TL;DR Nuclear physics in H2 Physics (9478) tests three interlinked skills: balancing decay equations, applying the exponential decay law, and reasoning about binding energy per nucleon to explain why fission and fusion release energy. This guide walks through each with worked calculations and exam-technique tips so you can solve nuclear problems with confidence rather than rote memory.
1 Where nuclear physics sits in the H2 Physics syllabus
Nuclear physics sits in the Modern Physics portion of the 2026 H2 Physics syllabus (9478). It covers the structure of the nucleus, radioactive decay, the decay law, mass-energy equivalence, and nuclear reactions (fission and fusion). The topic draws on energy conservation from mechanics and the photon concept from quantum physics, making it an excellent test of whether a student can connect ideas across the syllabus.
2 Nuclear structure - the essentials
2.1 Nucleon number, proton number, and isotopes
Every nucleus is described by two integers:
Proton numberZ - the number of protons (defines the element)
Nucleon numberA - the total number of protons and neutrons
Isotopes are nuclei with the same Z but different A. They are chemically identical (same electron configuration) but may differ in nuclear stability. For example, carbon-12 (612C) is stable, while carbon-14 (614C) is radioactive.
2.2 Nuclear notation
The standard notation is ZAX, where X is the chemical symbol. In every nuclear equation, bothA and Z must balance on each side - this is conservation of nucleon number and conservation of charge.
3 Radioactive decay modes
Radioactive decay is spontaneous and random. "Spontaneous" means it is not triggered by external conditions (temperature, pressure, chemical state). "Random" means we cannot predict which particular nucleus will decay next - only the probability of decay per unit time.
3.1 Alpha decay
An alpha particle is a helium-4 nucleus 24He. In alpha decay:
ZAX;→;Z−2A−4Y;+;24α
The parent loses 4 nucleons and 2 protons. Alpha particles are heavy, highly ionising, and stopped by a few centimetres of air or a sheet of paper.
3.2 Beta-minus decay
A neutron inside the nucleus converts into a proton, emitting an electron and an electron antineutrino:
n;→;p;+;e−;+;νˉe
In nuclear notation:
ZAX;→;Z+1AY;+;−10e;+;νˉe
A stays the same; Z increases by 1. The antineutrino carries away a share of the kinetic energy, which is why beta particles have a continuous energy spectrum (a fact that historically proved the neutrino's existence).
3.3 Beta-plus decay
A proton converts into a neutron, emitting a positron and an electron neutrino:
ZAX;→;Z−1AY;+;+10e;+;νe
A stays the same; Z decreases by 1. Beta-plus decay occurs in proton-rich nuclei and is the basis of PET (positron emission tomography) scans.
3.4 Gamma emission
After alpha or beta decay, the daughter nucleus is often in an excited energy state. It drops to its ground state by emitting a gamma-ray photon:
ZAX∗;→;ZAX;+;γ
Neither A nor Z changes. Gamma rays are the most penetrating (requiring thick lead or concrete to attenuate) but the least ionising.
3.5 Properties comparison
Property
Alpha
Beta-minus
Gamma
Composition
24He nucleus
Electron e−
Photon
Charge
+2e
-1e
0
Ionising ability
Very high
Moderate
Low
Penetrating power
Low (paper stops)
Moderate (few mm Al)
High (cm-thick Pb)
Deflection in E/B fields
Deflected (low q/m)
Deflected (high q/m, opposite direction)
Not deflected
Exam tip: Ionising ability and penetrating power are inversely related. Alpha particles interact strongly with matter (high ionisation), so they lose energy quickly and cannot penetrate far.
4 The decay law and half-life
4.1 The exponential model
The number of undecayed nuclei N decreases exponentially:
N=N0,e−λt
where λ is the decay constant (probability of decay per unit time, in s−1). Because activity A=λN, activity also decays exponentially:
A=A0,e−λt
4.2 Half-life
The half-lifet1/2 is the time for N (or A) to fall to half its initial value:
t1/2=λln2≈λ0.693
This is the single most-tested equation in the nuclear topic. It can be rearranged to find λ from a known half-life, or combined with the decay equation to solve for elapsed time.
4.3 Decay-law variable checkpoint
Before substituting values, decide whether the question is asking about nuclei count, activity, half-life, or elapsed time. The algebra is similar, but the first variable choice prevents most wrong starts.
Given clue
First variable to write
Next move
Common trap
"Number of undecayed nuclei remaining"
N=N0e−λt
Substitute N, N0, and t or solve for the missing variable.
Using activity symbols even when the question gives nuclei count.
"Activity falls from..."
A=A0e−λt
Use the same exponential form because A=λN.
"Half-life is..."
λ=ln2/t1/2
Convert the half-life into the same time unit used for t.
Mixing days and seconds inside the exponent.
"Find the time taken for a fraction to remain"
N0N=e−λt or A0A=e−λt
Worked check: If activity falls to one-quarter of its initial value, write A/A0=0.25, not 25. Since 0.25=(1/2)2, the elapsed time is two half-lives. If the half-life is 8.0 days, the time is 16 days before any exponential calculation is needed.
4.4 Worked example - activity calculation
Problem: A radioactive sample has an initial activity of 4.0×106 Bq and a half-life of 8.0 days. Find the activity after 20 days.
Step 1 - Find λ:
λ=t1/2ln2=8.0×864000.693=1.003×10−6;s−1
Step 2 - Apply the decay law:
A=A0,e−λt=4.0×106×e−(1.003×10−6)(20×86400)
A=4.0×106×e−1.733=4.0×106×0.1768=7.1×105;Bq
Quick check: 20 days is 2.5 half-lives, so A≈A0/22.5=4.0×106/5.66≈7.1×105 Bq. ✓
Exam technique: Always do the quick-check using N=N0(1/2)t/t1/2. If your exponential answer disagrees, you have an arithmetic error.
5 Mass-energy equivalence and binding energy
5.1 Einstein's relation
The total energy of a particle at rest is:
E=mc2
In nuclear physics, masses are often given in unified atomic mass units (u), where 1;u=1.661×10−27;kg. The energy equivalent of 1 u is 931.5;MeV.
5.2 Mass defect
When nucleons bind together to form a nucleus, the total mass of the nucleus is less than the sum of the individual nucleon masses. This "missing mass" is the mass defectΔm:
Δm=Zmp+(A−Z)mn−mnucleus
The direction is crucial: the mass of the separated nucleons is greater than the mass of the bound nucleus. The mass defect has been converted into binding energy via E=Δm,c2.
5.3 Binding energy and binding energy per nucleon
Binding energy (BE) is the energy required to completely separate a nucleus into its individual protons and neutrons. A higher binding energy means a more tightly bound (more stable) nucleus.
However, total BE increases with A simply because there are more nucleons. The meaningful measure of stability is binding energy per nucleon (BE/A).
The famous BE/A curve peaks at iron-56 (2656Fe), which has the highest BE/A at approximately 8.8 MeV per nucleon. This peak determines the direction of energy-releasing nuclear reactions:
Nuclei lighter than Fe-56 can increase their BE/A by fusing together - fusion releases energy.
Nuclei heavier than Fe-56 can increase their BE/A by splitting apart - fission releases energy.
Misconception check: Energy release does not mean the products are less tightly bound. In an energy-releasing fission or fusion reaction, the products have a higher total binding energy than the reactants. The missing mass-energy appears mainly as kinetic energy of the reaction products and emitted particles.
Induced fission occurs when a heavy nucleus (typically 92235U) absorbs a slow (thermal) neutron and splits into two medium-mass fragments plus additional neutrons:
92235U+01n;→;56141Ba+3692Kr+3,01n
The products sit closer to the peak of the BE/A curve, so the total binding energy increases. The increase in binding energy is released as kinetic energy of the fragments and neutrons.
Chain reaction: Each fission event releases 2 - 3 neutrons. If at least one of these causes another fission, the reaction is self-sustaining. Critical mass is the minimum mass of fissile material needed for a sustained chain reaction.
6.2 Nuclear fusion
Fusion combines light nuclei to form a heavier nucleus with higher BE/A. The most important example is the proton-proton chain that powers the Sun, ultimately converting four hydrogen nuclei into one helium-4 nucleus:
4,11H;→;24He+2,+10e+2νe+energy
Why fusion is difficult on Earth: The nuclei are all positively charged, so they experience strong Coulomb repulsion. To overcome this barrier, the fuel must be heated to temperatures of tens of millions of kelvin (plasma state) and confined at high density long enough for collisions to occur. This is why controlled fusion remains an engineering challenge.
Energy comparison: Fusion releases more energy per unit mass of fuel than fission, because the jump in BE/A for light nuclei (moving toward Fe-56) is steeper than the drop for heavy nuclei.
7 Common exam mistakes
Mistake 1 - Mass defect direction
Students sometimes write Δm=mnucleus−Σmnucleons, getting a negative number and becoming confused. Always define mass defect as the positive quantity:
Δm=Σmnucleons−mnucleus>0
The separated nucleons are heavier than the bound nucleus.
Mistake 2 - Unit conversion before E=mc2
If masses are in atomic mass units, you must either:
Convert to kg first (1;u=1.661×10−27;kg) and use c=3.00×108;m s−1 to get energy in joules, or
Use the shortcut 1;u≡931.5;MeV directly.
Mixing units (u with c in m/s) without proper conversion will produce nonsensical answers.
Mistake 3 - Total BE vs BE per nucleon
A common error is comparing total binding energies of different nuclei to determine stability. A uranium-238 nucleus has a very large total BE, but its BE per nucleon is lower than iron-56. Stability is determined by BE/A, not total BE.
Mistake 4 - Decay equation conservation
Always verify that both A and Z balance in every nuclear equation. In beta-minus decay, the electron is written as −10e - forgetting the −1 subscript leads to an unbalanced equation.
Mistake 5 - Ignoring the neutrino
Beta decay equations in H2 Physics must include the neutrino (or antineutrino). Omitting it loses marks. The neutrino also explains the continuous beta energy spectrum - without it, all beta particles would have the same kinetic energy.
8 Practice questions
Question 1 - Half-life algebra
A radioactive isotope has a half-life of 5.0 hours. A sample initially contains 6.0×1020 atoms.
(a) Calculate the decay constant λ. (b) Find the initial activity of the sample. (c) How many atoms remain after 15 hours? (d) After what time will the activity fall to 10% of its initial value?
Hint: For (d), set A/A0=0.10 and solve for t using logarithms.
Question 2 - Mass defect and binding energy
The mass of a nitrogen-14 nucleus is 13.99923 u. Given mp=1.00728;u and mn=1.00867;u:
(a) Calculate the mass defect of nitrogen-14. (b) Find the binding energy in MeV. (c) Determine the binding energy per nucleon and comment on the stability of nitrogen-14 compared to iron-56 (BE/A ≈8.8 MeV).
Hint: Nitrogen-14 has 7 protons and 7 neutrons.
Question 3 - Decay identification
An unknown nucleus 84210X undergoes two successive decays to form 82206Y.
(a) Identify the two decay modes that occurred. Justify your answer by showing how A and Z change. (b) Write the full decay equations, including any neutrinos.
Hint: Consider the total change: ΔA=−4, ΔZ=−2. Could this be achieved by a single alpha decay? Check carefully.
9 Further reading
Quantum Physics (Topic 19) - the preceding topic in the Modern Physics arc, covering the photon model and wave-particle duality.
Topic 20 summary notes - a more concise overview of the same content for quick revision.
