Nuclei and Nuclear Energy notes: ISC Class 12 Physics

1. Composition of the nucleus; isotopes, isobars and isotones

A nucleus contains protons (charge +e) and neutrons (uncharged), together called nucleons. Atomic number Z = number of protons; neutron number N; mass number A = Z + N. A nuclide is written $^{A}_{Z}\text{X}$ .
Isotopes: same Z, different A and N (same element). Isobars: same A, different Z. Isotones: same N, different Z.
The protons repel one another electrically; the strong nuclear force (short-range, attractive, the same between any two nucleons) holds the nucleus together.
The unified atomic mass unit u is 1/12 of the mass of a neutral carbon-12 atom: 1 u = 1.6605 × 10⁻²⁷ kg.

Formulas

Mass number
$A = Z + N$

Key terms

Nucleon: A proton or a neutron inside a nucleus.
Isotopes: Nuclides with the same atomic number but different mass numbers.
Isobars: Nuclides with the same mass number but different atomic numbers.
Isotones: Nuclides with the same number of neutrons but different numbers of protons.
Unified atomic mass unit: One twelfth of the mass of a neutral carbon-12 atom, 1.6605 × 10⁻²⁷ kg.

2. Size and density of the nucleus

Nuclear radius: $R = R_0A^{1/3}$ with $R_0 \approx 1.2$ fm (1 fm = 10⁻¹⁵ m). Radii: about 1.9 fm for ⁴He, 4.8 fm for ⁶⁴Zn, about 7.4 fm for ²³⁸U; a nucleus is about 10⁵ times smaller across than its atom.
Volume $\dfrac{4}{3}\pi R^3 \propto A$ : every nucleon occupies the same volume.
Density $\rho = \dfrac{Am}{\frac{4}{3}\pi R_0^3A} = \dfrac{3m}{4\pi R_0^3}$ : independent of A. About 2.3 × 10¹⁷ kg m⁻³ (with R₀ = 1.2 fm and m = 1.67 × 10⁻²⁷ kg), some 10¹⁴ times the density of water.

R = R₀A⁽1/3), R₀ = 1.2 fm

R is proportional to A^(1/3): the volume of a nucleus is proportional to its number of nucleons.

nuclear density, the same for every A

Nuclear density does not depend on A: about 2.3 × 10¹⁷ kg m⁻³ for every nucleus.

Formulas

Nuclear radius
$R = R_0A^{1/3}$
Nuclear density, independent of A
$\rho = \dfrac{3m}{4\pi R_0^3}$

Key terms

Nuclear density: The mass per unit volume of nuclear matter; about 2.3 × 10¹⁷ kg m⁻³, the same for every nucleus.

3. Mass–energy relation, mass defect and binding energy

Mass–energy relation: a mass m is equivalent to energy $E = mc^2$ . 1 u = 931.5 MeV/c² (this value is used throughout; 1 MeV = 1.6 × 10⁻¹³ J).
Mass defect: $\Delta m = Zm_p + (A - Z)m_n - m_{\text{nucleus}}$ . With atomic masses: $\Delta m = Zm_H + (A - Z)m_n - M_{\text{atom}}$ (the electron masses cancel).
Binding energy: $B = \Delta m\,c^2$ = the energy needed to separate the nucleus into free nucleons = the energy released when it forms. $B$ (MeV) $= \Delta m$ (u) $\times\ 931.5$ . Binding energy per nucleon $= B/A$ .
Example: ²⁰Ne (19.992440 u): Δm = 0.17246 u, B = 160.6 MeV, B/A = 8.03 MeV.

Formulas

Mass–energy relation
$E = mc^2$
Mass defect from atomic masses
$\Delta m = Zm_H + (A - Z)m_n - M$
Binding energy, Δm in u
$B = \Delta m \times 931.5\ \text{MeV}$

Key terms

Mass defect: The amount by which the mass of a nucleus is less than the total mass of its separate nucleons.
Binding energy: The energy equivalent of the mass defect: the energy needed to separate a nucleus into its nucleons.

4. Pair production and annihilation

Pair production: a gamma photon of energy $h\nu \ge 2m_ec^2 = 1.022$ MeV becomes an electron and a positron, in the field of a nucleus (which takes up momentum). Energy above 1.022 MeV is the kinetic energy of the pair.
Annihilation: an electron and a positron meeting (at rest) change completely into two photons of 0.511 MeV each, moving in opposite directions (so that momentum is conserved); wavelength $\lambda = hc/E = 2.43$ pm.
Both processes are direct evidence that mass and energy are equivalent.

Formulas

Threshold for pair production
$h\nu \ge 2m_ec^2 = 1.022\ \text{MeV}$

Key terms

Pair production: The change of a gamma photon into an electron and a positron near a nucleus; it needs at least 1.022 MeV.
Annihilation: The change of an electron and a positron into radiation (two photons of 0.511 MeV when at rest).

5. Binding energy per nucleon curve; stability; why fission and fusion release energy

B/A rises steeply for light nuclei (1.1 MeV for ²H; 7.07 MeV for ⁴He), has a maximum of about 8.8 MeV near A = 56–62 (iron, nickel), then falls slowly to about 7.6 MeV for uranium. Between A about 30 and 170 it is nearly constant.
The middle-order nuclei are the most tightly bound and the most stable. The short-range strong force explains the flat part; the long-range repulsion of the protons explains the fall at large A.
Both fission and fusion move nuclei towards the maximum: heavy nuclei split into middle-order fragments, light nuclei join into heavier ones. Higher B/A means the total binding energy increases, and the difference is released.
Estimate for fission: energy released = A × (B/A of the fragments − B/A of the heavy nucleus). For A = 236 splitting into two halves with 7.6 and 8.5 MeV per nucleon: 236 × 0.9 ≈ 212 MeV.

Formulas

Energy released from binding energies
$Q = \left(\sum B\right)_{\text{after}} - \left(\sum B\right)_{\text{before}}$

Key terms

Middle-order nuclei: Nuclei with mass numbers of about 50 to 80, with the highest binding energy per nucleon (about 8.8 MeV).

6. Nuclear fission, chain reaction, reactor and bomb

Fission: a heavy nucleus (such as ²³⁵U) absorbs a slow neutron and splits into two middle-order fragments, with 2–3 neutrons and about 200 MeV. Example: ¹₀n + ²³⁵₉₂U → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n, Q = 173.3 MeV (A and Z balance).
Chain reaction: the neutrons cause further fissions. Uncontrolled (more than one further fission each): explosive, a nuclear bomb. Controlled (one on average): steady, a reactor.
Reactor: fuel (enriched uranium); moderator (heavy water, graphite) slows neutrons so they cause fission; control rods (cadmium, boron) absorb neutrons, moved in or out to regulate the rate; coolant carries the heat to a heat exchanger and turbine; shielding (concrete, steel) absorbs neutrons and gamma rays.
A reactor cannot explode like a bomb: its fuel is only slightly enriched and the control rods keep the chain reaction controlled.
Not examined (removed in ISC): details of the chain reaction and quantitative reactor operation.

Formulas

A fission reaction
$^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n} + Q$

Key terms

Nuclear fission: The splitting of a heavy nucleus into two middle-order nuclei, with the release of neutrons and a large amount of energy.
Moderator: A material of light nuclei (such as heavy water or graphite) that slows fast neutrons so they can cause fission.
Control rods: Rods of a neutron absorber (cadmium, boron) moved in or out of the core to regulate the chain reaction.

7. Nuclear fusion; the Sun and stars; the hydrogen bomb

Fusion: two light nuclei join into a heavier one, raising B/A and releasing energy. Example: ²₁H + ³₂He → ⁴₂He + ¹₁H, Q = 18.35 MeV.
High temperature: both nuclei are positive and repel; to get within about 10⁻¹⁵ m they must climb the Coulomb barrier (0.60 MeV for two touching protons), so the gas must be at millions of kelvin: thermonuclear reactions. At 1.5 × 10⁷ K the mean energy is only about 2 keV; fusion proceeds because the fastest nuclei in the hot gas have far more than the mean.
Sun and stars: the proton–proton chain turns four hydrogen nuclei into one helium nucleus, releasing about 26.7 MeV and keeping the star shining for billions of years.
Hydrogen bomb: a fission bomb supplies the temperature to start uncontrolled fusion of hydrogen isotopes.

Formulas

Energy released in a reaction, masses in u
$Q = (m_{\text{reactants}} - m_{\text{products}}) \times 931.5\ \text{MeV}$

Key terms

Thermonuclear energy: Energy released by fusion reactions in a gas at a temperature of millions of kelvin, as in the Sun and stars.
Coulomb barrier: The electrostatic repulsion between two nuclei, which they must overcome to come within the range of the strong force.

Exam tips

Use atomic masses with $m_H$ and keep six decimals; convert Δm to MeV with 931.5 only at the end.
Show that A cancels when proving constant nuclear density.
Balance A and Z in every nuclear equation; Q = (mass lost) × 931.5 MeV.
Explain "why both release energy" with the maximum of the B/A curve and the direction in which each type of nucleus moves.
Each reactor part gets one function; fusion needs a high temperature because of the Coulomb repulsion.

1. Composition of the nucleus; isotopes, isobars and isotones

Formulas

Key terms

2. Size and density of the nucleus

Formulas

Key terms

3. Mass–energy relation, mass defect and binding energy

Formulas

Key terms

4. Pair production and annihilation

Formulas

Key terms

5. Binding energy per nucleon curve; stability; why fission and fusion release energy

Formulas

Key terms

6. Nuclear fission, chain reaction, reactor and bomb

Formulas

Key terms

7. Nuclear fusion; the Sun and stars; the hydrogen bomb

Formulas

Key terms

Exam tips

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