§-Syllabus dot point

NSWPhysicsSyllabus dot point

Inquiry Question 4: How is it known that human understanding of matter is still being refined?

Account for the energy released in nuclear fission and fusion in terms of mass defect and binding energy, using E = mc^2 and the binding energy curve

A focused answer to the HSC Physics Module 8 dot point on nuclear energy. Mass defect Delta m = Z m_p + N m_n - m_nucleus, binding energy Delta m c^2, the binding-energy-per-nucleon curve with its iron peak, energy release in fission (heavy nuclei split) and fusion (light nuclei combine), and worked examples for both.

Generated by Claude Opus 4.814 min answerUpdated 2026-07-01

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context

What this dot point is asking

NESA wants you to define mass defect and binding energy, calculate them from atomic mass data using $E = m c^2$ (with $1 \text{ u} \to 931.5$ MeV), describe the shape of the binding-energy-per-nucleon curve, and use it to explain why both nuclear fission of heavy nuclei and nuclear fusion of light nuclei release energy.

The answer

Mass defect and binding energy

The mass of a nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons. The difference is the mass defect:

\Delta m = Z m_p + N m_n - m_{\text{nuc}}

By $E = m c^2$ , this "missing" mass corresponds to the binding energy of the nucleus:

E_B = \Delta m \, c^2

This is the energy required to separate the nucleus into free protons and neutrons. Equivalently, it is the energy released when free nucleons assemble into the bound nucleus.

A useful unit conversion: $1 \text{ u} \cdot c^2 = 931.5$ MeV, so a mass defect in atomic mass units converts directly to a binding energy in MeV.

Binding energy per nucleon

The bound state of a nucleus is more meaningful when normalised by the number of nucleons:

\frac{E_B}{A}

This is the average energy needed to remove one nucleon. Plotted against mass number $A$ , it produces the famous binding-energy-per-nucleon curve:

$A = 1$ (hydrogen-1): 0 (single proton, no binding).
$A = 2$ (deuterium): 1.11 MeV per nucleon.
$A = 3$ (helium-3, tritium): 2.6-2.8 MeV per nucleon.
$A = 4$ (helium-4): 7.07 MeV per nucleon (a local peak, very tightly bound).
$A = 12$ (carbon-12): 7.7 MeV per nucleon.
$A = 56$ (iron-56): 8.79 MeV per nucleon. Near the maximum.
$A = 235$ (uranium-235): 7.6 MeV per nucleon. Declining.

The curve rises steeply for light nuclei, peaks near $A = 56$ , and falls slowly for heavy nuclei.

Rule for releasing energy

A nuclear process releases energy if the products have higher binding energy per nucleon than the reactants. Geometrically, this means the reaction moves the nucleons "uphill" on the binding-energy-per-nucleon curve, toward the iron peak.

Light side of the peak. Combining light nuclei (fusion) moves up toward iron: energy is released.
Heavy side of the peak. Splitting heavy nuclei (fission) moves up toward iron: energy is released.
At the peak (iron). Neither fission nor fusion releases energy. Iron is the end point of stellar nucleosynthesis (see the related dot point on stars).

Nuclear fission

A heavy unstable nucleus splits into two medium-mass fragments and a few neutrons. Example, induced fission of uranium-235 by a thermal neutron:

^{235}_{92}\text{U} + n \to ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3n

The barium and krypton fragments have $E_B / A \approx 8.5$ MeV per nucleon, while uranium has 7.6 MeV per nucleon. The increase of about 0.9 MeV per nucleon over 235 nucleons gives about 200 MeV released per fission event, carried away as kinetic energy of the fragments and neutrons plus prompt and delayed gamma radiation. Of this, roughly 173 MeV is the immediate (prompt) energy from the mass defect at the instant of splitting; the balance up to $\approx 200$ MeV comes from the subsequent radioactive (beta and gamma) decay of the fission products, so the two figures are consistent rather than contradictory.

A chain reaction can occur if the released neutrons trigger further fissions. In a controlled reactor, a moderator (water, graphite) slows neutrons to thermal energies for efficient capture by $^{235}$ U, and control rods absorb excess neutrons to keep the reaction critical (one fission triggers exactly one more). In a fission weapon, no control is used.

Nuclear fusion

Two light nuclei combine into a heavier one. Example, deuterium-tritium fusion (the easiest practical fusion reaction):

^2_1\text{H} + ^3_1\text{H} \to ^4_2\text{He} + n

Deuterium has 1.11 MeV per nucleon and tritium 2.83 MeV per nucleon. Helium-4 has 7.07 MeV per nucleon. The energy released is $\Delta m \, c^2 = 17.6$ MeV per reaction. Even though only 5 nucleons rearrange, the change per nucleon is much larger than for fission, so fusion is more energy-dense per kilogram of fuel.

Fusion requires temperatures of $\sim 10^8$ K to overcome the Coulomb repulsion between the positively charged nuclei, plus high densities and confinement times. In stars (Sun and similar), the proton-proton chain fuses four protons into a helium-4 nucleus, releasing 26.7 MeV per cycle, the energy source that sustains stellar luminosity.

On Earth, achieving net-energy-positive fusion is the long-running goal of magnetic-confinement (tokamak) and inertial-confinement (laser-driven) research.

Worked examples

Model 1 - deuterium-tritium fusion energy. Calculate the energy released by $^2_1\text{H} + {}^3_1\text{H} \to {}^4_2\text{He} + n$ . Atomic masses: $m(^2\text{H}) = 2.01410\,\text{u}$ , $m(^3\text{H}) = 3.01605\,\text{u}$ , $m(^4\text{He}) = 4.00260\,\text{u}$ , $m(n) = 1.00867\,\text{u}$ ; $1\,\text{u} = 931.5\,\text{MeV}/c^2$ .

Reactant mass: $2.01410 + 3.01605 = 5.03015\,\text{u}$ .

Product mass: $4.00260 + 1.00867 = 5.01127\,\text{u}$ .

Mass defect: $\Delta m = 5.03015 - 5.01127 = 0.01888\,\text{u}$ .

Energy released: $E = \Delta m \times 931.5 = 0.01888 \times 931.5 = 17.6\,\text{MeV}$ .

Marker's note: sum reactant and product masses SEPARATELY before subtracting - subtracting term-by-term without first summing each side is the most common way to drop a mass and get the wrong $\Delta m$ .

Model 2 - uranium-235 fission energy. Calculate the energy released per fission for $^{235}_{92}\text{U} + n \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3n$ . Atomic masses: $m(^{235}\text{U}) = 235.0439\,\text{u}$ , $m(^{141}\text{Ba}) = 140.9144\,\text{u}$ , $m(^{92}\text{Kr}) = 91.9262\,\text{u}$ , $m(n) = 1.0087\,\text{u}$ .

Reactant mass: $235.0439 + 1.0087 = 236.0526\,\text{u}$ .

Product mass: $140.9144 + 91.9262 + 3(1.0087) = 232.8406 + 3.0261 = 235.8667\,\text{u}$ .

Mass defect: $\Delta m = 236.0526 - 235.8667 = 0.1859\,\text{u}$ .

Energy released: $E = 0.1859 \times 931.5 = 173\,\text{MeV}$ . This is the immediate (prompt) energy from the mass defect at the instant of splitting; the commonly quoted $\approx 200\,\text{MeV}$ total additionally includes the delayed energy from the subsequent radioactive (beta and gamma) decay of the fission products, which this single mass-balance does not capture - so the two numbers are consistent, not contradictory.

Marker's note: state explicitly that the extra neutrons on the product side (here $3n$ ) must be included in the product mass sum - a very common lost mark is comparing $m(^{235}\text{U})$ directly against the two fragment masses without the neutrons on both sides.

In the D-T fusion of Model 1, most of the $17.6\,\text{MeV}$ is carried by the neutron (about $14.1\,\text{MeV}$ ) because the helium-4 nucleus is heavier and recoils more slowly. In a power reactor the neutron would deposit its energy in a blanket of lithium to breed more tritium and heat a working fluid. For comparison, burning a single carbon atom in a chemical reaction releases only a few eV, so a fission or fusion event releases roughly $10^8$ times more energy per reacting atom - why nuclear processes can power cities and weapons from kilograms of fuel.

Try it: Mass-energy calculator to convert between mass defects (in u or kg) and energy releases (in MeV or J) for any nuclear reaction.

Common traps

Forgetting that nuclear processes are governed by binding energy per nucleon, not total binding energy. A heavy nucleus has a larger total binding energy than a light one (more nucleons), but the energetic question is the energy per nucleon, the depth of the well each nucleon sits in.

Treating mass defect as a loss of nucleons: All nucleons are still present after the reaction; the mass loss is in the bound system as a whole, converted to kinetic energy and radiation.
Mixing up fission and fusion: Fission = heavy nucleus splits; fusion = light nuclei combine. Both move nucleons toward the iron peak and both release energy.
Using non-relativistic kinetic energy at nuclear scales: Fission fragments and fusion products often have several MeV of kinetic energy, well within the non-relativistic regime for the heavy fragments, but the relativistic energy budget $E = m c^2$ is the consistent accounting tool.
Saying fusion takes "no fuel": Fusion fuel (deuterium, tritium, lithium) is needed; tritium in particular must be bred in the reactor blanket from neutron capture on lithium.
Forgetting a particle in the mass balance: Leaving out a released neutron (fission) or not including the neutron product (D-T fusion) from one side of the equation silently changes $\Delta m$ - always sum every particle on both sides before subtracting.

Exam technique

"Calculate" wants a full numeric chain: sum the free-nucleon (or reactant) masses, subtract to get $\Delta m$ , multiply by $931.5$ (or by $c^2$ if working in kg), then state the final answer to three significant figures with its unit (MeV or J). "Explain" wants the general rule (energy is released when $E_B/A$ increases) applied with named numbers from the curve, not just "the curve goes up."

Always sum ALL reactant masses and ALL product masses separately (including any free neutrons) before subtracting - do not subtract term by term.

When reading the binding-energy graph, remember the vertical axis is $E_B/A$ in MeV, not total binding energy; a gradient on this graph has no simple physical meaning (unlike the $r$ - $v$ graph in Module 6), but the height of a point tells you that nucleus's binding energy per nucleon directly.

In plain English

Think of every nucleus as a group of people holding hands in a circle - the tighter and more evenly they hold on, the more energy it would take to pull any one of them away. That "tightness per person" is the binding energy per nucleon. A very small group (like hydrogen-2, just two nucleons) can only hold on loosely, and a very large group (like uranium, 235 nucleons) gets a bit crowded and the grip loosens slightly too. The best, tightest grip of all belongs to a medium-sized group around the size of iron-56. So if you take two tiny, loosely-linked groups and merge them (fusion), or take one huge, slightly-loose group and split it into two medium ones (fission), every member of the new groups ends up holding on more tightly than before. That extra "tightening" has to come from somewhere - it comes out as released energy, following Einstein's rule that a little bit of missing mass is really energy in disguise, $E = mc^2$ . Whether you split a big group or merge two small ones, moving the group size closer to iron always releases energy; that is the whole secret behind both nuclear reactors and the Sun.

Examples in context

Example 1. Binding energy per nucleon for $^{56}$ Fe via Lucas Heights mass-spectrometer data. ANSTO Lucas Heights mass spectrometers measure the atomic mass of $^{56}$ Fe at $m = 55.9349 \text{ u}$ . Because that is an atomic (nucleus-plus-electrons) mass, we build the constituents from hydrogen atoms so the 26 electron masses cancel: $26\, m(^1\text{H}) + 30 m_n = 26 \times 1.00783 + 30 \times 1.00867 = 56.4637 \text{ u}$ . Mass defect $\Delta m = 56.4637 - 55.9349 = 0.5288 \text{ u}$ . Binding energy $E_b = \Delta m \times 931.5 \text{ MeV} = 493 \text{ MeV}$ . Per nucleon: $E_b / A = 493 / 56 = 8.79 \text{ MeV/nucleon}$ , the peak of the binding-energy curve. This is why $^{56}$ Fe is the cosmic "ash" of stellar fusion - no further energy can be released by fusing or splitting iron nuclei.

Example 2. Energy budget of a $^{235}$ U fission at OPAL. A typical OPAL fission $^{235}_{92}\text{U} + n \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3n$ has reactant mass $235.0439 + 1.0087 = 236.053 \text{ u}$ . Products: $140.914 + 91.926 + 3 \times 1.0087 = 235.866 \text{ u}$ . Mass defect $\Delta m = 236.053 - 235.866 = 0.187 \text{ u} = 174 \text{ MeV}$ . Including delayed energy from fission-product decay, total per event is $\sim 200 \text{ MeV}$ . Per kg of $^{235}$ U: $E = 200 \times 1.6 \times 10^{-13} \times 6.02 \times 10^{23} / 0.235 = 8.2 \times 10^{13} \text{ J/kg}$ , equal to $\sim 2 \times 10^6$ kg of coal.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC5 marksCalculate the binding energy per nucleon of helium-4. Atomic masses (in u): m(He-4) = 4.0026, m(proton) = 1.00728, m(neutron) = 1.00867. (1 u = 931.5 MeV/c^2.)

Show worked answer →

Helium-4 has 2 protons and 2 neutrons.

Sum of free nucleon masses:

$2 m_p + 2 m_n = 2 \times 1.00728 + 2 \times 1.00867 = 4.03190$ u.

Mass defect:

$\Delta m = 4.03190 - 4.00260 = 0.02930$ u.

Binding energy:

$E_B = \Delta m \times 931.5 = 0.02930 \times 931.5 = 27.3$ MeV.

Binding energy per nucleon:

$E_B / A = 27.3 / 4 = 6.83$ MeV per nucleon.

Note on inputs: this $6.83$ MeV per nucleon comes from the ATOMIC masses given in the question (which include electron masses), whereas the canonical $7.07$ MeV per nucleon quoted elsewhere for helium-4 comes from the NUCLEAR masses (protons and neutrons only). Both are correct for their respective inputs; use whichever masses the question supplies.

Markers reward the sum of free masses, mass defect, conversion to energy via 931.5 MeV/u, and division by $A = 4$ .

2020 HSC4 marksExplain why fission of heavy nuclei (such as uranium-235) and fusion of light nuclei (such as deuterium and tritium) both release energy. Refer to the binding energy per nucleon curve.

Show worked answer →

The binding energy per nucleon $E_B / A$ varies with mass number $A$ . It is small for very light nuclei (around 1-3 MeV per nucleon for hydrogen-2 and helium-3), rises steeply through helium-4 (about 7 MeV per nucleon) and the light elements, peaks around iron-56 at about 8.8 MeV per nucleon, then declines slowly for heavier nuclei to about 7.5 MeV per nucleon at uranium.

Nuclear processes release energy when the products have higher $E_B / A$ than the reactants. Fission of uranium-235 (about 7.6 MeV per nucleon) into two medium-mass fragments such as barium and krypton (around 8.5 MeV per nucleon) increases $E_B / A$ by roughly 1 MeV per nucleon. With 235 nucleons rearranged, this is about 200 MeV per fission event.

Fusion of deuterium and tritium ( $\sim 1-3$ MeV per nucleon) into helium-4 (7.07 MeV per nucleon) increases $E_B / A$ by a much larger amount per nucleon, releasing about 17.6 MeV per fusion event with only 5 nucleons rearranged.

In both cases the gain in binding energy is the kinetic energy of the products, by $E = (\Delta m) c^2$ where $\Delta m$ is the mass defect of the reaction.

Markers reward the shape of the curve (low at extremes, peak at iron), the rule "products have higher $E_B / A$ ", and the specific values for fission (~200 MeV) and fusion (~17.6 MeV).

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksState the equation for mass defect

\Delta m

of a nucleus with

Z

protons and

N

neutrons and nucleus mass

m_{\text{nuc}}

, and explain in one sentence what the missing mass becomes.

Show worked solution →

Equation. $\Delta m = Z m_p + N m_n - m_{\text{nuc}}$ : the mass defect is the sum of the free-nucleon masses minus the actual mass of the bound nucleus.

What happens to it. The missing mass corresponds, by $E = \Delta m \, c^2$ , to the binding energy released when the nucleons assembled into the nucleus (equivalently, the energy needed to pull the nucleus apart again).

Marks: one for the correct equation, one for correctly identifying the missing mass as (released as / equivalent to) binding energy via $E = \Delta m c^2$ .

foundation3 marksCalculate the binding energy per nucleon of carbon-12. Take

m(^{12}\text{C}) = 12.00000

u exactly,

m_p = 1.00728

m_n = 1.00867

u, and

1\ \text{u} = 931.5\ \text{MeV}/c^2

Show worked solution →

Carbon-12 has $Z = 6$ protons and $N = 6$ neutrons.

Sum of free nucleon masses: $6 m_p + 6 m_n = 6(1.00728) + 6(1.00867) = 12.09570$ u.

Mass defect: $\Delta m = 12.09570 - 12.00000 = 0.09570$ u.

Binding energy: $E_B = 0.09570 \times 931.5 = 89.1$ MeV.

Binding energy per nucleon: $E_B / A = 89.1 / 12 = 7.43$ MeV per nucleon.

Marks: one for the correct sum of free-nucleon masses, one for $\Delta m = 0.09570$ u converted to $E_B = 89.1$ MeV, one for the correct division by $A = 12$ giving $7.43$ MeV per nucleon.

foundation2 marksState whether each of the following moves a nucleus toward or away from the iron-56 peak on the binding-energy-per-nucleon curve, and whether energy is released or absorbed: (a) fusion of two hydrogen-2 nuclei; (b) fission of uranium-238.

Show worked solution →

(a) Hydrogen-2 lies well to the light side of the peak, so fusing two of them moves the resulting nucleus toward iron-56, i.e. up the curve. Binding energy per nucleon increases, so energy is released.

(b) Uranium-238 lies on the heavy side of the peak, so splitting it into medium-mass fragments moves the fragments toward iron-56, i.e. up the curve. Binding energy per nucleon increases, so energy is released.

Marks: one mark for each part correctly stating "toward the peak" and "energy released."

core4 marksLithium-7 has atomic mass

7.01600

u. Using

m_p = 1.00728

m_n = 1.00867

u and

1\ \text{u} = 931.5\ \text{MeV}/c^2

, calculate (a) the mass defect and (b) the binding energy per nucleon of lithium-7 (

Z = 3

N = 4

Show worked solution →

(a) Mass defect. Sum of free-nucleon masses: $3 m_p + 4 m_n = 3(1.00728) + 4(1.00867) = 3.02184 + 4.03468 = 7.05652$ u.

$\Delta m = 7.05652 - 7.01600 = 0.04052$ u.

(b) Binding energy per nucleon. $E_B = \Delta m \times 931.5 = 0.04052 \times 931.5 = 37.7$ MeV.

$E_B / A = 37.7 / 7 = 5.39$ MeV per nucleon.

Marks: one for the correct free-nucleon mass sum, one for $\Delta m = 0.04052$ u, one for $E_B = 37.7$ MeV, one for the correctly divided $E_B/A = 5.39$ MeV per nucleon. Note lithium-7's value sits below helium-4's ( $7.07$ MeV per nucleon) - the curve is not perfectly smooth at very low $A$ .

core4 marksThe figure shows the binding-energy-per-nucleon curve. **(a)** Using the graph, estimate the binding energy per nucleon of iron-56 and of uranium-235, and hence the increase in

E_B/A

if uranium-235 fissioned into fragments as tightly bound as iron. **(b)** Explain, referring to the shape of the curve, why fusion of light nuclei releases far more energy per nucleon than fission of heavy nuclei.

Show worked solution →

(a) Reading the peak of the curve at iron-56 gives $E_B/A \approx 8.8$ MeV per nucleon; reading the curve at uranium-235 gives $E_B/A \approx 7.6$ MeV per nucleon. The increase, if uranium fragments were as tightly bound as iron, would be about $8.8 - 7.6 = 1.2$ MeV per nucleon (in practice real fission fragments such as barium and krypton reach only about $8.5$ MeV per nucleon, so the real increase is closer to $0.9$ MeV per nucleon).

(b) The curve rises very steeply on the light side (from $0$ at hydrogen-1 up to about $7$ - $9$ MeV per nucleon by carbon-12) but falls only gently on the heavy side (from the $8.8$ MeV peak down to $7.6$ MeV per nucleon at uranium). A light nucleus fusing therefore climbs a much steeper part of the curve, gaining several MeV per nucleon, whereas a heavy nucleus fissioning only descends the shallow far side of the peak, gaining roughly $1$ MeV per nucleon. This is why D-T fusion releases $17.6$ MeV from just $5$ nucleons rearranging, while uranium fission releases about $200$ MeV but from $235$ nucleons rearranging - fusion's energy release per nucleon is far larger.

Marks: one for reading $E_B/A \approx 8.8$ MeV per nucleon at iron-56 from the graph, one for reading $E_B/A \approx 7.6$ MeV per nucleon at uranium-235, one for the correct qualitative gradient comparison (steep light side, shallow heavy side), one for linking that gradient difference to the larger per-nucleon energy release in fusion.

core3 marksThe reaction

^2_1\text{H} + {}^3_1\text{H} \to {}^4_2\text{He} + n

releases

17.6

MeV. Calculate the mass defect of this reaction in kilograms. (

1\ \text{u} = 1.661 \times 10^{-27}

kg,

c = 3.00 \times 10^8

m/s.)

Show worked solution →

From $E = \Delta m \, c^2$ , $\Delta m = \dfrac{E}{c^2}$ .

Convert the energy to joules: $E = 17.6 \times 1.602 \times 10^{-13} = 2.82 \times 10^{-12}$ J.

$\Delta m = \dfrac{2.82 \times 10^{-12}}{(3.00 \times 10^8)^2} = \dfrac{2.82 \times 10^{-12}}{9.00 \times 10^{16}} = 3.13 \times 10^{-29}$ kg.

(Check: $\Delta m$ in u is $17.6 / 931.5 = 0.0189$ u $= 0.0189 \times 1.661 \times 10^{-27} = 3.14 \times 10^{-29}$ kg - consistent.)

Marks: one for converting the given MeV to joules, one for correctly applying $\Delta m = E/c^2$ with $c$ squared, one for the final answer to three significant figures with the correct unit (kg).

exam6 marksAnalyse how the shape of the binding-energy-per-nucleon curve accounts for the energy released in both nuclear fission and nuclear fusion, using specific values from the curve in your answer.

Show worked solution →

Band-6 plan. (1) Define binding energy per nucleon and state the general rule (energy released when products have higher $E_B/A$ ). (2) Describe the curve's shape with real values (steep rise, iron-56 peak at $8.8$ MeV/nucleon, gentle fall). (3) Apply the rule to fission with real numbers (U-235 to Ba/Kr, $\sim 200$ MeV). (4) Apply the rule to fusion with real numbers (D-T, $17.6$ MeV). (5) Synthesise: both move nucleons "uphill" toward iron, but from opposite sides, which is why fusion is more energy-dense per nucleon.

Model answer. The binding energy per nucleon, $E_B/A$ , measures how tightly a nucleus is bound; a nuclear process releases energy whenever the products end up with a higher $E_B/A$ than the reactants, the released mass defect appearing as kinetic energy of the products via $E = \Delta m \, c^2$ . Plotted against mass number $A$ , this quantity rises very steeply for light nuclei (from $0$ at hydrogen-1 through about $7.1$ MeV per nucleon at helium-4 and $7.7$ MeV per nucleon at carbon-12), reaches a maximum of about $8.8$ MeV per nucleon at iron-56, and then falls slowly for heavier nuclei to about $7.6$ MeV per nucleon at uranium-235.

For fission, a heavy nucleus such as uranium-235 sits on the gently falling, heavy side of the peak at $7.6$ MeV per nucleon. When it splits into two medium-mass fragments such as barium-141 and krypton-92 (each closer to $8.5$ MeV per nucleon), every one of the $235$ nucleons moves to a slightly more tightly bound state, an increase of roughly $0.9$ MeV per nucleon. Multiplied across $235$ nucleons this yields about $200$ MeV released per fission event.

For fusion, light nuclei such as deuterium ( $1.11$ MeV per nucleon) and tritium ( $2.83$ MeV per nucleon) sit on the very steep, light side of the curve. Combining them into helium-4 ( $7.07$ MeV per nucleon) produces a much larger jump in $E_B/A$ per nucleon than fission achieves, releasing $17.6$ MeV from only $5$ nucleons rearranging.

In both cases the products end up closer to the iron-56 peak than the reactants were - fission approaches from the heavy side, fusion from the light side - so both release energy, consistent with the single rule that nuclear reactions release energy whenever they increase $E_B/A$ . Because the curve is far steeper on the light side than the heavy side, fusion releases far more energy per nucleon of fuel than fission does, which is why stars are powered by fusion and why fusion fuel is the more energy-dense (per kilogram) nuclear resource.

Marker's note: the top band states the general rule explicitly, quotes real $E_B/A$ values from both sides of the curve (not just "fission releases 200 MeV, fusion releases 17.6 MeV" without linking them to the curve's shape), and closes with the comparative judgement about gradient steepness explaining the difference in energy density. A response that treats fission and fusion as two disconnected facts, without the unifying "both move toward the iron-56 peak" argument, caps in the middle band.

exam7 marksEvaluate the claim that "fusion is a better energy source than fission because it releases more energy." In your answer refer to mass defect, binding energy per nucleon, and the amount of fuel required.

Show worked solution →

Band-6 plan. Thesis: the claim is imprecise - fission releases more energy per REACTION EVENT, but fusion releases more energy per unit MASS of fuel; a fair evaluation must distinguish these. (1) Show the reaction data (fission $\sim 200$ MeV per event vs fusion $17.6$ MeV per event) which appears to support "fission releases more". (2) Reframe using binding energy per nucleon and nucleon count to show fusion is more energy-dense per kilogram. (3) Weigh both readings and give a final, qualified judgement.

Model answer. Taken literally, the claim is false: a single uranium-235 fission event, in which the mass defect between reactants and fragments converts about $200$ MeV to kinetic energy via $E = \Delta m \, c^2$ , releases far more energy than a single deuterium-tritium fusion event, which releases only $17.6$ MeV. By this reading, fission is the larger single-event energy source.

However, the more physically meaningful comparison is energy released per unit mass (or per nucleon) of fuel, because it is fuel mass that must be mined, purified and paid for. Here the binding-energy-per-nucleon curve is decisive. Uranium-235 starts at about $7.6$ MeV per nucleon and its fragments end near $8.5$ MeV per nucleon, a gain of roughly $0.9$ MeV per nucleon spread across all $235$ nucleons. Deuterium and tritium start much lower on the steep part of the curve (about $1$ - $3$ MeV per nucleon) and end at helium-4's $7.07$ MeV per nucleon, a gain of several MeV per nucleon spread across only $5$ nucleons. Because the light side of the curve is so much steeper than the heavy side, fusion converts a far larger fraction of its rest mass into energy: roughly $0.4\%$ of the D-T fuel mass compared with roughly $0.09\%$ of the uranium-235 mass. Per kilogram of fuel, fusion therefore releases several times more energy than fission.

Weighing these, the claim "fusion releases more energy" is correct only if "energy" means energy per kilogram of fuel, which is the metric that matters for how much fuel a power source needs; it is incorrect if "energy" means energy per individual reaction event, where fission dominates because so many more nucleons rearrange at once. A precise version of the claim would be: fusion is more energy-dense per unit mass of fuel than fission, because the binding-energy-per-nucleon curve is steeper on the light side of the iron-56 peak than on the heavy side.

Marker's note: the top band does not simply agree or disagree - it identifies the ambiguity in "more energy" (per event vs per unit mass), supports both readings with correct numbers ( $200$ MeV vs $17.6$ MeV per event; the steeper light-side gradient giving greater energy density per kilogram), and ends with an explicit, qualified judgement. A response that only restates the per-event or only the per-mass comparison, without resolving the tension between them, caps in the middle band.