Fission fusion mass-energy and nuclear energy

In a nuclear reaction the charge number and the total number of nucleons are conserved, but the sum of the rest masses of the products differs from that of the initial nuclei. This mass difference $\Delta m$ is converted into energy by Einstein's relation $E=\Delta m c^2$; in practice $\Delta m$ is measured in atomic mass units and energy in megaelectronvolts, where $1\ \text{u}\cdot c^2\approx 931.5\ \text{MeV}$. The binding energy of a nucleus, $E_b=\Delta m c^2$, is the energy that must be supplied to break it into separate nucleons; to compare different nuclei it is more convenient to use the per-nucleon value $E_b/A$. The curve of $E_b/A$ rises from light nuclei to a maximum near $^{56}\text{Fe}$ and falls slowly towards heavy nuclei, so energy is released when a system moves towards the peak of the curve: fusion of light nuclei and fission of heavy ones. The energy release of a reaction equals the difference of binding energies, $Q=E_b(\text{products})-E_b(\text{initial nuclei})$, and for $Q>0$ this energy is carried away as kinetic energy of the products — this is how nuclear power works.

Find the energy released in the fusion of deuterium and tritium:

$${}^{2}\text{H}+{}^{3}\text{H}\rightarrow{}^{4}\text{He}+n$$

Rest masses: $m(^{2}\text{H})=2.014102\ \text{u}$, $m(^{3}\text{H})=3.016049\ \text{u}$, $m(^{4}\text{He})=4.002602\ \text{u}$, $m_n=1.008665\ \text{u}$; conversion factor $931.5\ \text{MeV/u}$.

One channel of uranium fission by a thermal neutron:

$$n+{}^{235}\text{U}\rightarrow{}^{141}\text{Ba}+{}^{92}\text{Kr}+3n$$

Masses: $m(^{235}\text{U})=235.043930\ \text{u}$, $m(^{141}\text{Ba})=140.914411\ \text{u}$, $m(^{92}\text{Kr})=91.926156\ \text{u}$, $m_n=1.008665\ \text{u}$.

(a) Find the energy released per fission. (b) Estimate the energy released by complete fission of $1\ \text{g}$ of uranium-235.

In stars carbon can fuse: $2\,{}^{12}\text{C}\rightarrow{}^{24}\text{Mg}$. No mass table is at hand, but the binding energies per nucleon are known: $E_b/A=7.68\ \text{MeV}$ for $^{12}\text{C}$ and $E_b/A=8.26\ \text{MeV}$ for $^{24}\text{Mg}$. Estimate the energy release of the reaction and explain why this calculation gives the same result as the mass formula.

Safe self-study plan (calculation and simulation only — no radioactive sources, no exposure).

1. Open an online nuclear-fission simulation (for example PhET "Nuclear Fission") or a $Q$-value calculator. Work in sandbox mode: the chain reaction and irradiation are simulated, there are no real materials.
2. Take three reactions from a data table: (i) fusion $^{2}\text{H}+{}^{3}\text{H}$, (ii) fusion $^{2}\text{H}+{}^{2}\text{H}$, (iii) fission of $^{235}\text{U}$. For each, compute $\Delta m$ and $Q$ in megaelectronvolts.
3. Mark each reaction's position on the $E_b/A$ curve and explain in one paragraph why the sign of $Q$ matches the direction of motion towards the peak of the curve.
4. Check yourself: for $1\ \text{g}$ of a substance with mass number $A$, the number of nuclei is $N=\frac{1}{A}N_A$, not simply $N_A$ (Avogadro's number is the number of particles per mole, not per gram). Multiply $Q$ by $N$ and compare the resulting energy with the heat of combustion of $1\ \text{g}$ of coal ($\approx 3\cdot 10^{4}\ \text{J}$); for example, for $^{235}\text{U}$ this gives $N\approx 2.56\cdot 10^{21}$ and $E\approx 7.1\cdot 10^{10}\ \text{J}$.

Do all arithmetic on paper or in a spreadsheet and record the units at every step.

To show that fission and fusion are two routes towards the same peak of the binding-energy-per-nucleon curve, and thereby explain the sign and the size of the energy release $Q$.

A sketch graph: the $X$-axis is the mass number $A$ (logarithmic scale from $1$ to $240$), the $Y$-axis is the binding energy per nucleon $E_b/A$ in $\text{MeV}$ (from $0$ to $9$). The curve rises steeply from $^{2}\text{H}$ ($\approx 1.1\ \text{MeV}$) to a maximum at $^{56}\text{Fe}$ ($\approx 8.8\ \text{MeV}$), then falls slowly towards $^{235}\text{U}$ ($\approx 7.6\ \text{MeV}$). Points are marked on the curve for $^{2}\text{H}$, $^{3}\text{H}$, $^{4}\text{He}$, $^{12}\text{C}$, $^{24}\text{Mg}$, $^{56}\text{Fe}$, $^{141}\text{Ba}$, $^{92}\text{Kr}$, $^{235}\text{U}$. An arrow on the left leads from the light nuclei up to the peak, labelled "fusion, $Q>0$"; an arrow on the right leads from $^{235}\text{U}$ up to the medium-mass region, labelled "fission, $Q>0$". Below is a short legend: the height gained along the curve is proportional to the energy released.

A power-plant reactor splits uranium-235 nuclei as $^{235}\mathrm{U} + n \rightarrow {}^{141}\mathrm{Ba} + {}^{92}\mathrm{Kr} + 3n$, releasing about $200\ \mathrm{MeV} = 3.2\times10^{-11}\ \mathrm{J}$ per fission. A plant producing $1\ \mathrm{GW}$ of electricity (roughly $3\ \mathrm{GW}$ of thermal power at 33 % efficiency) therefore fissions $\frac{3\times10^{9}}{3.2\times10^{-11}} \approx 9\times10^{19}$ nuclei every second — less than $40\ \mathrm{mg}$ of uranium-235 per second. One gram of uranium-235 yields about $8\times10^{10}\ \mathrm{J}$, as much as burning roughly $3\ \mathrm{t}$ of coal.

This uses the mass–energy relation $E = \Delta m c^{2}$ through the mass defect: the fragments $^{141}\mathrm{Ba}$, $^{92}\mathrm{Kr}$ and three neutrons together are about $0.2\ \mathrm{u}$ lighter than the original $^{235}\mathrm{U}$ nucleus plus the neutron, and that missing mass becomes the kinetic energy of the flying fragments. At the same time the conservation laws of nuclear reactions are checked: nucleon number $235+1 = 141+92+3 = 236$ and charge $92 = 56+36$ match on both sides.

Every second the Sun turns about $600\times10^{6}\ \mathrm{t}$ of hydrogen into $595.7\times10^{6}\ \mathrm{t}$ of helium, so $\Delta m \approx 4.3\times10^{6}\ \mathrm{t}$ of mass disappears and appears as energy $E = \Delta m c^{2} \approx 3.9\times10^{26}\ \mathrm{J}$ per second — this is the Sun's luminosity $3.85\times10^{26}\ \mathrm{W}$. The net reaction $4{}^{1}\mathrm{H} \rightarrow {}^{4}\mathrm{He} + 2e^{+} + 2\nu$ releases $26.7\ \mathrm{MeV} = 4.3\times10^{-12}\ \mathrm{J}$. At Earth's orbit we receive $1.36\ \mathrm{kW/m^{2}}$ of this flow as sunlight.

The scenario uses the same mass defect as fission but in the direction of fusing light nuclei: four protons have mass $4\times1.007276\ \mathrm{u} = 4.029104\ \mathrm{u}$, while the $^{4}\mathrm{He}$ nucleus has $4.001506\ \mathrm{u}$; the difference $0.0276\ \mathrm{u}$ gives $25.7\ \mathrm{MeV}$, and annihilation of the two positrons adds another $1.02\ \mathrm{MeV}$. Energy is released because the binding energy per nucleon rises from $0$ in $^{1}\mathrm{H}$ to $7.07\ \mathrm{MeV}$ in $^{4}\mathrm{He}$.

In a tokamak, deuterium and tritium nuclei fuse as $^{2}\mathrm{H} + {}^{3}\mathrm{H} \rightarrow {}^{4}\mathrm{He} + n$, releasing $17.6\ \mathrm{MeV} = 2.8\times10^{-12}\ \mathrm{J}$ per reaction. The alpha particle carries $3.5\ \mathrm{MeV}$ and heats the plasma, the neutron carries $14.1\ \mathrm{MeV}$. Producing $500\ \mathrm{MW}$ of fusion power for $400\ \mathrm{s}$, that is $2\times10^{11}\ \mathrm{J}$, requires $\frac{2\times10^{11}}{2.8\times10^{-12}} \approx 7\times10^{22}$ reactions — only about $0.6\ \mathrm{g}$ of D–T fuel at a plasma temperature near $1.5\times10^{8}\ \mathrm{K}$.

The mass defect is visible directly in the tabulated masses: $2.014102\ \mathrm{u} + 3.016049\ \mathrm{u} = 5.030151\ \mathrm{u}$ before the reaction and $4.002602\ \mathrm{u} + 1.008665\ \mathrm{u} = 5.011267\ \mathrm{u}$ after it; the difference $0.018884\ \mathrm{u}$ gives $Q = \Delta m c^{2} = 17.6\ \mathrm{MeV}$. Conservation of charge and nucleon number ($2+3 = 5 = 4+1$) lets us write the reaction, while the high temperature is needed to overcome the Coulomb repulsion of the two positively charged nuclei.