Nuclei — Concepts, Formulas & Solved Numericals

The Nucleus: Tiny, Dense, and Absolutely Fascinating

The nucleus sits at the heart of every atom, occupying roughly $10^{-15}$ m — about $\frac{1}{100000}$ the size of the atom itself. Yet it contains more than 99.9% of the atom’s mass. That extreme density is what makes nuclear physics so different from everything else in Class 12.

In this chapter, we deal with two fundamental nuclear processes: radioactive decay (unstable nuclei breaking apart) and nuclear reactions (nuclei fusing or splitting). Both release enormous energy — far beyond anything chemical reactions can manage — because of how Einstein’s $E = mc^2$ plays out at the nuclear scale.

For NEET, this chapter typically contributes 2–3 questions. For JEE Main, expect 1–2 questions often involving numerical calculations on half-life, Q-value, or binding energy. CBSE boards usually ask one 5-mark question with a decay chain or a numerical on activity. Knowing where to focus your energy matters.

The key insight that ties everything together: nuclei are stable because of the strong nuclear force, which overcomes the electrostatic repulsion between protons. When that balance breaks down, the nucleus decays. When we force nuclei together or split them, we can extract the binding energy difference as usable energy.

Key Terms & Definitions

Atomic number (Z): Number of protons in the nucleus. Defines the element.

Mass number (A): Total number of protons + neutrons. Also called nucleon number.

Neutron number (N): $N = A - Z$ .

Nuclide: A specific nucleus characterized by both Z and A. Written as $^A_Z X$ . For example, $^{238}_{92}\text{U}$ is uranium-238.

Isotopes: Same Z, different A. Example: $^{235}_{92}\text{U}$ and $^{238}_{92}\text{U}$ are both uranium but have different numbers of neutrons.

Isobars: Same A, different Z. Example: $^{40}_{18}\text{Ar}$ and $^{40}_{20}\text{Ca}$ .

Isotones: Same N, different Z. Example: $^{13}_6\text{C}$ and $^{14}_7\text{N}$ (both have N = 7).

Nuclear radius: $R = R_0 A^{1/3}$ , where $R_0 = 1.2 \times 10^{-15}$ m = 1.2 fm.

R = R_0 A^{1/3}, \quad R_0 = 1.2 \times 10^{-15} \text{ m}

This gives nuclear volume $\propto A$ , meaning nuclear density is constant for all nuclei — a remarkable experimental fact.

Binding energy (BE): The energy needed to completely separate a nucleus into its individual protons and neutrons. Higher BE per nucleon = more stable nucleus.

Mass defect ( $\Delta m$ ): The difference between the sum of individual nucleon masses and the actual nuclear mass.

\Delta m = Z \cdot m_p + N \cdot m_n - M_{\text{nucleus}}

Binding energy per nucleon (BE/A): The graph of BE/A vs A is the single most important graph in this chapter. Know its shape cold.

Radioactivity: Spontaneous emission of radiation from unstable nuclei.

Activity (A): Number of decays per second. SI unit: Becquerel (Bq) = 1 decay/s. Older unit: Curie (Ci) = $3.7 \times 10^{10}$ Bq.

Half-life ( $T_{1/2}$ ): Time for half the nuclei in a sample to decay.

Q-value: Energy released in a nuclear reaction. If Q > 0, the reaction is exothermic (energy released). If Q < 0, the reaction requires energy input.

Core Concepts

1. Binding Energy and the BE/A Curve

The mass defect converts directly to binding energy via:

\text{BE} = \Delta m \cdot c^2

In practical units: $\text{BE (in MeV)} = \Delta m \text{ (in u)} \times 931.5 \text{ MeV/u}$

The BE/A vs A graph peaks around iron (Fe-56) at about 8.75 MeV/nucleon. This is why:

Fission (heavy nuclei like U-235 splitting) releases energy — products have higher BE/A than the parent.
Fusion (light nuclei like H combining) releases energy — products have higher BE/A than the reactants.

The BE/A curve question appears in NEET almost every year. Remember: light nuclei (A < 20) have low BE/A, it peaks near Fe, then slowly decreases for heavy nuclei. Carbon-12 has BE/A ≈ 7.68 MeV/nucleon.

2. Radioactive Decay Law

The number of undecayed nuclei decreases exponentially:

N(t) = N_0 e^{-\lambda t}

A(t) = A_0 e^{-\lambda t} = \lambda N(t)

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}

\tau = \frac{1}{\lambda} = \frac{T_{1/2}}{0.693}

where $\lambda$ = decay constant, $\tau$ = mean lifetime.

Important relationship: $\tau = 1.443 \times T_{1/2}$ . Mean lifetime is always greater than half-life.

After $n$ half-lives: $N = \frac{N_0}{2^n}$

3. Types of Radioactive Decay

Alpha ( $\alpha$ ) decay: Nucleus emits $^4_2\text{He}$ .

^A_Z X \rightarrow ^{A-4}_{Z-2}Y + ^4_2\text{He}

A decreases by 4, Z decreases by 2.

Beta-minus ( $\beta^-$ ) decay: A neutron converts to a proton. Emits electron + antineutrino.

^A_Z X \rightarrow ^A_{Z+1}Y + e^- + \bar{\nu}_e

A unchanged, Z increases by 1.

Beta-plus ( $\beta^+$ ) decay: A proton converts to a neutron. Emits positron + neutrino.

^A_Z X \rightarrow ^A_{Z-1}Y + e^+ + \nu_e

A unchanged, Z decreases by 1.

Gamma ( $\gamma$ ) decay: Nucleus in excited state drops to ground state, emitting a photon. No change in A or Z.

Students often confuse beta decay changing A. It doesn’t — A stays the same in both $\beta^-$ and $\beta^+$ decay. Only the Z changes.

4. Q-value Calculations

For any nuclear reaction $a + b \rightarrow c + d$ :

Q = (m_a + m_b - m_c - m_d) \times 931.5 \text{ MeV/u}

If Q > 0: exothermic (energy released, appears as KE of products). If Q < 0: endothermic (threshold energy required).

5. Nuclear Fission and Fusion

Fission: A heavy nucleus (like $^{235}_{92}\text{U}$ ) absorbs a neutron and splits into two medium-mass fragments + 2–3 neutrons + energy (~200 MeV per fission).

^{235}_{92}\text{U} + ^1_0n \rightarrow ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3^1_0n + \text{energy}

The released neutrons trigger more fissions — this is the chain reaction. Controlled in nuclear reactors, uncontrolled in nuclear bombs.

Fusion: Light nuclei combine at extremely high temperatures (10⁷ K — needed to overcome Coulomb repulsion). Powers the sun.

^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0n + 17.6 \text{ MeV}

Fusion releases more energy per unit mass than fission. The technical challenge is achieving the required temperature and confinement.

Solved Examples

Example 1 — Easy (CBSE Level)

Find the binding energy of $^{16}_8\text{O}$ . Given: mass of O-16 nucleus = 15.9949 u, $m_p$ = 1.00728 u, $m_n$ = 1.00866 u.

Step 1: Calculate mass defect.

\Delta m = 8(1.00728) + 8(1.00866) - 15.9949

= 8.05824 + 8.06928 - 15.9949 = 0.13262 \text{ u}

Step 2: Convert to MeV.

\text{BE} = 0.13262 \times 931.5 = 123.5 \text{ MeV}

Step 3: BE per nucleon.

\frac{\text{BE}}{A} = \frac{123.5}{16} = 7.72 \text{ MeV/nucleon}

This is a typical CBSE 3-mark numerical. Lay out each step clearly for full marks.

Example 2 — Medium (JEE Main / NEET Level)

The half-life of a radioactive sample is 30 minutes. What fraction of the sample remains after 2 hours?

2 hours = 120 minutes = $\frac{120}{30} = 4$ half-lives.

\frac{N}{N_0} = \frac{1}{2^4} = \frac{1}{16}

Answer: $\frac{1}{16}$ of the original sample remains.

Always convert time to number of half-lives first. Don’t jump into the exponential formula unless the time isn’t a clean multiple of $T_{1/2}$ .

Example 3 — Medium (NEET 2023 Pattern)

A radioactive nucleus X decays to Y with a half-life of 693 s. Find the decay constant and the activity of a 1 μg sample if the molar mass of X is 100 g/mol.

Step 1: Decay constant.

\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{693} = 10^{-3} \text{ s}^{-1}

Step 2: Number of nuclei in 1 μg.

N = \frac{10^{-6}}{100} \times 6.022 \times 10^{23} = 6.022 \times 10^{15} \text{ nuclei}

Step 3: Activity.

A = \lambda N = 10^{-3} \times 6.022 \times 10^{15} = 6.022 \times 10^{12} \text{ Bq}

Example 4 — Hard (JEE Main 2024 Shift 1 Pattern)

Find the Q-value of the alpha decay: $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^4_2\text{He}$ . Given atomic masses: U-238 = 238.05079 u, Th-234 = 234.04363 u, He-4 = 4.00260 u.

Why we use atomic masses: Atomic masses include electron masses. In alpha decay, both sides have the same number of electrons (92 on each side when we include the alpha’s 2 electrons), so electron masses cancel perfectly. We can use atomic masses directly.

Q = (m_U - m_{Th} - m_{He}) \times 931.5

= (238.05079 - 234.04363 - 4.00260) \times 931.5

= (0.00456) \times 931.5

= 4.25 \text{ MeV}

Since Q > 0, this decay is spontaneous. The energy appears as kinetic energy of the alpha particle and the recoiling Th nucleus.

In JEE Main 2024 Shift 1, a similar Q-value question appeared with beta decay. For $\beta^-$ decay using atomic masses: $Q = (m_X - m_Y) \times 931.5$ (electron masses cancel). For $\beta^+$ decay: $Q = (m_X - m_Y - 2m_e) \times 931.5$ — the extra $2m_e$ accounts for the positron creation. This distinction costs many students marks.

Example 5 — Hard (JEE Advanced Pattern)

A sample contains two radioactive isotopes A and B with half-lives 2 hours and 3 hours respectively. Initially, the ratio of number of atoms $N_A : N_B = 4 : 1$ . Find the time at which the activities of both isotopes are equal.

Let initial atoms be $4N_0$ and $N_0$ .

Activity of A: $A_A = \lambda_A \cdot 4N_0 e^{-\lambda_A t} = \frac{0.693}{2} \cdot 4N_0 e^{-\lambda_A t}$

Activity of B: $A_B = \lambda_B \cdot N_0 e^{-\lambda_B t} = \frac{0.693}{3} \cdot N_0 e^{-\lambda_B t}$

Setting $A_A = A_B$ :

\frac{0.693}{2} \cdot 4 e^{-\lambda_A t} = \frac{0.693}{3} \cdot e^{-\lambda_B t}

2 e^{-\lambda_A t} = \frac{1}{3} e^{-\lambda_B t}

6 = e^{(\lambda_A - \lambda_B)t} = e^{(0.3465 - 0.231)t} = e^{0.1155t}

t = \frac{\ln 6}{0.1155} = \frac{1.7918}{0.1155} \approx 15.5 \text{ hours}

Exam-Specific Tips

CBSE Class 12 Board Exams

The chapter carries around 6–8 marks in the board exam. The 5-mark question is typically:

A decay chain problem (find mass number and atomic number after a series of α and β decays)
A numerical on half-life or activity
A descriptive question on nuclear fission vs fusion

For decay chains: each α decay: A → A−4, Z → Z−2. Each β⁻ decay: A unchanged, Z → Z+1. Draw a table if there are multiple steps.

CBSE marking scheme is generous — partial marks for correct formula setup even if the numerical answer is wrong. Show your work.

JEE Main

JEE Main weightage for Nuclei: 1–2 questions per session, typically 4–8 marks. From 2021–2024, the most repeated topics are: (1) half-life and activity calculations, (2) binding energy and mass defect, (3) Q-value of alpha/beta decay. The BE/A curve conceptual question appears roughly once per year.

Focus on numerical speed. The half-life questions have clean numbers — if your answer isn’t coming out clean, recheck your arithmetic before the method.

NEET

NEET has 2–3 questions from this chapter. At least one is conceptual (BE/A curve, types of decay, properties of radiation). The numerical is usually straightforward — half-life or activity.

Know that:

Alpha particles have the lowest penetrating power but highest ionizing power.
Gamma rays have the highest penetrating power but lowest ionizing power.
Alpha particles are deflected least in a magnetic field (heaviest).

SAT (Physics Subject Test)

Nuclear physics is lightly tested. Focus on the conceptual understanding of half-life and the difference between fission and fusion.

Common Mistakes to Avoid

Mistake 1: Confusing atomic mass with nuclear mass. When calculating mass defect, the problem might give atomic masses (which include electron masses). For alpha and beta-minus decay Q-values using atomic masses, the electron masses cancel — but not for beta-plus decay. Always check what masses are given.

Mistake 2: Using $T_{1/2}$ as the time for complete decay. After one half-life, half remains. It never reaches exactly zero — it approaches zero asymptotically. Saying “the sample fully decays after 2 half-lives” is wrong and will cost you marks.

Mistake 3: Mixing up decay constant $\lambda$ and activity $A$ . Activity $A = \lambda N$ (in Bq). The decay constant $\lambda$ has units s⁻¹. Many students write $A = \lambda$ which makes no dimensional sense.

Mistake 4: Wrong unit conversion. 1 u = 931.5 MeV/c², NOT 931.5 MeV. The binding energy in MeV = $\Delta m$ (in u) × 931.5. Write this conversion line explicitly in board exams.

Mistake 5: Getting the BE/A curve shape wrong. Students often draw it as monotonically increasing. It’s not — it rises steeply for light nuclei, peaks near Fe-56, then gradually decreases for heavier nuclei. Iron is at the bottom of the “energy valley” — it’s the most stable nucleus.

Practice Questions

Q1. (CBSE Level) The half-life of $^{90}_{38}\text{Sr}$ is 28 years. What is the disintegration rate of 15 mg of this isotope? (Avogadro number = $6.023 \times 10^{23}$ )

N = (15 × 10⁻³ / 90) × 6.023 × 10²³ = 1.004 × 10²⁰ nuclei

λ = 0.693 / (28 × 3.156 × 10⁷) = 7.85 × 10⁻¹⁰ s⁻¹

Activity = λN = 7.85 × 10⁻¹⁰ × 1.004 × 10²⁰ = 7.88 × 10¹⁰ Bq

Q2. (NEET Level) In a radioactive decay, $^{232}_{90}\text{Th}$ undergoes 6 alpha decays and 4 beta-minus decays. What is the resulting nucleus?

Each α: A → A−4, Z → Z−2. Six α decays: A decreases by 24, Z decreases by 12.

Each β⁻: A unchanged, Z → Z+1. Four β⁻ decays: Z increases by 4.

Final: A = 232 − 24 = 208, Z = 90 − 12 + 4 = 82

Result: $^{208}_{82}\text{Pb}$ (Lead-208)

Q3. (JEE Main Level) Two radioactive elements X and Y have half-lives 3 hours and 6 hours respectively. If initially both have the same number of atoms, after how many hours will the ratio $N_X : N_Y = 1 : 8$ ?

$\frac{N_X}{N_Y} = \frac{N_0 / 2^{t/3}}{N_0 / 2^{t/6}} = 2^{t/6 - t/3} = 2^{-t/6}$

Setting this equal to 1/8 = 2⁻³:

$-t/6 = -3 \Rightarrow t = 18$ hours

Q4. (NEET Level) The binding energy of $^2_1\text{H}$ (deuteron) is 2.23 MeV. What is the mass defect in atomic mass units?

$\Delta m = \text{BE} / 931.5 = 2.23 / 931.5 = 0.002394$ u ≈ 0.0024 u

Q5. (JEE Main Level) A radioactive nucleus has a half-life of 1 year. After how many years will $\frac{7}{8}$ of a given sample have decayed?

If 7/8 has decayed, 1/8 remains.

$\frac{N}{N_0} = \frac{1}{8} = \frac{1}{2^3}$

So 3 half-lives have passed = 3 years.

Q6. (CBSE / NEET) In $\beta^+$ decay, the Q-value formula using atomic masses is different from $\beta^-$ decay. Derive why there’s an extra $2m_e c^2$ term.

In β⁺ decay: $^A_Z X \rightarrow ^A_{Z-1}Y + e^+ + \nu_e$

Using nuclear masses: $Q = (M_{nuc,X} - M_{nuc,Y} - m_e) c^2$

Atomic mass = nuclear mass + Z × $m_e$ (ignoring binding energy of electrons).

So $M_{nuc,X} = M_{at,X} - Z m_e$ and $M_{nuc,Y} = M_{at,Y} - (Z-1)m_e$

Substituting: $Q = [(M_{at,X} - Z m_e) - (M_{at,Y} - (Z-1)m_e) - m_e]c^2$ $= [M_{at,X} - M_{at,Y} - Z m_e + Z m_e - m_e - m_e]c^2$ $= (M_{at,X} - M_{at,Y} - 2m_e)c^2$

Hence the extra $2m_e c^2 \approx 1.022$ MeV must be subtracted.

Q7. (JEE Main 2023 Pattern) The activity of a radioactive sample decreases from $8000$ Bq to $1000$ Bq in 9 days. Calculate the half-life.

$\frac{A}{A_0} = \frac{1000}{8000} = \frac{1}{8} = \frac{1}{2^3}$

Three half-lives in 9 days → $T_{1/2} = 9/3 = \mathbf{3}$ days.

Q8. (JEE Advanced Level) The radius of $^{27}_{13}\text{Al}$ is 3.6 fm. Find the radius of $^{125}_{52}\text{Te}$ using the relation $R \propto A^{1/3}$ .

$\frac{R_{Te}}{R_{Al}} = \left(\frac{125}{27}\right)^{1/3} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}$

$R_{Te} = 3.6 \times \frac{5}{3} = \mathbf{6.0 \text{ fm}}$

FAQs

Why is nuclear density constant for all nuclei?

Because $R \propto A^{1/3}$ , nuclear volume $V \propto R^3 \propto A$ . Since mass $\propto A$ (roughly — each nucleon ≈ 1 u), density = mass/volume ≈ constant. Every nucleus, from hydrogen to uranium, has roughly the same density: about $2.3 \times 10^{17}$ kg/m³. That’s about 10¹⁴ times denser than water.

Why does the BE/A curve peak at iron?

Iron represents the most stable configuration of nucleons. Moving toward iron from either direction (by fusion for light nuclei, or fission for heavy nuclei) releases energy. This is why iron is called the “endpoint” of stellar nucleosynthesis — stars can’t extract energy by fusing iron.

What’s the difference between decay constant and half-life?

Decay constant $\lambda$ is the probability of a single nucleus decaying per unit time. Half-life $T_{1/2} = \ln 2 / \lambda$ is the time for half a large sample to decay. They measure the same thing from different angles. A large $\lambda$ means fast decay = short half-life.

Why does mean lifetime $\tau$ differ from half-life?

Half-life is a “median” (50% mark), while mean lifetime is the average time a nucleus survives. Because the exponential distribution has a long tail, the average is pulled higher: $\tau = T_{1/2}/\ln 2 \approx 1.44 \, T_{1/2}$ . NEET has asked this distinction as a conceptual question.

Can Q-value be negative?

Yes. A negative Q-value means the reaction is endothermic — energy must be supplied to make it happen. The minimum kinetic energy needed (threshold energy) is slightly more than |Q| because momentum must also be conserved. Threshold energy = $|Q|(1 + m_{\text{projectile}}/m_{\text{target}})$ .

Why is gamma radiation more penetrating than alpha?

Alpha particles ( $^4_2\text{He}$ ) are heavy, doubly charged, and interact strongly with matter — they transfer energy rapidly and stop within a few centimetres of air. Gamma rays are electromagnetic waves (photons) with no charge and no rest mass — they interact weakly and can penetrate several centimetres of lead. Penetrating power is inversely related to ionizing power.

In nuclear fission, why are exactly 2–3 neutrons released?

Heavy nuclei like U-235 have a higher neutron-to-proton ratio than stable mid-mass nuclei. When U-235 splits into two mid-mass fragments, those fragments have fewer neutrons than they need. The excess neutrons are released promptly. The exact number varies by specific reaction — it averages 2.47 for U-235 fission, but we round to “2–3” in practice.

What makes a nucleus radioactive — can we predict which isotopes decay?

Stability depends on the neutron-to-proton ratio and the balance between nuclear binding energy and Coulomb repulsion. Stable nuclei fall in a “valley of stability” on the N-Z plot. Nuclei outside this valley are unstable. Nuclei with Z > 83 (bismuth) are all unstable — above a certain proton count, no amount of extra neutrons can compensate for Coulomb repulsion.