What is deuteron binding energy?

m = (m_p + m_n) - m_d = (1.00728 + 1.00866) - 2.01355 = 2.39 × 10^-3 u. E_b = 2.39 × 10^-3 × 931.5 = 2.23 MeV. Per nucleon: 1.11 MeV.

What is iron-56 binding energy?

m ≈ 0.528 u. E_b ≈ 492 MeV. Per nucleon: ≈ 8.79 MeV - the peak of the binding-energy curve.

What is fission energy release?

When U-235 captures a neutron and fissions into Ba-141 + Kr-92 + 3 neutrons, the mass defect is approximately 0.215 u, giving about 200 MeV per fission event. A reactor running at 1 GW thermal fissions about 3 × 10^19 U-235 atoms per second.

§-Syllabus dot point

NSWPhysicsSyllabus dot point

Inquiry Question 3: What evidence supports the relativistic model of the universe?

Derive and apply the mass-energy equivalence E = mc^2, including the calculation of mass defect and binding energy in nuclear reactions

A focused answer to the HSC Physics Module 7 dot point on mass-energy equivalence. The total relativistic energy E = gamma m c^2, the rest energy E_0 = mc^2, mass defect Delta m in nuclear binding, and worked examples for fission, fusion and the deuteron binding energy.

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

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

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What this dot point is asking
The answer
Examples in context

What this dot point is asking

NESA wants you to state and apply $E = m c^2$ , link it to the rest energy of a particle and the total relativistic energy, calculate mass defect and binding energy for simple nuclear reactions, and use the unified atomic mass unit conversion $1$ u $= 931.5$ MeV/ $c^2$ .

The answer

The famous equation

In Einstein's 1905 special relativity, the total energy of a free particle of rest mass $m$ moving at speed $v$ is:

E = \gamma m c^2

where $\gamma = 1 / \sqrt{1 - v^2 / c^2}$ . When the particle is at rest ( $v = 0$ , $\gamma = 1$ ), this reduces to the rest energy:

\boxed{E_0 = m c^2}

This is the famous "mass-energy equivalence": mass is a form of energy, and the conversion factor is $c^2 \approx 9.0 \times 10^{16}$ m $^2$ s $^{-2}$ - an enormous number. Even a few grams of mass converted to energy yields a colossal output.

The kinetic energy in special relativity

Splitting the total energy gives kinetic energy as:

KE = E - E_0 = (\gamma - 1) m c^2

In the non-relativistic limit $v \ll c$ , a Taylor expansion gives $KE \approx \tfrac{1}{2} m v^2$ , recovering classical mechanics. Near $c$ , $KE$ diverges, which is why no massive object can reach $c$ (infinite energy would be required).

Unit conventions

For atomic and nuclear calculations, the unified atomic mass unit is convenient:

1 \text{ u} = 1.66054 \times 10^{-27} \text{ kg}

In energy units ( $E = m c^2$ ):

1 \text{ u} \cdot c^2 = 931.494 \text{ MeV} \approx 931.5 \text{ MeV}

Particle masses are often quoted in MeV/ $c^2$ : $m_e c^2 = 0.511$ MeV, $m_p c^2 = 938.3$ MeV, $m_n c^2 = 939.6$ MeV. Energy and mass are interconvertible currencies.

Mass defect and binding energy

The mass of a bound nucleus is less than the sum of the masses of its free constituents (protons and neutrons). The difference is the mass defect:

\Delta m = \sum m_{\text{free constituents}} - m_{\text{nucleus}}

The corresponding energy:

E_b = \Delta m \cdot c^2

is the binding energy, the energy that was released when the nucleus formed (or equivalently, the energy that must be supplied to dissociate it back into free nucleons).

Dividing by the number of nucleons gives the binding energy per nucleon, which peaks near iron-56 at about $8.8$ MeV per nucleon. Light nuclei (below iron) can release energy by fusion (smaller systems combine into more strongly bound systems). Heavy nuclei (above iron) can release energy by fission (large systems split into more strongly bound systems).

Three worked examples

Deuteron binding energy: $\Delta m = (m_p + m_n) - m_d = (1.00728 + 1.00866) - 2.01355 = 2.39 \times 10^{-3}$ u. $E_b = 2.39 \times 10^{-3} \times 931.5 = 2.23$ MeV. Per nucleon: $1.11$ MeV.
Iron-56 binding energy: $\Delta m \approx 0.528$ u. $E_b \approx 492$ MeV. Per nucleon: $\approx 8.79$ MeV - the peak of the binding-energy curve.
Fission energy release: When U-235 captures a neutron and fissions into Ba-141 + Kr-92 + 3 neutrons, the mass defect is approximately $0.215$ u, giving about $200$ MeV per fission event. A reactor running at $1$ GW thermal fissions about $3 \times 10^{19}$ U-235 atoms per second.

Pair production and annihilation

The cleanest demonstrations of $E = m c^2$ are at particle level:

A gamma photon of at least $1.022$ MeV ( $= 2 m_e c^2$ ) can convert in the field of a nucleus into an electron-positron pair, creating mass out of pure radiation energy.
The reverse: an electron and a positron annihilate to produce two $0.511$ -MeV gamma photons (or three for a parallel-spin state).

These processes are routine in particle physics; they conserve energy, momentum and charge while showing mass-energy conversion in both directions.

Energy in a chemical bond

For comparison, chemical-bond energies are of order eV per molecule, six orders of magnitude smaller than nuclear binding energies. That is why nuclear reactions release millions of times more energy per atom than chemical reactions.

The must-know equations

Rest energy: $E_0 = mc^2$ . Total relativistic energy: $E = \gamma mc^2$ , with $\gamma = 1/\sqrt{1 - v^2/c^2}$ .

Mass defect: $\Delta m = \sum m_{\text{free constituents}} - m_{\text{nucleus}}$ . Binding energy: $E_b = \Delta m \cdot c^2$ .

Conversion: $1\ \text{u} = 931.5\ \text{MeV}/c^2 = 1.661 \times 10^{-27}\ \text{kg}$ ; $c = 3.00 \times 10^8\ \text{m s}^{-1}$ .

Binding energy per nucleon $E_b / A$ peaks near iron-56 (about $8.8$ MeV per nucleon) - the most stable region of the nuclide chart.

Worked examples

Model 1 - binding energy of an alpha particle. A $^4_2\text{He}$ nucleus (an alpha particle) has a measured nuclear mass of $4.001506\ \text{u}$ . Calculate its binding energy and its binding energy per nucleon. ( $m_p = 1.007276\ \text{u}$ , $m_n = 1.008665\ \text{u}$ .)

Sum the free-nucleon masses: $2m_p + 2m_n = 2(1.007276) + 2(1.008665) = 2.014552 + 2.017330 = 4.031882\ \text{u}$ .

Mass defect: $\Delta m = 4.031882 - 4.001506 = 0.030376\ \text{u}$ .

Binding energy: $E_b = \Delta m \times 931.5 = 0.030376 \times 931.5 = 28.3\ \text{MeV}$ .

Per nucleon: $\dfrac{E_b}{A} = \dfrac{28.3}{4} = 7.07\ \text{MeV}$ per nucleon.

Marker's note: show the free-nucleon sum as a separate line before subtracting the nucleus mass - markers reward the explicit $\Delta m$ step, not just a final number. State units (u, then MeV) at every stage.

Model 2 - energy released per second by a fusion reactor. A D-T fusion reaction releases $17.6\ \text{MeV}$ per event. Estimate the fusion rate needed to generate a thermal power output of $500\ \text{MW}$ .

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

Power is energy released per second, so the reaction rate is $R = \dfrac{P}{E} = \dfrac{5.00 \times 10^8}{2.82 \times 10^{-12}} = 1.77 \times 10^{20}\ \text{reactions s}^{-1}$ .

Marker's note: the key step is converting MeV to joules before dividing by the (SI) power - a common slip is dividing $500 \times 10^6$ by $17.6$ directly, which gives the wrong units entirely.

Common traps

Quoting $E = m c^2$ when the particle is moving. That formula is only the rest energy. The total energy of a moving particle is $E = \gamma m c^2$ .

Using kilograms with $c$ in km/s, or u with $c$ in m/s. Stick to SI throughout, or use the $1$ u $= 931.5$ MeV shortcut.

Confusing mass defect with the masses themselves: $\Delta m$ is a small difference, often $10^{-3}$ u or less; the binding energy in MeV is the headline result.
Saying mass is destroyed in a nuclear reaction: Mass and energy are equivalent forms; what is conserved is total relativistic energy (which includes rest energy). It is more accurate to say "rest mass is converted to other forms of energy".
Treating chemical and nuclear energy releases as the same scale: Chemical reactions release of order eV per atom; nuclear reactions release of order MeV per nucleon - a factor of one million larger.

Exam technique

State the relation before substituting: write " $E_b = \Delta m \cdot c^2$ " (or " $E_b = \Delta m \times 931.5$ " in MeV) before you put numbers in - a stated formula earns a method mark even if arithmetic slips later.

Watch your units: if $\Delta m$ is in kilograms, use $c^2$ in $\text{m}^2\ \text{s}^{-2}$ and the answer is in joules; if $\Delta m$ is in u, use the $931.5\ \text{MeV}$ shortcut and the answer is in MeV directly. Do not mix the two conventions in one calculation.

For "calculate" or "estimate", show every intermediate value with units and finish with sensible significant figures (usually two or three, matching the given data). For "explain" or "justify" (e.g. why fusion releases energy), name the binding-energy-per-nucleon curve and the direction of mass change - a bare " $E = mc^2$ " restates the formula without answering the question.

When reading the binding-energy-per-nucleon graph, the vertical axis gives stability (higher = more tightly bound), not total energy of the nucleus - do not confuse it with the total binding energy, which requires multiplying by $A$ .

In plain English

Picture stacking LEGO bricks into a model: once they are clicked together, the model as a whole feels a tiny bit "lighter" than the loose pile of bricks would suggest, because some of that pile's potential energy went into the click that locked them together. Atomic nuclei do the same trick, just on an energy scale millions of times larger. When separate protons and neutrons snap together into a nucleus, the finished nucleus weighs very slightly less than the sum of its separate parts - and that missing mass has not vanished, it has been radiated away as energy, exactly as much energy as Einstein's $E = mc^2$ predicts. The tighter the nucleus is bound (think iron, the most tightly-packed nucleus there is), the more mass "disappeared" as energy when it formed - which is exactly why splitting a heavy nucleus like uranium (fission) or fusing light nuclei like hydrogen (fusion) both release huge amounts of energy: either way, the nucleons end up more tightly bound than before, and the mass difference pays the energy bill.

Examples in context

Example 1. Energy released in an ANSTO Lucas Heights $^{235}$ U fission event. A typical fission of $^{235}$ U yields a mass defect $\Delta m = 0.215 \text{ u} = 0.215 \times 1.66 \times 10^{-27} = 3.57 \times 10^{-28} \text{ kg}$ . Energy released: $E = \Delta m c^2 = 3.57 \times 10^{-28} \times (3.0 \times 10^8)^2 = 3.21 \times 10^{-11} \text{ J} = 200 \text{ MeV}$ per fission. At ANSTO's OPAL research reactor running at $20 \text{ MW}$ , the fission rate is $P / E = 2.0 \times 10^7 / 3.21 \times 10^{-11} = 6.2 \times 10^{17} \text{ fissions/s}$ . Over a year ( $3.15 \times 10^7 \text{ s}$ ), this consumes $1.96 \times 10^{25}$ uranium nuclei $= 7.6 \text{ kg}$ of $^{235}$ U.

Example 2. Pair production at the Australian Synchrotron. A $\gamma$ -ray photon of energy $E = 1.50 \text{ MeV}$ can create an electron-positron pair near a heavy nucleus. The threshold energy is $E_{\text{th}} = 2 m_e c^2 = 2 \times 0.511 = 1.022 \text{ MeV}$ ; the excess $1.50 - 1.022 = 0.478 \text{ MeV}$ becomes shared kinetic energy of the electron and positron ( $0.239 \text{ MeV}$ each, assuming symmetric momenta). Total rest mass created is $2 \times 9.11 \times 10^{-31} = 1.82 \times 10^{-30} \text{ kg}$ , drawn from photon energy via $E = mc^2$ . This is the inverse of electron-positron annihilation used in PET scanners at NSW hospitals.

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.

2022 HSC5 marksThe deuteron consists of one proton (mass 1.00728 u) and one neutron (mass 1.00866 u), with a measured deuteron mass of 2.01355 u. Calculate the mass defect of the deuteron and the binding energy in MeV. Explain why the deuteron mass is less than the sum of its constituent masses.

Show worked answer →

Mass defect:

$\Delta m = (m_p + m_n) - m_d = (1.00728 + 1.00866) - 2.01355 = 2.01594 - 2.01355 = 2.39 \times 10^{-3}$ u.

Converting to energy. The conversion factor is $1$ u $= 931.5$ MeV/ $c^2$ :

$E_b = \Delta m \cdot c^2 = 2.39 \times 10^{-3} \times 931.5 = 2.23$ MeV.

Explanation: when the proton and neutron bind to form the deuteron, the system releases $2.23$ MeV of energy (a gamma photon in the case of free deuteron formation). By mass-energy equivalence, the bound system has correspondingly less mass than the free constituents. The released energy is the binding energy, the same amount of energy that must be supplied to break the deuteron back into a free proton and neutron.

Markers reward correct mass defect, correct unit conversion using $1$ u $\approx 931.5$ MeV, and a clear statement that the difference equals the binding energy.

2018 HSC4 marksIn a fusion reaction inside the Sun, four protons combine through the proton-proton chain to form one helium-4 nucleus (mass 4.00151 u) plus other products totalling 4 protons (mass 1.00728 u each). Estimate the total energy released per helium-4 formed, ignoring the masses of the neutrinos and the kinetic energies of the products.

Show worked answer →

Mass defect (per helium-4 produced):

$\Delta m = 4 m_p - m_{\text{He}} = 4 \times 1.00728 - 4.00151 = 4.02912 - 4.00151 = 0.02761$ u.

Energy released (using $1$ u $= 931.5$ MeV):

$E = \Delta m \cdot c^2 = 0.02761 \times 931.5 = 25.7$ MeV.

In joules: $E = 25.7 \times 10^6 \times 1.602 \times 10^{-19} = 4.12 \times 10^{-12}$ J per reaction.

This is the dominant source of solar luminosity; the energy is shared among gamma rays, positrons, neutrinos and kinetic energy of the products. Markers reward the correct mass defect, conversion to MeV (and optionally to joules), and recognition that this energy powers the Sun.

A small correction: the proton-proton chain also releases neutrinos that carry off about $2$ to $5\%$ of the energy, so the energy deposited in the Sun is closer to $26$ MeV minus neutrino losses, but $25.7$ MeV is the standard answer.

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 mass-energy equivalence relation for a particle at rest, and give the energy equivalent of

1\ \text{u}

in MeV.

Show worked solution →

The rest energy of a particle of mass $m$ is $E_0 = mc^2$ .

By definition, $1\ \text{u} = 931.5\ \text{MeV}/c^2$ , so $1\ \text{u}$ of mass is equivalent to $931.5\ \text{MeV}$ of energy.

Marks: one for correctly stating $E_0 = mc^2$ , one for the conversion $1\ \text{u} \equiv 931.5\ \text{MeV}$ .

foundation3 marksThe binding energy of a

^4_2\text{He}

nucleus (an alpha particle) is

28.3\ \text{MeV}

. Calculate the mass defect in atomic mass units, then convert it to kilograms. (

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

Show worked solution →

Rearrange $E_b = \Delta m \cdot 931.5$ (in MeV, with $\Delta m$ in u):

$\Delta m = \dfrac{E_b}{931.5} = \dfrac{28.3}{931.5} = 0.0304\ \text{u}$ .

Converting to kilograms: $\Delta m = 0.0304 \times 1.661 \times 10^{-27} = 5.05 \times 10^{-29}\ \text{kg}$ .

Marks: one for rearranging to $\Delta m = E_b / 931.5$ , one for $\Delta m = 0.0304\ \text{u}$ , one for the correct conversion to $5.05 \times 10^{-29}\ \text{kg}$ with the unit.

foundation3 marksA nuclear reaction releases

17.6\ \text{MeV}

of energy. Calculate the corresponding mass defect in kilograms directly using

E = \Delta m\, c^2

(do not use the u-to-MeV shortcut). (

c = 3.00 \times 10^8\ \text{m s}^{-1}

1\ \text{MeV} = 1.602 \times 10^{-13}\ \text{J}

Show worked solution →

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

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

Marks: one for converting MeV to joules, one for correctly rearranging $\Delta m = E/c^2$ , one for $\Delta m = 3.13 \times 10^{-29}\ \text{kg}$ with the unit.

core4 marksA carbon-12 nucleus (

^{12}_6\text{C}

) has a nuclear mass of exactly

12.00000\ \text{u}

and is made of

6

protons (

m_p = 1.00728\ \text{u}

) and

6

neutrons (

m_n = 1.00866\ \text{u}

). Calculate (a) the mass defect and (b) the binding energy per nucleon in MeV.

Show worked solution →

(a) Mass defect. $\Delta m = (6 m_p + 6 m_n) - m_{\text{C-12}} = (6 \times 1.00728 + 6 \times 1.00866) - 12.00000$ .

$= (6.04368 + 6.05196) - 12.00000 = 12.09564 - 12.00000 = 0.09564\ \text{u}$ .

(b) Binding energy per nucleon. $E_b = \Delta m \times 931.5 = 0.09564 \times 931.5 = 89.1\ \text{MeV}$ .

Per nucleon: $\dfrac{E_b}{A} = \dfrac{89.1}{12} = 7.42\ \text{MeV per nucleon}$ .

Marks: one for the correct mass defect $0.09564\ \text{u}$ , one for $E_b = 89.1\ \text{MeV}$ , one for dividing by $A = 12$ , one for $7.42\ \text{MeV}$ per nucleon with the unit.

core5 marksThe figure shows binding energy per nucleon against mass number

A

for several nuclides. **(a)** Using the points for

^4\text{He}

(

A = 4

7.1\ \text{MeV}

) and

^{56}\text{Fe}

(

A = 56

8.8\ \text{MeV}

), state which nucleus is more tightly bound per nucleon and explain what this means physically. **(b)** Read the approximate binding energy per nucleon of

^{238}\text{U}

from the graph and use it to estimate its total binding energy. **(c)** Explain, using the shape of the curve, why both fusion of light nuclei and fission of heavy nuclei release energy.

Show worked solution →

(a) $^{56}\text{Fe}$ is more tightly bound per nucleon ( $8.8\ \text{MeV}$ versus $7.1\ \text{MeV}$ for $^4\text{He}$ ). A higher binding energy per nucleon means more energy would be needed to pull each nucleon out of the nucleus, so the iron-56 nucleus is more stable.

(b) Reading from the graph, $^{238}\text{U}$ has a binding energy per nucleon of about $7.6\ \text{MeV}$ . Total binding energy $\approx 7.6 \times 238 = 1.81 \times 10^3\ \text{MeV} \approx 1810\ \text{MeV}$ .

(c) The curve rises steeply from light nuclei up to the peak near iron-56, then falls slowly for heavier nuclei. Both fusing light nuclei together and splitting heavy nuclei apart move the products toward the peak, i.e. toward a higher binding energy per nucleon. A higher binding energy per nucleon means the products are more tightly bound (lower mass per nucleon) than the reactants, so mass is lost and, by $E = \Delta m\, c^2$ , energy is released.

Marks: one for identifying $^{56}\text{Fe}$ as more tightly bound, one for correctly interpreting binding energy per nucleon as a stability measure, one for reading $\approx 7.6\ \text{MeV}$ for $^{238}\text{U}$ and one for the total binding energy $\approx 1810\ \text{MeV}$ , one for explaining that both processes move nuclei toward the peak (higher binding energy per nucleon, lower mass, released energy).

exam6 marksEvaluate the claim that mass-energy equivalence means 'mass is destroyed' in a nuclear fission reaction. In your answer, refer to a specific reaction and to what is actually conserved.

Show worked solution →

Band-6 plan. (1) State the claim and why it is a common misconception. (2) Give the correct physics: total relativistic energy (including rest energy) is conserved, not rest mass alone. (3) Use a concrete fission example with numbers to show where the "lost" mass goes. (4) Conclude with a precise restatement of what happens.

Model answer. It is a common but imprecise claim that mass is "destroyed" in fission. What is actually conserved in any nuclear reaction is the total relativistic energy of the system, which includes both the rest energy $mc^2$ of the particles and their kinetic energy. Rest mass is not independently conserved, because rest mass is just one form that energy can take.

Consider the fission of $^{235}_{92}\text{U}$ after neutron capture into (typically) $^{141}_{56}\text{Ba}$ , $^{92}_{36}\text{Kr}$ and $3$ neutrons. The summed rest mass of the products is measured to be about $0.215\ \text{u}$ less than the rest mass of the reactants. Using $E_b = \Delta m \times 931.5 = 0.215 \times 931.5 \approx 200\ \text{MeV}$ , this "missing" rest mass reappears as $200\ \text{MeV}$ of kinetic energy of the fragments and neutrons, plus gamma radiation. No mass has vanished from the universe; rest mass has been converted into kinetic and radiant energy, and the total energy (rest energy plus kinetic energy) before and after the reaction is exactly equal.

The precise statement is therefore not "mass is destroyed" but "rest mass is converted into other forms of energy, while total relativistic energy is conserved". The products end up with less rest mass but more kinetic energy than the reactants, in an amount fixed exactly by $E = \Delta m\, c^2$ .

Marker's note: the top band names the correct conserved quantity (total relativistic energy, not rest mass alone), applies it to a specific reaction with a numeric mass defect and energy release, and explicitly corrects the "destroyed" language to "converted". A response that only asserts " $E = mc^2$ so mass becomes energy" without the conservation argument caps in the middle band.

exam6 marksThe Sun converts hydrogen to helium via the proton-proton chain, releasing energy that is radiated as sunlight. Assess the statement that mass-energy equivalence is 'the reason stars shine', explaining the role of binding energy per nucleon in this process and estimating the Sun's rate of mass loss given a luminosity of

3.85 \times 10^{26}\ \text{W}

Show worked solution →

Band-6 plan. (1) Confirm the statement is essentially correct and explain why (fusion moves nucleons up the binding-energy curve). (2) Link binding energy per nucleon to the mass defect and to $E = mc^2$ . (3) Compute the mass-loss rate from the given luminosity. (4) Close with a judgement on the statement's accuracy/precision.

Model answer. The statement is essentially correct, though it is more precise to say that mass-energy equivalence describes how the Sun shines rather than why fusion occurs. Inside the Sun's core, protons fuse via the proton-proton chain to eventually form helium-4, a nucleus with a higher binding energy per nucleon ( $7.1\ \text{MeV}$ ) than the isolated protons it came from. Because the fused helium nucleus is more tightly bound, its rest mass is measurably less than the total rest mass of the four protons that combined to form it.

This mass defect $\Delta m$ is converted into energy according to $E = \Delta m\, c^2$ , carried away as gamma radiation, the kinetic energy of the products, and neutrinos. It is this released energy, ultimately radiated from the Sun's surface as sunlight, that we observe as luminosity.

To estimate the mass-loss rate, rearrange $E = \Delta m\, c^2$ using power $P = \dfrac{\Delta m}{\Delta t}c^2$ , so $\dfrac{\Delta m}{\Delta t} = \dfrac{P}{c^2} = \dfrac{3.85 \times 10^{26}}{(3.00 \times 10^8)^2} = \dfrac{3.85 \times 10^{26}}{9.00 \times 10^{16}} = 4.28 \times 10^9\ \text{kg s}^{-1}$ .

This is an enormous rate of mass loss by everyday standards (about $4.3$ million tonnes per second), yet it is a negligible fraction of the Sun's total mass ( $\sim 2 \times 10^{30}\ \text{kg}$ ) even over billions of years, which is why the Sun can shine steadily for its roughly $10$ -billion-year main-sequence lifetime. The statement is therefore accurate: mass-energy equivalence is precisely the mechanism, via nuclear binding energy, by which fusion reactions supply the Sun's luminosity.

Marker's note: the top band correctly frames binding energy per nucleon as the physical reason mass is lost (not a vague "mass turns into energy"), performs the mass-loss-rate calculation correctly with units, and gives a specific numerical judgement (tonnes per second vs total solar mass) rather than a generic conclusion.