx 1.2 \times 10^{-15} \text{ m})$$
2
Mass Defect ($\Delta m$)
Formula
The mass of a stable nucleus is always strictly less than the sum of the masses of its constituent protons and neutrons. This missing mass is the mass defect.
$$\Delta m = [Z m_p + (A-Z) m_n] – M$$
3
Binding Energy ($BE$)
Formula
The energy equivalent of the mass defect. It is the minimum energy required to completely disassemble a nucleus into its separate protons and neutrons.
$$BE = \Delta m \cdot c^2$$
$$\text{Shortcut: } BE = \Delta m (\text{in u}) \times 931.5 \text{ MeV}$$
4
Binding Energy per Nucleon ($BE/A$)
Formula
The total binding energy divided by the mass number. It is the true measure of nuclear stability. The higher the $BE/A$, the more tightly bound and stable the nucleus is.
$$\text{Stability} \propto \frac{BE}{A}$$
5
Strong Nuclear Force
Concept
The strongest force in nature, binding nucleons together. It is strictly short-range (operates only up to ~3 fm), rapidly becomes repulsive below 0.7 fm, and is completely independent of electric charge ($n-n$, $p-p$, and $n-p$ forces are roughly equal).
$$F_{\text{strong}} \gg F_{\text{electrostatic}} \gg F_{\text{gravity}}$$
6
Radioactive Decay Law
Formula
The rate of radioactive disintegration is directly proportional to the number of undecayed nuclei present at that instant. This leads to an exponential decay curve.
$$-\frac{dN}{dt} = \lambda N \implies N = N_0 e^{-\lambda t}$$
7
Half-Life ($T_{1/2}$) & Mean Life ($\tau$)
Formula
Half-life is the time required for half the active nuclei in a sample to decay. Mean life is the average lifespan of all nuclei in the sample.
$$T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda} \quad | \quad \tau = \frac{1}{\lambda}$$
8
Alpha ($\alpha$) and Beta ($\beta$) Decay
Case Formulas
$\alpha$-decay: Emits a Helium nucleus ($_{2}^{4}\text{He}$). Mass number drops by 4; atomic number by 2.
$\beta^-$-decay: A neutron turns into a proton, emitting an electron ($e^-$) and an antineutrino ($\bar{\nu}$). Atomic number increases by 1.
$\beta^-$-decay: A neutron turns into a proton, emitting an electron ($e^-$) and an antineutrino ($\bar{\nu}$). Atomic number increases by 1.
$$_{Z}^{A}\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\text{He} + Q$$
$$_{Z}^{A}\text{X} \rightarrow _{Z+1}^{A}\text{Y} + e^- + \bar{\nu} + Q$$
9
Nuclear Fission
Concept
The splitting of a heavy, unstable nucleus (like Uranium-235) into two intermediate-mass nuclei. It releases huge energy because the product nuclei have a higher Binding Energy per nucleon.
$$_{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow _{56}^{141}\text{Ba} + _{36}^{92}\text{Kr} + 3_{0}^{1}\text{n} + Q$$
10
Nuclear Fusion
Concept
The combining of two extremely light nuclei (like Hydrogen isotopes) to form a heavier nucleus. It requires millions of degrees of temperature to overcome electrostatic repulsion, which is why it powers the stars.
$$_{1}^{2}\text{H} + _{1}^{2}\text{H} \rightarrow _{2}^{3}\text{He} + n + Q$$
Concept Deep Dive
01
The Constant Density of Nuclei
Why the nucleus is like a drop of liquidUnlike atoms (which are mostly empty space), the nucleus is incredibly dense. How dense? If you calculate the density $\rho = \text{Mass} / \text{Volume}$ using $V = \frac{4}{3}\pi R^3$ and $R = R_0 A^{1/3}$, the mass number $A$ cancels out completely!
This means the nuclear density is a universal constant for all elements—from Carbon to Uranium. It is roughly $2.3 \times 10^{17} \text{ kg/m}^3$. A single teaspoon of nuclear matter would weigh over a billion tons on Earth!
This means the nuclear density is a universal constant for all elements—from Carbon to Uranium. It is roughly $2.3 \times 10^{17} \text{ kg/m}^3$. A single teaspoon of nuclear matter would weigh over a billion tons on Earth!
$$\rho = \frac{m A}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{3m}{4\pi R_0^3} = \text{Constant}$$
02
Pauli’s Neutrino Hypothesis
The ghost particle that saved physicsDuring Beta decay, scientists observed that electrons were emitted with varying kinetic energies, forming a continuous spectrum. This seemingly violated the Law of Conservation of Energy, as the energy released in a nuclear transition should be a fixed, discrete value.
To save the laws of physics, Wolfgang Pauli proposed a radical idea: a “ghost” particle was carrying away the missing energy. This particle had no charge, nearly zero mass, and barely interacted with matter. He called it the neutrino (little neutral one). Decades later, its existence was experimentally proven, proving that conservation laws hold true!
To save the laws of physics, Wolfgang Pauli proposed a radical idea: a “ghost” particle was carrying away the missing energy. This particle had no charge, nearly zero mass, and barely interacted with matter. He called it the neutrino (little neutral one). Decades later, its existence was experimentally proven, proving that conservation laws hold true!
Compare & Contrast
✗ Nuclear Fission
- Splitting of a heavy nucleus ($A > 170$).
- Can be triggered easily at room temperature by bombarding with slow thermal neutrons.
- Produces highly radioactive waste fragments.
- Energy released per nucleon is relatively small ($\sim 0.9 \text{ MeV}$).
- Used in current nuclear power plants.
✓ Nuclear Fusion
- Fusing of two very light nuclei ($A < 30$).
- Requires massive extreme temperatures ($>10^7 \text{ K}$) and high pressure to overcome Coulomb repulsion.
- Produces clean, non-radioactive products (like Helium).
- Energy released per nucleon is enormous ($\sim 6.5 \text{ MeV}$).
- Powers the Sun and hydrogen bombs.
Common Mistakes to Avoid
Mistake 1
Miscalculating the Q-Value: Q-value is the energy released in a reaction. It is $Q = (\text{Mass of Reactants} – \text{Mass of Products}) \times c^2$. A positive $Q$ means energy is released (Exothermic). Students frequently reverse this and do Products – Reactants, yielding a negative value and a wrong conclusion!
Mistake 2
Confusing Binding Energy with BE per Nucleon: Uranium has a massive Total Binding Energy because it has 235 nucleons. But Iron-56 has a much higher Binding Energy per Nucleon ($BE/A$). Iron is fundamentally more stable. Always check if the question is asking for Total BE or $BE/A$.
Mistake 3
Using the Wrong Mass in Mass Defect: The mass given in exam tables is usually Atomic Mass (nucleus + electrons), not Nuclear Mass. When doing $\beta$-decay or precise mass defect calculations, ensure you account for the mass of the electrons if the formula requires purely nuclear masses.
Exam Tips
Tip 1
The “n Half-Lives” Shortcut: To find the amount of radioactive substance left after a time $t$, use the formula $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$, where $n = \frac{t}{T_{1/2}}$ is the number of half-lives that have passed. This is far faster than using the exponential $e^{-\lambda t}$ equation.
Tip 2
Identifying the Missing Particle: In any nuclear reaction, both Mass Number ($A$) and Atomic Number ($Z$) must be conserved. If the left side has total $Z=92$ and $A=235$, the right side must sum to exactly the same. Use this to instantly deduce the $A$ and $Z$ of an unknown particle in a decay equation.
Expected Exam Questions
SQ
Board Pattern Questions
Class 12 · Nuclei · CBSE Exam1
A radioactive isotope has a half-life of 5 years. How long will it take for the activity to reduce to $3.125\%$ of its original value? [2 marks]
Answer
25 years
📝
Explanation
Using the shortcut: $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$.
Given remaining fraction is $3.125\% = \frac{3.125}{100} = \frac{1}{32}$.
$\frac{1}{32} = \left(\frac{1}{2}\right)^n \implies \left(\frac{1}{2}\right)^5 = \left(\frac{1}{2}\right)^n$.
Therefore, $n = 5$ half-lives have passed.
Total time $t = n \times T_{1/2} = 5 \times 5 = 25 \text{ years}$.
2
Calculate the binding energy per nucleon (in MeV) for an Iron nucleus ($_{26}^{56}\text{Fe}$).
(Given: Mass of $_{26}^{56}\text{Fe}$ = $55.9349 \text{ u}$, $m_p = 1.0078 \text{ u}$, $m_n = 1.0087 \text{ u}$, $1\text{ u} = 931.5 \text{ MeV}$) [3 marks]
(Given: Mass of $_{26}^{56}\text{Fe}$ = $55.9349 \text{ u}$, $m_p = 1.0078 \text{ u}$, $m_n = 1.0087 \text{ u}$, $1\text{ u} = 931.5 \text{ MeV}$) [3 marks]
Answer
$\approx 8.79 \text{ MeV/nucleon}$
📝
Explanation
1. Identify Nucleons: $Z = 26$ protons, $A – Z = 56 – 26 = 30$ neutrons.
2. Mass of Nucleons:
Mass of 26 protons $= 26 \times 1.0078 = 26.2028 \text{ u}$
Mass of 30 neutrons $= 30 \times 1.0087 = 30.2610 \text{ u}$
Total mass $= 56.4638 \text{ u}$.
3. Mass Defect ($\Delta m$): $56.4638 – 55.9349 = 0.5289 \text{ u}$.
4. Total Binding Energy: $BE = 0.5289 \times 931.5 = 492.67 \text{ MeV}$.
5. BE per Nucleon: $\frac{492.67}{56} \approx 8.79 \text{ MeV/nucleon}$.
(This confirms Iron-56 is at the peak of the stability curve!).
3
A nucleus with mass number $A=240$ and binding energy per nucleon of $7.6 \text{ MeV}$ splits into two fragments of $A=120$, each with a binding energy per nucleon of $8.5 \text{ MeV}$. Calculate the total energy released ($Q$) in this fission process. [3 marks]
Answer
$216 \text{ MeV}$
📝
Explanation
Energy released ($Q$) is the difference between the Total Binding Energy of the products and the reactants.
1. Initial Total BE: $240 \times 7.6 = 1824 \text{ MeV}$.
2. Final Total BE: The two fragments have a total of 240 nucleons.
Total final BE $= 240 \times 8.5 = 2040 \text{ MeV}$.
3. Energy Released: $Q = \text{Final BE} – \text{Initial BE}$
$Q = 2040 – 1824 = 216 \text{ MeV}$.
Concept Map
Nuclei connects to →
Modern Physics
Atoms (The core of the Bohr model)
Dual Nature (Energy-mass equivalence)
Thermodynamics (Stellar fusion)
Related Posts
- Wave Optics – Concept Booster | Class 12 Physics CBSE
- Semiconductor Electronics – Concept Booster | Class 12 Physics CBSE
- Ray Optics and Optical Instruments – Concept Booster | Class 12 Physics CBSE
- Nuclei – Concept Booster | Class 12 Physics CBSE
- Moving Charges & Magnetism – Concept Booster | Class 12 Physics CBSE
- Magnetism and Matter – Concept Booster | Class 12 Physics CBSE
- Electromagnetic Waves – Concept Booster | Class 12 Physics CBSE
- Electromagnetic Induction – Concept Booster | Class 12 Physics CBSE
- Electric Potential Energy & Potential – Concept Booster | Class 12 Physics CBSE
- Electric Charges and Fields – Concept Booster | Class 12 Physics CBSE
- Dual Nature of Radiation and Matter – Concept Booster | Class 12 Physics CBSE
- Current Electricity – Concept Booster | Class 12 Physics CBSE
- Capacitance – Concept Booster | Class 12 Physics CBSE
- Atoms – Concept Booster | Class 12 Physics CBSE
- Alternating Current – Concept Booster | Class 12 Physics CBSE
