nuclear reactions fission fusion and radiation safety

Nuclear reactions are written as balanced equations: the sums of $A$ and $Z$ on both sides must be equal, because nucleon number and electric charge are conserved.

Fission is the splitting of a heavy nucleus: for example $^{235}_{92}\mathrm{U}$ captures a neutron and breaks into two medium-mass fragments plus two or three neutrons, while fusion is the joining of light nuclei, such as $^{2}_{1}\mathrm{H}$ and $^{3}_{1}\mathrm{H}$ into $^{4}_{2}\mathrm{He}$ plus a neutron.

Both processes can release energy because the products sit closer to the maximum of the binding-energy-per-nucleon curve near $^{56}\mathrm{Fe}$; the released energy equals the mass defect multiplied by $c^2$:

$$E=\Delta m c^2,\qquad 1\,\mathrm{u}\approx 931.5\,\mathrm{MeV}/c^2.$$

Fission needs only a neutron to start and is controlled in a reactor by moderator, control rods and coolant, whereas fusion demands temperatures high enough to overcome the Coulomb barrier, which is why it powers stars and is still hard to sustain on Earth.

The same energetic particles and photons — alpha, beta, gamma and neutrons — ionise tissue, so radiation protection follows the ALARA principle: minimise time near the source, maximise distance (dose rate falls as $1/r^2$), and place a suitable shield between yourself and the source.

Alpha particles are stopped by a sheet of paper or by skin, beta particles by a few millimetres of plastic or aluminium, and gamma rays and neutrons need dense or hydrogen-rich materials, so the choice of shield depends on the type and energy of the radiation.

Balance the fission equation and identify the missing fragment:

$$^{235}_{92}\mathrm{U} + ^{1}_{0}\mathrm{n} \to ^{141}_{56}\mathrm{Ba} + ^{A}_{Z}\mathrm{Kr} + 3\,^{1}_{0}\mathrm{n}.$$

Give $A$ and $Z$ and name the fragment.

In the deuterium–tritium fusion reaction

$$^{2}_{1}\mathrm{H} + ^{3}_{1}\mathrm{H} \to ^{4}_{2}\mathrm{He} + ^{1}_{0}\mathrm{n}$$

the atomic masses are $2.014102\,\mathrm{u}$, $3.016049\,\mathrm{u}$, $4.002603\,\mathrm{u}$ and $1.008665\,\mathrm{u}$. Find the mass defect and the energy released in MeV.

A point gamma source produces a dose rate of $400\,\mu\mathrm{Sv/h}$ at $1.0\,\mathrm{m}$. A worker must work for $30\,\mathrm{min}$ at a distance of $2.5\,\mathrm{m}$ with no shield.

(a) What dose will the worker receive?

(b) If the local rule limits a single entry to $20\,\mu\mathrm{Sv}$, how long may the worker stay, and what other simple change would help?

A safe self-study plan (never handle radioactive sources — make observations only in simulations).

1. Open a nuclear-fission simulation (for example PhET “Nuclear Fission” or an equivalent) and start a chain reaction. Record how inserting absorbing control rods changes the number of fissions per second, and connect this to criticality.

2. In the same or another online simulation, build a reaction chain and balance at least three fission and fusion equations, checking conservation of $A$ and $Z$.

3. Using a table of atomic masses, compute $\Delta m$ and $E=\Delta m c^2$ for one fission reaction and one fusion reaction, and compare the energy per nucleon.

4. Test the inverse-square law with a safe analogue: a lamp plus a light meter (or a phone lux-meter app) at distances $0.5$, $1.0$, $1.5$, $2.0\,\mathrm{m}$. Plot illuminance against distance and confirm that it falls as $1/r^2$; transfer the conclusion to gamma dose rate.

5. Write a short safety protocol: for alpha, beta, gamma and neutron radiation, state a suitable shield, the role of distance and time, and express the ALARA rule in your own words.

Show why both fission and fusion release energy: each process moves the products toward the maximum of binding energy per nucleon, linking the binding-energy curve to real reactions.

A sketch graph: $x$-axis is mass number $A$ (from $1$ to $240$), $y$-axis is binding energy per nucleon $E_B/A$, peaking near $8.8\,\mathrm{MeV}$ at $^{56}\mathrm{Fe}$. An arrow to the left from the $^{235}\mathrm{U}$ point is labelled “fission”, an arrow to the right from the $^{2}\mathrm{H}$ and $^{3}\mathrm{H}$ points is labelled “fusion”; both point toward the top of the curve.

A smoke detector is mounted on the ceiling. Inside there is a tiny capsule containing about $0.3\,\mu\text{g}$ of americium-241 (activity $\sim 37\,\text{kBq}$). A $^{241}\text{Am}$ nucleus emits an $\alpha$-particle with energy about $5.5\,\text{MeV}$, becoming $^{237}\text{Np}$. The alpha particles ionize the air in the chamber, creating a small current. When smoke enters the chamber, ions attach to smoke particles, the current drops, and the alarm triggers.

Here, $\alpha$-decay operates — one type of nuclear reaction. The $\alpha$-particle energy $5.5\,\text{MeV}$ is the binding energy released due to the mass difference between the $^{241}\text{Am}$ nucleus and the decay products (mass defect). The range of $\alpha$-particles in air is only a few centimeters, so the capsule is safe: radiation does not escape the housing.

At a nuclear power plant, fuel assemblies contain ceramic uranium dioxide enriched to $3-5\%$ in $^{235}\text{U}$. When a thermal neutron is captured, a $^{235}\text{U}$ nucleus splits into two fragments (e.g., $^{141}\text{Ba}$ and $^{92}\text{Kr}$) and $2-3$ neutrons, releasing about $200\,\text{MeV}$ of energy. Complete fission of $1\,\text{g}$ of $^{235}\text{U}$ ($N = \frac{1}{235}N_A \approx 2.56\times10^{21}$ nuclei) yields $Q = N \cdot 200\,\text{MeV} \approx 8.2\times10^{10}\,\text{J}$ — equivalent to burning $2.5$ tonnes of coal.

The energy release in fission is explained by the binding energy curve: the $^{235}\text{U}$ nucleus has lower binding energy per nucleon than mid-mass fragments, so the mass difference $\Delta m$ converts into kinetic energy of fragments and neutrons via $E=\Delta m c^2$. The chain reaction is sustained by emitted neutrons and controlled by boron or cadmium rods that absorb neutrons.

In the Sun's core at about $15\,\text{million K}$ and $2.5\times10^{11}\,\text{atm}$, the proton-proton cycle proceeds: $4\,^1\text{H} \rightarrow ^4\text{He} + 2e^+ + 2\nu_e + \gamma$, releasing $26.7\,\text{MeV}$ per helium nucleus. The ITER project plans the reaction $D + T \rightarrow ^4\text{He} + n + 17.6\,\text{MeV}$. One kilogram of deuterium-tritium mixture could yield $\sim 3.4\times10^{14}\,\text{J}$ — like $10,000$ tonnes of oil.

Fusion of light nuclei is also based on the mass defect: the mass of $^4\text{He}$ is less than the sum of four protons, and the difference $\Delta m$ is released as energy via $E=\Delta m c^2$. The binding energy per nucleon of helium-4 is maximal among light nuclei, making fusion energetically favorable. Neutrons in the D-T reaction carry $14.1\,\text{MeV}$ and require radiation shielding (concrete, water), so fusion facilities have biological shielding.