Is 931.5 MeV on the HSC data sheet?

Yes. The 2019 data sheet lists the atomic mass unit as 1 u = 1.661 × 10^−27 kg or 931.5 MeV/c^2. It also has E = mc^2, Nt = N0e^−λt and λ = ln 2 ÷ t1/2, and the masses of the proton, neutron and electron in kg.

What is the difference between mass defect and binding energy?

The mass defect is a mass: how much less a nucleus weighs than its separate protons and neutrons. The binding energy is the energy equivalent of that mass, E = Δmc^2: the energy needed to pull the nucleus apart. 1 u of mass defect corresponds to 931.5 MeV.

What is the difference between fission and fusion?

Fission splits a heavy nucleus (such as uranium-235) into two middle-sized nuclei plus neutrons; fusion joins light nuclei (such as hydrogen isotopes) into a heavier one. Both release energy because the products have more binding energy per nucleon, and so less mass, than the starting nuclei.

Why does fusion need such high temperatures?

Nuclei are positively charged and repel each other. They must be moving fast enough to get within about 10^−15 m, where the attractive strong force takes over. That needs temperatures of millions of kelvin, like the Sun's core, plus high density so collisions happen often.

Should I use atomic masses or nuclear masses?

Use whatever the question gives, consistently. Atomic masses include the electrons, which cancel when the electron count is the same on both sides (that is why hydrogen-1 atoms, not bare protons, go with an atomic mass of helium). Don't mix an atomic mass on one side with bare-nucleus masses on the other.

Why do both the fission of uranium-235 and the fusion of hydrogen release energy?

The products have a higher binding energy per nucleon than the reactants.

The products have more mass than the reactants.

Both reactions move nuclei away from iron-56 on the binding energy curve.

Both reactions destroy some protons and neutrons.

HSC Physics · Module 8 · From the Universe to the Atom · 14 min read

Nuclear physics: decay, half-life and binding energy

Some nuclei are unstable. Sooner or later they break down and give out radiation, and nobody can say when a particular nucleus will go, only how quickly a large sample shrinks. Other reactions, splitting heavy nuclei or joining light ones, release millions of times more energy per kilogram than burning fuel. HSC nuclear physics is about both: the kinds of decay and how fast they happen, and where the energy comes from.

This guide covers alpha, beta and gamma radiation and their decay equations, half-life with N_t = N₀e^−λt, mass defect and binding energy (using 931.5 MeV per u), the binding energy per nucleon curve, and why fission and fusion both release energy. Then come the common mistakes, a quick quiz and a full worked example.

What you need to know

For unstable nuclei, analyse how they decay on their own, and the properties of each radiation they give off: alpha, beta and gamma.
Use half-life as a model of radioactive decay, with N_t = N₀e^−λt and λ = ln 2 ÷ t_1/2, to predict a sample's activity or the amount left.
Explain and model nuclear fission, covering chain reactions (controlled or uncontrolled), and account for the energy it releases.
Analyse how mass–energy is conserved when nuclei change, spontaneously or artificially, including fission, fusion, alpha and beta decay.
Account for why nuclear fusion releases energy.
Calculate the energy released in a decay or other nuclear reaction, fission and fusion included, using conservation of energy, mass defect, binding energy and E = mc².

The equations (from the 2019 HSC data sheet unless marked):

N_t = N₀e^−λtNumber of undecayed nuclei after time t. The activity follows the same law, A_t = A₀e^−λt.

λ = ln 2 ÷ t_1/2Decay constant from the half-life. Its unit is 'per time' (s⁻¹, day⁻¹, y⁻¹): match it to the unit of t.

E = mc²Mass and energy are equivalent. With Δm in kg and c = 3.00 × 10⁸ m s⁻¹, E is in joules.

1 u = 1.661 × 10⁻²⁷ kg = 931.5 MeV/c²From the data sheet's constants. So E (MeV) = Δm (u) × 931.5.

Δm = m_before − m_after not on the data sheetMass defect of a reaction. For binding energy: mass of the separate nucleons minus the mass of the nucleus.

Dot points: PH-M8-15, PH-M8-16, PH-M8-17, PH-M8-18, PH-M8-19, PH-M8-20 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

Alpha, beta and gamma

A nucleus is unstable when its mix of protons and neutrons is wrong for its size: too many neutrons, too few, or simply too big. It becomes more stable by emitting radiation. The three types behave very differently:

Alpha (α): a helium-4 nucleus (2 protons, 2 neutrons), charge +2. Heavy and slow, so it is strongly ionising but stopped by a sheet of paper or a few centimetres of air. Mostly from heavy nuclei.
Beta-minus (β⁻): a fast electron made when a neutron turns into a proton (an antineutrino is also emitted). Charge −1, moderately ionising, stopped by a few millimetres of aluminium. Beta-plus (β⁺) is a positron from a proton turning into a neutron (with a neutrino).
Gamma (γ): a high-energy photon, no charge and no mass. Weakly ionising and very penetrating: thick lead or concrete reduces it but never stops all of it. It often follows α or β decay, as the new nucleus drops from an excited state.
In a field: α and β bend in opposite directions (β much more, being far lighter); γ is not deflected.

Penetration: α is stopped by paper, β by a few mm of aluminium, and γ is only reduced by thick lead (dashed: weaker beyond it). Ionising ability runs the other way: α is the most ionising.

Writing decay equations

In every nuclear equation the mass numbers (top) and the atomic numbers (bottom) must add to the same totals on both sides. That is conservation of nucleon number and of charge. An alpha particle is ⁴₂He, a beta-minus particle is ⁰₋₁e and a positron is ⁰₊₁e. Gamma changes neither number.

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂HeAlpha: A drops by 4, Z by 2.

¹⁴₆C → ¹⁴₇N + ⁰₋₁e + antineutrinoBeta-minus: A unchanged, Z up by 1.

²²₁₁Na → ²²₁₀Ne + ⁰₊₁e + neutrinoBeta-plus: A unchanged, Z down by 1.

⁶⁰₂₈Ni* → ⁶⁰₂₈Ni + γGamma: the excited nucleus (*) loses energy; A and Z unchanged.

Half-life

Decay is random for each nucleus, but a large sample shrinks in a completely predictable way: in every half-life t_1/2, half of the nuclei that are left decay. After one half-life ½ remains, after two ¼, after three ⅛. The curve is exponential, never a straight line, and it never quite reaches zero.

The decay constant λ = ln 2 ÷ t_1/2 is the probability per unit time that any one nucleus decays, so a larger λ means a shorter half-life and faster decay. For any time, N_t = N₀e^−λt. Iodine-131, used in medicine, has t_1/2 = 8.02 days, so λ = 0.693 ÷ 8.02 = 0.0864 day⁻¹. After 20 days, N_t ÷ N₀ = e^{−0.0864 × 20} = 0.178: about 18% is left. Keep λ and t in the same time unit. Activity (decays per second) is proportional to N, so it follows the same curve.

Each half-life halves what is left: N₀ → N₀/2 → N₀/4 → N₀/8. Read t_1/2 from any point: find the time for that value to halve.

Mass defect and binding energy

A nucleus has less mass than the separate protons and neutrons that make it. The difference is the mass defect, and the energy equivalent of it, E = Δmc², is the binding energy: the energy you would have to supply to pull the nucleus completely apart (and the energy released when it forms).

Helium-4, using atomic masses (rounded to six decimal places from standard nuclear data tables): two hydrogen-1 atoms, 2 × 1.007825 u, plus two neutrons, 2 × 1.008665 u, make 4.032980 u. A helium-4 atom is 4.002603 u. So Δm = 0.030377 u and the binding energy is 0.030377 × 931.5 = 28.3 MeV, or 7.07 MeV per nucleon. Using hydrogen-1 atoms rather than bare protons makes the two electrons in the helium atom cancel.

The same idea gives the energy of any reaction: add up the masses before and after. If the products have less mass, the difference has been released as energy (mostly kinetic energy of the products), so E = Δm × 931.5 MeV.

Binding energy per nucleon: why fission and fusion work

Dividing the binding energy by the number of nucleons shows how tightly each nucleon is held. The curve rises steeply for light nuclei, peaks near iron-56 (about 8.8 MeV per nucleon, the most stable nuclei), then falls slowly for heavy ones.

Any reaction that moves nuclei towards the peak raises the binding energy per nucleon, lowers the total mass and releases energy. Fusion joins light nuclei (moving up the steep left side); fission splits a heavy nucleus into two middle-sized ones (moving up the gentle right side). Both move towards iron, from opposite ends.

Iron-56 sits at the peak. Fusion (light nuclei joining) and fission (heavy nuclei splitting) both move towards it, raising the binding energy per nucleon and releasing energy.

Fission, chain reactions and fusion

Fission. A uranium-235 nucleus absorbs a slow neutron, becomes unstable uranium-236 and splits, for example: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n. Each fission releases about 200 MeV and 2 or 3 neutrons, which can cause more fissions: a chain reaction. In a reactor it is controlled: a moderator slows the neutrons and control rods absorb some, so on average exactly one neutron from each fission causes another. In a weapon it is uncontrolled: more than one neutron per fission goes on to split another nucleus, so the rate grows exponentially. A chain reaction needs at least a critical mass of fuel, or too many neutrons escape.

Fusion. Light nuclei join, such as hydrogen into helium in the Sun's core. The nuclei are both positive and repel strongly, so they only get close enough for the short-range strong force to take over if they are moving very fast: that takes temperatures of about 10⁷ K and above, plus high density to make collisions frequent. Per kilogram of fuel, fusion releases several times more energy than fission.

Common mistakes

Treating a mass in u as if it were an energy, or multiplying u by c².Either multiply Δm in u by 931.5 to get MeV, or convert Δm to kg (× 1.661 × 10⁻²⁷) and use E = mc² for joules.
Rounding the masses before subtracting, so most of a small mass defect disappears.Keep every decimal place the question gives until after the subtraction. The mass defect is often only the third or fourth decimal place.
Writing a decay equation where the top or bottom numbers don't balance, or mixing up α and β changes.Check both totals. α: A − 4, Z − 2. β⁻: Z + 1. β⁺: Z − 1. γ: no change.
Thinking that after two half-lives the whole sample has decayed.Each half-life halves what is left: after two half-lives a quarter remains, after three an eighth. The decay is exponential, not linear.
Using λ in y⁻¹ with t in seconds, or confusing the decay constant λ with a wavelength.Keep λ and t in the same time unit so λt has no unit. Write 'decay constant' next to λ so you don't mix it up with λ for wavelength.
Saying a nucleus with a bigger binding energy releases more energy, or that fission and fusion move away from iron.Stability depends on binding energy per nucleon. Fission and fusion both move towards iron-56, increasing binding energy per nucleon; the increase is the energy released.

Worked example

A proposed fusion reactor uses the deuterium–tritium reaction: ²₁H + ³₁H → ⁴₂He + ¹₀n.

Atomic masses: ²₁H = 2.014102 u, ³₁H = 3.016049 u, ⁴₂He = 4.002603 u; neutron = 1.008665 u. Calculate the energy released in one reaction, in MeV and in joules.

(4 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. From binding energy per nucleon: helium-4 is about 7 MeV × 4 nucleons ≈ 28 MeV, deuterium about 1 MeV × 2 ≈ 2 MeV, and tritium about 3 MeV × 3 ≈ 8 MeV. The gain in binding energy is about 28 − 2 − 8 ≈ 18 MeV. I predict roughly 17–18 MeV, which is a few × 10⁻¹² J.

K · Knowns

I sketch the reaction with each particle's mass written under it. These are atomic masses, so before using them I check the electrons balance: one electron in each hydrogen atom on the left, two in the helium atom on the right, none with the neutron. 2 = 2, so they cancel.

m(²H) = 2.014102 u, m(³H) = 3.016049 u
m(⁴He) = 4.002603 u, m(n) = 1.008665 u
1 u = 931.5 MeV/c², 1 eV = 1.602 × 10⁻¹⁹ J

Knowns on the sketch: the four masses in u. Unknown: the energy released per reaction.

N · Not known

One energy, asked for in two units: MeV first (natural with u), then joules.

E = ? MeV and ? J

O · Organise

Mass before minus mass after is the mass defect. It must come out positive if energy is released. Then 931.5 MeV per u turns it into energy, and 1 MeV = 10⁶ × 1.602 × 10⁻¹⁹ J gives joules.

Δm = [m(²H) + m(³H)] − [m(⁴He) + m(n)]
E (MeV) = Δm (u) × 931.5
E (J) = E (MeV) × 10⁶ × 1.602 × 10⁻¹⁹

W · Work it out

I add each side first and keep all six decimal places, because the difference is small. Only then do I subtract and convert.

before: 2.014102 + 3.016049 = 5.030151 u
after: 4.002603 + 1.008665 = 5.011268 u
Δm = 5.030151 − 5.011268 = 0.018883 u
E = 0.018883 × 931.5 = 17.59 MeV
E = 17.59 × 10⁶ × 1.602 × 10⁻¹⁹ = 2.818 × 10⁻¹² J

U · Units

The masses are given to six decimal places, so the mass defect is good to about four or five significant figures; three or four in the answer is fair.

E ≈ 17.6 MeV ≈ 2.82 × 10⁻¹² J per reaction

D · Direction (sense check)

Energy has no direction, so I check it by the other route on the data sheet: the mass defect in kilograms with E = mc². The tiny difference is just rounding in the data-sheet constants.

Δm = 0.018883 × 1.661 × 10⁻²⁷ = 3.136 × 10⁻²⁹ kg
E = 3.136 × 10⁻²⁹ × (3.00 × 10⁸)² = 2.82 × 10⁻¹² J ✓
Δm > 0, so energy is released, as fusion of light nuclei should.

Each deuterium–tritium fusion releases about 17.6 MeV, which is 2.82 × 10⁻¹² J.

Check against the prediction. I predicted roughly 17–18 MeV from the binding energy per nucleon values, and the mass defect gives 17.6 MeV. The two routes to joules (MeV × eV, and kg × c²) agree at 2.82 × 10⁻¹² J.

HSC advice

What markers keep saying

Remember that a mass in u is still a mass. Convert it to energy with 931.5 MeV per u (or to kg and use E = mc²), and finish with the energy unit, MeV or J. Copy masses from a table carefully, and keep every decimal place until the end.NESA marking feedback, 2023, summarised
When comparing reactions, back your answer with a calculation and link the larger energy release to the larger mass defect (or the larger change in binding energy). Keep units consistent, and don't round masses in u part-way through.NESA marking feedback, 2022, summarised
If a decay graph is given, read the half-life from it (the time for any value to halve) rather than calculating it some other way. Then use λ = ln 2 ÷ t_1/2 and solve N_t = N₀e^−λt with natural logs, showing each substitution.NESA marking feedback, 2022, summarised
To show how to tell alpha from beta, give a method and the result you would expect: for example, paper stops alpha but not beta, or the two bend opposite ways in a field. 'Use a detector' alone earns little.NESA marking feedback, 2025, summarised
In alpha decay, use conservation of momentum: the alpha particle and the daughter nucleus move off with equal and opposite momentum, so the much lighter alpha takes most of the kinetic energy.NESA marking feedback, 2019, summarised
In a multi-step nuclear calculation, plan the path first, set out every step, label each quantity clearly (words or subscripts), reuse your answers from earlier parts, and check that the last line answers the question asked.NESA marking feedback, 2021, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2019 · 2021 · 2022 · 2023 · 2025.

How often it is examined

Nuclear physics: decay, half-life and binding energy questions were worth 40 marks across the seven HSC Physics exams from 2019 to 2025 (about 5.7 marks a paper), and appeared in 7 of 7 papers.

Marks per paper (Module 8, Properties of the Nucleus). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Is 931.5 MeV on the HSC data sheet?: Yes. The 2019 data sheet lists the atomic mass unit as 1 u = 1.661 × 10⁻²⁷ kg or 931.5 MeV/c². It also has E = mc², N_t = N₀e^−λt and λ = ln 2 ÷ t_1/2, and the masses of the proton, neutron and electron in kg.
What is the difference between mass defect and binding energy?: The mass defect is a mass: how much less a nucleus weighs than its separate protons and neutrons. The binding energy is the energy equivalent of that mass, E = Δmc²: the energy needed to pull the nucleus apart. 1 u of mass defect corresponds to 931.5 MeV.
What is the difference between fission and fusion?: Fission splits a heavy nucleus (such as uranium-235) into two middle-sized nuclei plus neutrons; fusion joins light nuclei (such as hydrogen isotopes) into a heavier one. Both release energy because the products have more binding energy per nucleon, and so less mass, than the starting nuclei.
Why does fusion need such high temperatures?: Nuclei are positively charged and repel each other. They must be moving fast enough to get within about 10⁻¹⁵ m, where the attractive strong force takes over. That needs temperatures of millions of kelvin, like the Sun's core, plus high density so collisions happen often.
Should I use atomic masses or nuclear masses?: Use whatever the question gives, consistently. Atomic masses include the electrons, which cancel when the electron count is the same on both sides (that is why hydrogen-1 atoms, not bare protons, go with an atomic mass of helium). Don't mix an atomic mass on one side with bare-nucleus masses on the other.

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