Nuclear Physics

Info: Board Coverage AQA Paper 2 | Edexcel CP6 | OCR (A) Paper 2 | CIE P4

In 1911, Geiger and Marsden (under Rutherford’s direction) fired alpha particles at a thin gold foil. Most passed straight through, some were deflected through small angles, and a few ($\sim 1$ in 8000) Were deflected through angles greater than $90^\circ$.

The results were incompatible with Thomson’s “plum pudding” model (in which positive charge is Diffusely spread through the atom). A diffuse charge distribution could not produce the large-angle Deflections observed.

Rutherford proposed that all positive charge and nearly all mass are concentrated in a tiny, dense nucleus. The large-angle deflections occur when an alpha particle approaches a nucleus head-on and Is repelled by the Coulomb force.

For a head-on collision, the alpha particle momentarily stops (all kinetic energy converted to Electric potential energy):

$\frac{1}{2}m_\alpha v^2 = \frac{Z_{\mathrm{Au}} \cdot 2e^2}{4\pi\varepsilon_0 d}$

$\boxed{d = \frac{Z_{\mathrm{Au}} e^2}{2\pi\varepsilon_0 E_k}}$

Where $d$ is the distance of closest approach. For 5.5 MeV alpha particles on gold ($Z = 79$):

$d = \frac{79 \times (1.60 \times 10^{-19})^2}{2\pi \times 8.85 \times 10^{-12} \times 5.5 \times 10^6 \times 1.60 \times 10^{-19}} \approx 4.1 \times 10^{-14}\ \mathrm{m}$

This gives an upper bound on the nuclear radius of gold ($\sim 10^{-14}$ m, compared to the atomic Radius of $\sim 10^{-10}$ m).

## 2. Nuclear Structure

The nucleus contains protons and neutrons (collectively, nucleons).

Property	Proton	Neutron	Electron
Charge	$+e$	$0$	$-e$
Mass (u)	1.00728	1.00867	0.00055
Location	Nucleus	Nucleus	Electron shells

Notation. A nuclide $\prescript{A}{Z}\mathrm{X}$ has mass number $A$ (total nucleons) and Atomic number $Z$ (protons). The neutron number is $N = A - Z$.

Isotopes have the same $Z$ but different $N$ (hence different $A$). Isotopes have nearly Identical chemical properties but different nuclear properties (stability, half-life, decay mode).

Isotones have the same $N$ but different $Z$. Isobars have the same $A$ but different $Z$.

The mass of a nucleus is less than the sum of the masses of its constituent nucleons. The Difference is the mass defect:

$\boxed{\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}}}$

$\boxed{E = mc^2}$

The mass defect corresponds to the binding energy — the energy released when the nucleus was formed From its constituent nucleons, or equivalently, the energy required to separate the nucleus into its Individual nucleons.

$\boxed{E_b = \Delta m\,c^2}$

The binding energy per nucleon is a measure of nuclear stability:

$\boxed{\frac{E_b}{A} = \frac{\Delta m\,c^2}{A}}$

Worked Example: Binding Energy of Helium-4

Calculate the binding energy per nucleon of $\prescript{4}{2}\mathrm{He}$. Given: $m_p = 1.00728$ u, $m_n = 1.00867$ u, $m_{\mathrm{He}} = 4.00151$ u, $1\ \mathrm{u} = 931.5$ MeV/c$^2$.

Answer. $\Delta m = 2 \times 1.00728 + 2 \times 1.00867 - 4.00151 = 2.01456 + 2.01734 - 4.00151 = 0.03039$ u.

$E_b = 0.03039 \times 931.5 = 28.3$ MeV.

$E_b/A = 28.3/4 = 7.08$ MeV per nucleon.

The binding energy per nucleon plotted against mass number shows:

• Light nuclei ($A \lt 20$): Low binding energy per nucleon, with peaks at $\prescript{4}{2}\mathrm{He}$, $\prescript{12}{6}\mathrm{C}$And $\prescript{16}{8}\mathrm{O}$ (magic numbers).
• Iron-56 ($\prescript{56}{26}\mathrm{Fe}$): Maximum binding energy per nucleon ($\sim 8.8$ MeV) — the most stable nucleus.
• Heavy nuclei ($A \gt 60$): Gradually decreasing binding energy per nucleon.

Implications:

• Fission of heavy nuclei ($A \gt 56$) releases energy because the products have higher binding energy per nucleon (the mass defect per nucleon increases).
• Fusion of light nuclei ($A \lt 56$) releases energy for the same reason.

Stable nuclei cluster around $N \approx Z$ for light nuclei, shifting to $N \gt Z$ for heavier Nuclei. The excess neutrons in heavy nuclei provide additional strong nuclear force to counteract the Increasing Coulomb repulsion between protons.

Why not all-neutron nuclei? The Pauli exclusion principle forces neutrons into progressively Higher energy states. Adding protons allows nucleons to occupy lower-energy states, reducing the Total energy. For light nuclei, the balance favours $N \approx Z$.

Condition	Dominant decay	Reason
$N/Z$ too large	$\beta^-$ decay	Neutron converts to proton
$N/Z$ too small	$\beta^+$ decay or electron capture	Proton converts to neutron
$A \gt 150$	Alpha decay	Reduces both $N$ and $Z$
Excited state	Gamma decay	Releases excess energy

Alpha decay occurs predominantly for $A \gt 150$ because the alpha particle is exceptionally Stable (high binding energy per nucleon of 7.08 MeV), making it energetically favourable to emit.

Nuclei with $Z$ or $N$ equal to 2, 8, 20, 28, 50, 82, or 126 are unusually stable, analogous to Noble gas electron configurations. These “magic numbers” arise from the shell structure of the Nucleus, predicted by the nuclear shell model (Mayer and Jensen, 1949).

An alpha particle ($\prescript{4}{2}\alpha = \prescript{4}{2}\mathrm{He}$) is emitted:

$\prescript{A}{Z}\mathrm{X} \to \prescript{A-4}{Z-2}\mathrm{Y} + \prescript{4}{2}\alpha$

Conservation: $A$ decreases by 4, $Z$ decreases by 2. Highly ionising, stopped by paper.

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

$\prescript{A}{Z}\mathrm{X} \to \prescript{A}{Z+1}\mathrm{Y} + \prescript{0}{-1}\beta^- + \bar{\nu}_e$

Conservation: $A$ unchanged, $Z$ increases by 1. The antineutrino was postulated (Pauli, 1930; Fermi, 1934) to conserve energy and momentum — the continuous electron energy spectrum requires a Third particle to carry away the remaining energy.

A proton converts to a neutron, emitting a positron and a neutrino:

$\prescript{A}{Z}\mathrm{X} \to \prescript{A}{Z-1}\mathrm{Y} + \prescript{0}{+1}\beta^+ + \nu_e$

This requires $m_{\mathrm{parent}} \gt m_{\mathrm{daughter}} + 2m_e$ (the positron mass must be Created).

Excited nucleus de-excites by emitting a high-energy photon:

$\prescript{A}{Z}\mathrm{X}^* \to \prescript{A}{Z}\mathrm{X} + \gamma$

No change in $A$ or $Z$. Weakly ionising, highly penetrating (requires thick lead or concrete).

## 6. Exponential Decay Law and Half-Life

The decay constant $\lambda$ is the probability per unit time that a single nucleus will decay. If There are $N$ nuclei:

$\frac{dN}{dt} = -\lambda N$

Separating variables and integrating from $N_0$ at $t = 0$ to $N$ at time $t$:

$\int_{N_0}^{N}\frac{dN'}{N'} = -\int_0^t \lambda\,dt'$

$\ln\left(\frac{N}{N_0}\right) = -\lambda t$

$\boxed{N = N_0 e^{-\lambda t}}$

$\boxed{A = \lambda N = -\frac{dN}{dt}}$

SI unit: becquerel (Bq). $1\ \mathrm{Bq} = 1\ \mathrm{decay\,s}^{-1}$.

Setting $N = N_0/2$ at $t = t_{1/2}$:

$\frac{1}{2} = e^{-\lambda t_{1/2}} \implies t_{1/2} = \frac{\ln 2}{\lambda}$

$\boxed{t_{1/2} = \frac{\ln 2}{\lambda}}$

A heavy nucleus ( $\prescript{235}_{92}\mathrm{U}$ or $\prescript{239}_{94}\mathrm{Pu}$) Absorbs a neutron, becoming unstable and splitting into two lighter nuclei (fission fragments) plus 2—3 neutrons and energy:

$\prescript{235}_{92}\mathrm{U} + \prescript{1}_{0}\mathrm{n} \to \prescript{236}_{92}\mathrm{U}^* \to \prescript{141}_{56}\mathrm{Ba} + \prescript{92}_{36}\mathrm{Kr} + 3\prescript{1}_{0}\mathrm{n} + \mathrm{energy}$

The binding energy per nucleon of the products ($\sim 8.5$ MeV) exceeds that of $\prescript{235}_{92}\mathrm{U}$ ($\sim 7.6$ MeV). The energy released per fission event is approximately 200 MeV, primarily as kinetic energy of the fission fragments.

Each fission event releases 2—3 neutrons, which can induce further fission events. For a self-sustaining chain reaction, the reproduction factor $k$ (average neutrons per fission that Cause another fission) must equal 1.

• $k \lt 1$: subcritical (reaction dies out).
• $k = 1$: critical (steady reaction — nuclear reactor).
• $k \gt 1$: supercritical (exponential growth — nuclear weapon).

Critical mass: The minimum mass of fissile material required to sustain a chain reaction. For $\prescript{235}_{92}\mathrm{U}$This is approximately 50 kg (sphere). The critical mass depends On geometry, density, and the presence of a neutron reflector.

Key components:

• Fuel rods: Enriched uranium ($\sim 3$—$5\%$\ $\prescript{235}_{92}\mathrm{U}$).
• Moderator: Graphite or heavy water — slows neutrons to thermal energies where the fission cross-section of $\prescript{235}_{92}\mathrm{U}$ is largest.
• Control rods: Boron or cadmium — absorb neutrons to regulate $k$.
• Coolant: Water, liquid sodium, or CO$_2$ — transfers heat from the reactor to the turbines.

Two light nuclei combine to form a heavier nucleus, releasing energy when the product has higher Binding energy per nucleon.

The positively charged nuclei must overcome their Coulomb repulsion to get within range of the strong Nuclear force ($\sim 10^{-15}$ m). This requires:

1. Very high temperatures ($\sim 10^8$ K) to give nuclei sufficient kinetic energy.
2. Very high densities to ensure sufficient collision rates.
3. Sufficient confinement time for reactions to occur.

The product of these three quantities is the Lawson criterion:

$n\tau \gt 10^{20}\ \mathrm{s\,m^{-3}}$

For deuterium—tritium fusion.

In the Sun’s core ($T \approx 1.5 \times 10^7$ K), hydrogen fuses to helium via the proton—proton Chain:

$4\prescript{1}{1}\mathrm{H} \to \prescript{4}{2}\mathrm{He} + 2\prescript{0}{+1}\beta^+ + 2\nu_e + 2\gamma$

Net energy release: $\sim 26.7$ MeV per helium-4 nucleus formed.

Achieving and confining a plasma at $10^8$ K is extraordinarily difficult. Two main approaches:

• Magnetic confinement (tokamak): Strong magnetic fields confine the plasma in a toroidal chamber. ITER is the largest current project.
• Inertial confinement: Laser pulses compress and heat a fuel pellet to fusion conditions (National Ignition Facility).

Worked Example: Energy from Fission

Calculate the energy released when a $\prescript{235}_{92}\mathrm{U}$ nucleus undergoes fission. Given: $m(\prescript{235}_{92}\mathrm{U}) = 235.044$ u, $m(\prescript{141}_{56}\mathrm{Ba}) = 140.914$ u, $m(\prescript{92}_{36}\mathrm{Kr}) = 91.926$ u, $m(\prescript{1}_{0}\mathrm{n}) = 1.00867$ u.

Answer. Mass of products: $140.914 + 91.926 + 3 \times 1.00867 = 235.866$ u.

Mass defect: $\Delta m = 235.044 + 1.00867 - 235.866 = 0.187$ u.

Energy released: $E = 0.187 \times 931.5 = 174$ MeV.

Problem 1

Calculate the distance of closest approach for a 7.7 MeV alpha particle scattered by a gold nucleus ($Z = 79$).

Answer. $d = \frac{Ze^2}{2\pi\varepsilon_0 E_k} = \frac{79 \times (1.60 \times 10^{-19})^2}{2\pi \times 8.85 \times 10^{-12} \times 7.7 \times 10^6 \times 1.60 \times 10^{-19}} = \frac{79 \times 1.60 \times 10^{-19}}{2\pi \times 8.85 \times 10^{-12} \times 7.7 \times 10^6} = 2.9 \times 10^{-14}$ m.

Problem 2

Calculate the binding energy per nucleon of $\prescript{56}_{26}\mathrm{Fe}$. Given: $m(\prescript{56}_{26}\mathrm{Fe}) = 55.9349$ u.

Answer. $\Delta m = 26 \times 1.00728 + 30 \times 1.00867 - 55.9349 = 26.1893 + 30.2601 - 55.9349 = 0.5145$ u.

$E_b = 0.5145 \times 931.5 = 479.3$ MeV. $E_b/A = 479.3/56 = 8.56$ MeV/nucleon.

Problem 3

Write the balanced equation for the beta-minus decay of $\prescript{14}{6}\mathrm{C}$.

Answer. $\prescript{14}{6}\mathrm{C} \to \prescript{14}{7}\mathrm{N} + \prescript{0}{-1}\beta^- + \bar{\nu}_e$.

Check: $A$: $14 = 14 + 0 + 0$. $Z$: $6 = 7 + (-1) + 0$. Both conserved.

Problem 4

A sample has activity 400 Bq and half-life 5.0 hours. Calculate the activity after 20 hours and the Number of nuclei present initially.

Answer. After 20 hours: $n = 20/5 = 4$ half-lives. $A = 400/2^4 = 25$ Bq.

$A_0 = \lambda N_0$, $\lambda = \ln 2/t_{1/2} = 0.693/18000 = 3.85 \times 10^{-5}$ s$^{-1}$. $N_0 = A_0/\lambda = 400/(3.85 \times 10^{-5}) = 1.04 \times 10^7$ nuclei.

Problem 5

Explain why energy is released in both nuclear fission and nuclear fusion, using the binding energy Curve.

Answer. The binding energy per nucleon curve has a maximum near $A = 56$ (iron). Fission splits Heavy nuclei ($A \gt 56$) into lighter fragments with higher binding energy per nucleon, so energy is Released. Fusion combines light nuclei ($A \lt 56$) into heavier products with higher binding energy Per nucleon, also releasing energy. In both cases, the products are closer to the peak of the curve Than the reactants, meaning mass is converted to energy via $E = \Delta m\,c^2$.

Problem 6

Calculate the energy released when two deuterium nuclei fuse to form helium-3 and a neutron: $\prescript{2}{1}\mathrm{H} + \prescript{2}_{1}\mathrm{H} \to \prescript{3}_{2}\mathrm{He} + \prescript{1}_{0}\mathrm{n}$. Given: $m(\prescript{2}_{1}\mathrm{H}) = 2.01410$ u, $m(\prescript{3}_{2}\mathrm{He}) = 3.01603$ u.

Answer. $\Delta m = 2 \times 2.01410 - 3.01603 - 1.00867 = 4.02820 - 4.02470 = 0.00350$ u.

$E = 0.00350 \times 931.5 = 3.26$ MeV.

Problem 7

Why must a fusion reactor achieve extremely high temperatures? Why is a moderator not needed?

Answer. Fusion requires overcoming the Coulomb repulsion between positively charged nuclei. Only at Very high temperatures ($\sim 10^8$ K) do nuclei have sufficient kinetic energy to approach within the Range of the strong nuclear force ($\sim 10^{-15}$ m).

A moderator slows neutrons down, which is needed for fission (thermal neutrons have larger fission Cross-sections). In fusion, the reactants are positively charged nuclei, not neutrons, and the Reaction requires them to be fast (high energy), not slow. A moderator would be counterproductive.

Problem 8

A radioactive sample contains $\prescript{131}_{53}\mathrm{I}$ (half-life 8.04 days) with initial Activity 800 Bq. How long until the activity falls to 50 Bq?

Answer. $A = A_0 e^{-\lambda t}$. $50 = 800 e^{-\lambda t}$. $e^{-\lambda t} = 1/16$. $\lambda t = \ln 16 = 4\ln 2$. $t = 4\ln 2/\lambda = 4t_{1/2} = 4 \times 8.04 = 32.2$ days.

1. Confusing atomic number (protons) with mass number (protons + neutrons).

2. Forgetting that radioactive decay is random and spontaneous. It cannot be predicted for individual nuclei.

3. Misunderstanding that half-life is constant regardless of the initial amount of substance.

4. Confusing displacement with distance, or velocity with speed, particularly in graphs and calculations.

5. Incorrectly applying $\vec{F} = m\vec{a}$when forces are not collinear. Resolve into components first.

6. Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

The universe operates through fundamental forces and energy transfers. Forces are pushes and pulls that change motion, energy is the currency that drives all processes, and waves transfer energy without transferring matter. These principles connect seemingly different phenomena - from the orbit of planets to the vibration of atoms - under unified explanations that reveal the elegant simplicity underlying nature’s complexity.

• Mechanics
• Waves
• Electricity
• Fields