Chapter 13: Nuclei

If Chapter 12 zoomed into the atom and discovered the nucleus, Chapter 13 zooms further — into the nucleus itself. A speck $10^4$ times smaller than the atom contains $99.9\%$ of the mass and stores energies a million times those of chemical bonds. We will assemble the nucleus out of protons and neutrons, weigh it (and find it lighter than its parts), recognize the mass defect as binding energy via $E=mc^2$, explore the strong nuclear force that holds nucleons together, then trace the statistics of radioactive decay, and finally see how fission and fusion power both reactors and stars.

Concept Map

Nucleus = Z protons + N neutrons (A = Z + N)
        │
        ├── Size: R = R₀ A^(1/3) → density constant ~10¹⁷ kg/m³
        │
        ├── Mass defect Δm → Binding energy BE = Δm c²
        │       │
        │       └── BE/A curve (peaks ~⁵⁶Fe) → fusion (light) / fission (heavy)
        │
        ├── Nuclear force: short range, charge-independent, saturating
        │
        ├── Radioactivity: α, β⁻, β⁺, γ
        │       │
        │       └── Decay law: N = N₀ e^(−λt), T₁/₂ = ln2/λ, τ = 1/λ
        │
        └── Reactions: fission (²³⁵U + n → fragments + 2-3 n)
                       fusion (4¹H → ⁴He + 2e⁺ + 2ν + γ, solar p-p)

13.1 Composition of the Nucleus

Definition

A nucleus is specified by two integers:

• Atomic number $Z$: the number of protons. Determines chemical identity.
• Mass number $A$: the total number of nucleons (protons + neutrons). The neutron number is $N = A - Z$.

We write a nuclide as $^{A}_{Z}\text{X}$, e.g., $^{12}_{6}\text{C}$, $^{235}_{92}\text{U}$.

Pitfalls

• $Z$ identifies the element; $A$ identifies the isotope.
• The mass of a neutron ($1.00866\ \text{u}$) is slightly greater than that of a proton ($1.00728\ \text{u}$); a free neutron $\beta$-decays with half-life $\sim 10.2$ min.

13.2 Atomic Masses and Mass-Energy Equivalence

Definition

The unified atomic mass unit (u) is defined as one-twelfth of the mass of a neutral $^{12}$C atom in its ground state:

By Einstein's mass-energy equivalence $E = mc^2$,

So $1\ \text{u} \leftrightarrow 931.5\ \text{MeV}/c^2$. Some standard nuclear masses (in u):

Derivation — $E = mc^2$ unit conversion

Start with $1\ \text{u} = 1.66054\times 10^{-27}\ \text{kg}$. Then

Convert to eV via $1\ \text{eV} = 1.602\times 10^{-19}\ \text{J}$:

Worked Example

The mass difference between a neutron and an antiproton is $0.00138\ \text{u}$. Convert to MeV.

$0.00138\times 931.5 = 1.29\ \text{MeV}.$ This is also the energy released in the $\beta$-decay of a free neutron (approximately, neglecting neutrino mass).

Pitfalls

• $931.5\ \text{MeV}$ corresponds to one full u of mass, not to one nucleon's rest energy.
• Tabulated atomic masses include electron masses; when computing nuclear processes involving $\beta$-decay, you must keep careful track of electrons.

13.3 Size of the Nucleus and Nuclear Density

Definition

Empirically (from electron-scattering experiments), the radius of a nucleus of mass number $A$ obeys

This implies the nuclear volume is proportional to $A$, hence each nucleon occupies the same volume:

Derivation — nuclear density

The mass of a nucleus is $\approx A\,m_n$ (treating nucleons as equal mass). Density:

Computing:

This is independent of $A$ — a startling experimental fact, and the smoking gun of the saturation property of the nuclear force. (For comparison: water $10^3$, white-dwarf $10^9$, neutron-star $10^{17}$ — neutron stars are essentially gigantic atomic nuclei.)

Worked Example

Find the radius of $^{197}_{79}$Au.

Pitfalls

• $R\propto A^{1/3}$, not $A$ — otherwise density would not be constant.
• Nuclear "radius" is not sharp; $R$ here is the half-density radius.
• Atomic radius $\sim 10^{-10}\ \text{m}$ ≫ nuclear radius $\sim 10^{-15}\ \text{m}$: ratio $10^5$.

13.4 Mass Defect and Binding Energy

Definition

The mass of a nucleus is less than the sum of the masses of its constituent free nucleons. The difference

is called the mass defect. The corresponding energy

is the binding energy — the energy that would be required to disassemble the nucleus into free nucleons. Equivalently, it is the energy released when free nucleons are assembled into the nucleus.

In atomic-mass tables (which include $Z$ electrons):

because the $Z$ electron rest energies cancel (electron binding energies of $\sim$ eV are negligible against MeV nuclear scales).

Derivation — binding energy of $^{4}_{2}$He

Atomic masses: $M(^{1}\text{H}) = 1.00783\ \text{u}$, $m_n = 1.00866\ \text{u}$, $M(^{4}\text{He}) = 4.00260\ \text{u}$.

He-4 is exceptionally tightly bound (closed-shell magic nucleus $Z=N=2$), which is why α-decay is so energetically favourable.

Worked Example — BE/A for $^{56}$Fe

For iron-56, $BE \approx 492\ \text{MeV}$, so $BE/A \approx 8.79\ \text{MeV/nucleon}$ — near the peak of the BE/A curve.

The BE/A vs $A$ Curve — Features

Two practical consequences:

1. Fusion (small $A$ → larger $A$ toward $^{56}$Fe) releases energy.
2. Fission (large $A$ → mid-$A$) also releases energy.

Both processes "climb" the BE/A curve, since higher $BE/A$ means a more tightly-bound (lower-energy) state.

Pitfalls

• Mass defect $\Delta m$ is positive for bound nuclei; binding energy $BE > 0$.
• Don't confuse "binding energy" (positive) with "total nuclear energy" (negative, like atomic levels).
• $BE/A$, not $BE$, indicates stability per nucleon. $^{238}$U has more total $BE$ than $^{4}$He, but lower $BE/A$.

13.5 Nuclear Force

Definition

The force that binds nucleons inside a nucleus, overcoming Coulomb repulsion between protons. Properties:

1. Strongest known force (within its range): ~$10^2$ times stronger than electromagnetism.
2. Short range: significant only for separations $\lesssim 2\ \text{fm}$. Beyond $\sim 3\ \text{fm}$, it is effectively zero.
3. Repulsive core: at very small distances ($r \lesssim 0.4\ \text{fm}$) the force becomes strongly repulsive — preventing collapse.
4. Charge independent: $F(p,p) \approx F(p,n) \approx F(n,n)$ (Coulomb aside). Proton and neutron are isospin "twins" of one nucleon.
5. Saturating: each nucleon interacts only with its few nearest neighbours, not with all others. (Otherwise $BE\propto A^2$, not $\propto A$.)
6. Spin-dependent: stronger between aligned spins (deuteron is bound, di-neutron is not).
7. Non-central: it has a tensor component.

The nuclear force is now understood as a residual effect of the more fundamental strong (color) force between quarks, mediated by gluons. The Yukawa picture (pion exchange) gives the form

The range $\lambda$ equals the Compton wavelength of the exchanged pion — confirming the meson-exchange origin.

Pitfalls

• The nuclear force is not gravity; it is not Coulomb. It is its own thing.
• Charge-independence means the force is the same between $pp$, $pn$, $nn$ aside from Coulomb.
• Saturation explains why $BE/A$ is roughly constant on the plateau.

13.6 Radioactivity — α, β, γ

Definition

In 1896 Becquerel discovered that uranium spontaneously emits penetrating radiation. The Curies isolated polonium and radium. Rutherford identified three components in a magnetic field:

α-decay equation

Example: $^{238}_{92}\text{U}\to ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$.

Energy released $Q = [M(\text{X}) - M(\text{Y}) - M(\text{He})]\,c^2$ is shared between the α (kinetic energy) and the recoiling daughter:

β$^-$-decay equation

A neutron inside the nucleus converts to a proton, emitting an electron and an antineutrino:

Example: $^{14}_{6}\text{C}\to ^{14}_{7}\text{N} + e^{-} + \bar\nu_e$.

Because the antineutrino carries away variable energy, the β-spectrum is continuous up to a maximum $Q$-value (this puzzled physicists until Pauli postulated the neutrino in 1930).

β$^+$-decay equation (positron emission)

Example: $^{22}_{11}\text{Na}\to ^{22}_{10}\text{Ne} + e^{+} + \nu_e$.

γ-decay

A nucleus in an excited state drops to lower state by emitting a photon — no change in $A$ or $Z$:

γ-emission usually follows α or β decay, since the daughter is often left excited.

Pitfalls

• α is doubly positive ($+2e$), not $+e$.
• β$^-$ has charge $-e$, β$^+$ has charge $+e$.
• The antineutrino in β$^-$ decay was predicted from energy conservation, not observed until 1956.

13.7 Law of Radioactive Decay — Derivation

Definition

The number of undecayed nuclei in a sample decreases exponentially in time:

where $\lambda$ is the decay constant (units s$^{-1}$).

Derivation

Postulate: the probability per unit time that a given nucleus decays is a constant $\lambda$, independent of how old the nucleus is. (Radioactive nuclei have no memory — pure quantum statistics.)

If $N(t)$ nuclei remain at time $t$, the expected number decaying in $dt$ is $\lambda N\,dt$, so

Separate variables and integrate:

Half-Life $T_{1/2}$

Time for $N$ to fall to $N_0/2$:

Mean Life $\tau$

The average lifetime of a nucleus, $\tau = \langle t\rangle$, is computed by

So $\tau = 1/\lambda = T_{1/2}/\ln 2 = 1.44\,T_{1/2}$.

Activity $A(t)$

The number of decays per second:

Units: $1\ \text{becquerel (Bq)} = 1$ decay/s; $1\ \text{curie (Ci)} = 3.7\times 10^{10}\ \text{Bq}$.

Worked Example — half-life of $^{14}$C

For radiocarbon, $T_{1/2} = 5730\ \text{yr}$. So

Mean life $\tau = 1/\lambda = 2.6\times 10^{11}\ \text{s}\approx 8270\ \text{yr}$.

Worked Example — activity of $1\ \mu\text{g}$ of $^{14}$C

Number of $^{14}$C atoms: $N = \dfrac{10^{-6}\ \text{g}}{14\ \text{g/mol}}\cdot N_A = \dfrac{10^{-6}}{14}\cdot 6.02\times 10^{23} = 4.30\times 10^{16}$.

Activity: $A = \lambda N = 3.84\times 10^{-12}\cdot 4.30\times 10^{16} = 1.65\times 10^5\ \text{Bq}$.

Pitfalls

• $T_{1/2}\ne \tau$ — they differ by a factor of $\ln 2 \approx 0.693$. The mean life is longer than the half-life.
• Half-life is independent of temperature, pressure, chemistry. (Pure quantum tunneling for α; weak interaction for β.)
• After $n$ half-lives, the fraction remaining is $(1/2)^n$.

13.8 Decay Equations — Examples

α-decay examples

β$^-$-decay examples

(The excited $^{60}\text{Ni}^*$ then emits two γ-photons of $1.17$ and $1.33\ \text{MeV}$.)

Worked Example — Q of $^{226}$Ra decay

Given $M(^{226}\text{Ra}) = 226.0254\ \text{u}$, $M(^{222}\text{Rn}) = 222.0176\ \text{u}$, $M(^{4}\text{He}) = 4.0026\ \text{u}$. Compute $Q$.

Pitfalls

• The $Q$-value must be positive for spontaneous decay. If $\Delta m < 0$, the reaction does not occur.
• In β$^-$ decay, the kinetic energy is shared continuously between $e^-$ and $\bar\nu_e$ — only the maximum-energy electron carries the full $Q$.

13.9 Nuclear Fission

Definition

A heavy nucleus (e.g., $^{235}$U) absorbs a slow (thermal) neutron, becomes a highly excited compound nucleus $^{236}$U$^*$, which splits into two medium-mass fragments and 2–3 prompt neutrons, releasing $\sim 200\ \text{MeV}$.

Canonical example:

Energy budget: $Q\approx 200\ \text{MeV}$ per fission, distributed approximately as:

Energy from BE/A reasoning

BE/A of $^{235}$U is $\sim 7.6\ \text{MeV}$; of mid-$A$ fragments $\sim 8.5\ \text{MeV}$. Per nucleon gain $\sim 0.9\ \text{MeV}$, multiplied by $A=235$ nucleons gives $\sim 211\ \text{MeV}$ — consistent.

Chain Reaction

Each fission produces 2–3 neutrons, which can trigger further fissions in nearby $^{235}$U nuclei. The multiplication factor $k$ is the average number of these that go on to produce another fission:

• $k < 1$: subcritical — chain dies.
• $k = 1$: critical — steady-state (reactor).
• $k > 1$: supercritical — exponential growth (bomb).

Reactor (controlled chain reaction)

Key components:

1. Fuel: enriched UO$_2$ ($3$–$5\%\ ^{235}$U) or natural uranium.
2. Moderator: light/heavy water or graphite, to slow fast (~MeV) prompt neutrons to thermal ($0.025\ \text{eV}$) energies where the $^{235}$U fission cross-section is largest.
3. Control rods: cadmium or boron carbide, strong neutron absorbers, inserted to lower $k$.
4. Coolant: water, CO$_2$, or liquid sodium, transfers heat to the steam turbine.
5. Shielding: thick concrete + lead.

In Indian reactors (PHWRs at Tarapur, Kakrapar, Kaiga), heavy water D$_2$O serves as both moderator and coolant; natural-U fuel works because D$_2$O absorbs fewer neutrons than light water.

Worked Example — energy from 1 kg of $^{235}$U

Number of nuclei in 1 kg: $N = \dfrac{1000\ \text{g}}{235\ \text{g/mol}}\cdot N_A = 2.56\times 10^{24}$.

Total energy: $N\times 200\ \text{MeV} = 2.56\times 10^{24}\times 200\times 1.602\times 10^{-13}\ \text{J} = 8.2\times 10^{13}\ \text{J}\approx 23\ \text{GWh}$ — about $2.5\times 10^6$ kg of coal equivalent.

Pitfalls

• Fission requires slow neutrons for $^{235}$U; fast neutrons largely scatter elastically.
• $^{238}$U does not undergo thermal fission; it captures neutrons to breed $^{239}$Pu (fissile).
• Delayed neutrons (~$0.65\%$ of total, half-life seconds) are what make reactor control possible — without them, response time would be milliseconds.

13.10 Nuclear Fusion — Solar p-p Cycle

Definition

Light nuclei combine to form heavier ones, with mass deficit released as kinetic energy of products. Fusion drives stars.

The Sun's principal energy-producing reaction is the proton-proton (p-p) chain, in which four protons effectively combine to make one $^{4}$He nucleus:

The chain proceeds in three sub-steps:

1. $^{1}\text{H} + ^{1}\text{H}\to ^{2}_{1}\text{H} + e^{+} + \nu_e$ (slow, weak)
2. $^{2}_{1}\text{H} + ^{1}\text{H}\to ^{3}_{2}\text{He} + \gamma$ (fast)
3. $^{3}_{2}\text{He} + ^{3}_{2}\text{He}\to ^{4}_{2}\text{He} + 2\,^{1}\text{H}$ (fast)

The first step's slowness (it requires weak p $\to$ n conversion) is what makes the Sun shine slowly enough to support life — without this bottleneck, the Sun would burn out in $\sim 10^6$ years.

Conditions for fusion

Coulomb barrier between two protons at $r\sim 1\ \text{fm}$ is

Yet the Sun's core is "only" $\sim 1.5\times 10^7\ \text{K}$ ($\sim 1\ \text{keV}$). Fusion still proceeds thanks to:

• Quantum tunneling through the Coulomb barrier (Gamow factor).
• Maxwell-Boltzmann tail — a tiny fraction of protons in the high-energy tail succeed.
• Enormous density and lifetime — there are $\sim 10^{56}$ protons in the core, and $10^9$ years available.

For terrestrial fusion (D-T):

Required: temperature $\sim 10^8\ \text{K}$ (plasma), confinement time $\sim 1\ \text{s}$, density $\sim 10^{20}\ \text{m}^{-3}$ (Lawson criterion $n\tau \ge 10^{20}\ \text{s/m}^3$).

Worked Example — energy of solar fusion

Per p-p chain, $Q = 26.7\ \text{MeV}$. The Sun's luminosity is $L_\odot = 3.83\times 10^{26}\ \text{W}$. Rate of $^4$He production:

Mass burnt per second: $R\cdot 4 m_p \approx 6\times 10^{11}\ \text{kg/s}$. The Sun has $\sim 10^{30}\ \text{kg}$ of H, enough for $\sim 10^{10}\ \text{yr}$ — about 5 Gyr remaining.

Pitfalls

• Fusion releases more energy per unit mass than fission ($\sim 7\ \text{MeV/nucleon}$ vs $\sim 0.9\ \text{MeV/nucleon}$).
• Solar fusion is not D-T; it is proton-proton.
• The "ignition" temperatures quoted assume tunneling — purely classical barrier-crossing temperatures would be $10^4$ times higher.

Solved Problems

Problem 1 — Nuclear radius of $^{16}$O and $^{56}$Fe

$R(^{16}\text{O}) = 1.2\times 16^{1/3} = 1.2\times 2.52 = 3.02\ \text{fm}$.

$R(^{56}\text{Fe}) = 1.2\times 56^{1/3} = 1.2\times 3.83 = 4.59\ \text{fm}$.

Ratio: $4.59/3.02 = 1.52 = (56/16)^{1/3} = 3.5^{1/3}$ ✓.

Problem 2 — Binding energy per nucleon of $^{56}$Fe

$M(^{56}\text{Fe}) = 55.9349\ \text{u}$. Then

$BE = 0.52848 \times 931.5 = 492.3\ \text{MeV}$, $BE/A = 8.79\ \text{MeV/nucleon}$. ✓

Problem 3 — Half-life from activity

A sample's activity drops from $A_0 = 1000\ \text{Bq}$ to $A = 125\ \text{Bq}$ in $3\ \text{hours}$. Find $T_{1/2}$.

$A/A_0 = 1/8 = (1/2)^3$, so 3 half-lives elapsed in 3 hours. $T_{1/2} = 1\ \text{hour}$.

Problem 4 — Age via radiocarbon dating

A bone shows $^{14}$C activity that is $1/4$ of fresh-bone activity. Estimate its age.

$1/4 = (1/2)^2 \Rightarrow 2\ T_{1/2} = 2\times 5730 = 11460\ \text{yr}$.

Problem 5 — α-decay kinematics

In the decay $^{226}\text{Ra}\to ^{222}\text{Rn} + ^{4}\text{He}$, $Q = 4.87\ \text{MeV}$. Find KE of α.

By momentum conservation: $p_\alpha = p_{\text{Rn}}$. KE inversely proportional to mass:

Daughter recoil: $T_{\text{Rn}} = 0.086\ \text{MeV}$.

Problem 6 — Energy from D-T fusion

How much D-T fuel is needed to provide the energy of $1\ \text{kg}$ of $^{235}$U fission?

$^{235}$U fission: $200\ \text{MeV}$ per nucleus, total $8.2\times 10^{13}\ \text{J/kg}$.

D-T: $17.6\ \text{MeV}$ per fusion of $5\ \text{u}$ of fuel.

Per kg of D-T mixture: $N = \dfrac{1000}{5}\cdot N_A = 1.20\times 10^{26}$ fusions, energy $= 1.20\times 10^{26}\times 17.6\times 1.6\times 10^{-13} = 3.38\times 10^{14}\ \text{J/kg}$.

So $1\ \text{kg}$ of D-T releases $\sim 4\times$ the energy of $1\ \text{kg}$ of $^{235}$U.

Problem 7 — Number of decays per second

A $^{60}$Co source has activity $1.0\ \text{Ci}$. How many $^{60}$Co nuclei are present? ($T_{1/2} = 5.27\ \text{yr}$.)

$\lambda = 0.693/(5.27\cdot 3.15\times 10^7) = 4.17\times 10^{-9}\ \text{s}^{-1}$.

$A = 1\ \text{Ci} = 3.7\times 10^{10}\ \text{Bq}$, so $N = A/\lambda = 8.87\times 10^{18}$ nuclei.

JEE/NEET Edge Cases

1. Successive decays — Bateman equation. For $A\to B\to C$ with constants $\lambda_A, \lambda_B$:

2. Secular equilibrium. If $\lambda_A \ll \lambda_B$, eventually $\lambda_A N_A = \lambda_B N_B$ (activities equal).
3. Neutron-to-proton ratio for stability. Stable nuclei: $N\approx Z$ for light, $N/Z$ rises to $\sim 1.5$ at heavy ($^{208}$Pb). β-decay drives $N/Z\to$ stable.
4. Magic numbers $2, 8, 20, 28, 50, 82, 126$: extra-stable nuclei (closed nuclear shells). $^{208}_{82}$Pb is doubly magic.
5. Gamow factor. Penetration probability $\sim e^{-2\pi\eta}$ with $\eta = Z_1 Z_2 e^2/(4\pi\varepsilon_0\hbar v)$ — explains 27-orders-of-magnitude spread in α-decay lifetimes.
6. Energy from mass:
   • $1\ \text{u} = 931.5\ \text{MeV}$
   • $m_e c^2 = 0.511\ \text{MeV}$
   • $m_p c^2 = 938.3\ \text{MeV}$
   • $m_n c^2 = 939.6\ \text{MeV}$
7. In β-decay, mass calculation traps. If using atomic masses (which include $Z$ electrons), in β$^-$ decay the extra electron is automatically accounted; in β$^+$ decay you must subtract $2m_e c^2 = 1.022\ \text{MeV}$.
8. $Q$-value formulas:
   • α: $Q = [M_X - M_Y - M_{\text{He}}]c^2$ (atomic masses, all electrons cancel).
   • β$^-$: $Q = [M_X - M_Y]c^2$ (atomic masses; the additional electron $M_Y$ has accounts for the emitted $e^-$).
   • β$^+$: $Q = [M_X - M_Y - 2m_e]c^2$.
9. Curie/Becquerel conversions. $1\ \text{Ci} = 3.7\times 10^{10}\ \text{Bq}$. The "curie" originally was the activity of 1 g of $^{226}$Ra.
10. Reactor neutron economy. The "four-factor formula" $k_\infty = \eta f p \varepsilon$ — JEE rarely asks, but useful intuition.

Quick Recap

• Nucleus: $^{A}_{Z}\text{X}$, $A = Z + N$. Isotopes (same $Z$), isobars (same $A$), isotones (same $N$).
• $R = R_0 A^{1/3}$, $R_0 = 1.2\ \text{fm}$. Nuclear density constant $\sim 2.3\times 10^{17}\ \text{kg/m}^3$.
• Mass-energy: $1\ \text{u} = 931.5\ \text{MeV}/c^2$.
• Mass defect $\Delta m$, binding energy $BE = \Delta m\,c^2$. BE/A peaks near $^{56}$Fe at $\sim 8.8\ \text{MeV}$.
• Nuclear force: short-range, charge-independent, saturating, repulsive core.
• Decay law: $N = N_0 e^{-\lambda t}$, $T_{1/2} = 0.693/\lambda$, $\tau = 1/\lambda = 1.44\,T_{1/2}$, $A = \lambda N$.
• α: $^{A}_{Z}\text{X}\to ^{A-4}_{Z-2}\text{Y} + ^{4}_{2}$He. β$^-$: $n\to p + e^- + \bar\nu$. γ: photon, no change in $A, Z$.
• Fission of $^{235}$U: $\sim 200\ \text{MeV}$, chain reaction.
• Fusion (Sun's p-p): $4\,^{1}\text{H}\to ^{4}\text{He} + 2e^+ + 2\nu + 2\gamma$, $26.7\ \text{MeV}$.

Formula Sheet