Mass-energy equivalence E = mc^2 and nuclear binding energy: HSC Physics Module 7

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.

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.

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 $1\ \text{u} \equiv 931.5\ \text{MeV}$.

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.

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.

(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.

(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).

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.

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.