Nuclear Reaction & Mass Defect Calculator

Nuclear Energy Guide

What do I need to know about Mass Defect and E=mc2?

Nuclear mass is always less than the sum of its protons and neutrons. The mass deficit converts to binding energy: E = delta-m times c2. 1 atomic mass unit = 1.66054 x 10 to the -27 kg. 1u times c2 = 931.5 MeV. Mass defect of Fe-56: 26 protons + 30 neutrons - actual mass = 0.52847u. Binding energy = 0.52847 x 931.5 = 492.3 MeV = 8.79 MeV/nucleon. Iron has the highest binding energy per nucleon — neither fission nor fusion of iron releases energy.

What do I need to know about Nuclear Fission?

U-235 + n gives Ba-141 + Kr-92 + 3n + energy. Energy per fission approx 173 MeV. Energy per kg U-235 = approximately 71 TJ. Compare: 1kg coal releases 30 MJ. Nuclear fission is about 2 million times more energy dense than chemical fuel.

What's the key thing to understand about Nuclear Fusion?

D-T fusion: deuterium + tritium gives helium-4 + neutron + 17.6 MeV. Energy per kg fuel approx 337 TJ. Requires 100 million degrees Celsius to overcome the Coulomb barrier. Stars achieve this through gravitational compression. ITER fusion reactor currently under construction in France.

What's the difference between Fission and Fusion Waste?

Fission waste: long-lived radioactive fission products (Cs-137 half-life 30 years, Sr-90 half-life 29 years) require secure storage for centuries. Fusion waste: primarily helium (inert) and neutron-activated reactor components (half-lives of decades, not centuries). This is one of the key advantages of fusion power over fission.

What should I know about Binding Energy Per Nucleon?

Binding energy per nucleon — the energy needed to pull a single proton or neutron out of a nucleus, divided by the total number of nucleons — explains why both nuclear fission and nuclear fusion release energy, even though one splits atoms and the other joins them. When plotted against atomic mass number, binding energy per nucleon rises steeply for light elements, peaks around iron-56 and nickel-62 (the most tightly bound nuclei known), and then gradually declines for heavier elements. This shape means that fusing light nuclei (like hydrogen isotopes into helium) moves them toward the peak, releasing energy, while splitting very heavy nuclei (like uranium-235) also moves the resulting fragments toward the peak, likewise releasing energy. Any nuclear reaction that moves nucleons closer to the iron/nickel peak of this curve is energetically favourable and releases energy, while a reaction that would move further away from the peak would require energy input instead. This single curve is the unifying explanation for why both the Sun's fusion reactions and a nuclear reactor's fission reactions are net energy sources.

What's the key thing to understand about Half-Life and Radioactive Decay?

Many products of nuclear reactions are radioactive and decay over time according to a predictable exponential pattern governed by their half-life — the time required for exactly half of a given quantity of radioactive atoms to decay. The decay follows N = N₀ × (1/2)^(t / T), where N₀ is the initial quantity, N is the remaining quantity after time t, and T is the half-life. For example, iodine-131 (a common fission product) has a half-life of about 8 days, so starting from 100g, about 50g remains after 8 days, 25g after 16 days, 12.5g after 24 days, and so on — the quantity never technically reaches exactly zero, only approaches it indefinitely. Half-lives vary enormously between isotopes, from fractions of a second to billions of years, and this variation is exactly why some nuclear waste products (like certain fission fragments) become safe relatively quickly, while others (like some actinides) require storage and isolation for thousands of years before their radioactivity decays to background levels.

What do I need to know about Conservation Laws in Nuclear Equations?

Every balanced nuclear equation must conserve both mass number (A, the total number of protons plus neutrons) and atomic number (Z, total charge/proton count) on both sides — alongside the usual conservation of momentum and total energy (including the energy equivalent of any mass change, via E=mc²). For example, in alpha decay ⁴²⁴²⁴⁴U → ²⁴⁰⁰₉₀Th + ⁴₂He, the mass numbers (238 = 234+4) and atomic numbers (92 = 90+2) both balance. Checking these two totals is the quickest way to verify any nuclear equation is written correctly, before worrying about energy released.

What should I know about Chain Reactions and Critical Mass?

A single fission event releases 2–3 free neutrons, and if enough of those neutrons go on to trigger further fissions in nearby nuclei, the result is a self-sustaining chain reaction. Whether a chain reaction grows, dies out, or holds steady depends on the effective multiplication factor k: the average number of neutrons from one fission that cause another fission. When k < 1 (subcritical), each generation produces fewer fissions than the last and the reaction fades; when k = 1 (critical), the reaction rate stays constant, which is the controlled, steady-state condition maintained inside nuclear power reactors; when k > 1 (supercritical), each generation produces more fissions than the last, and the reaction rate grows — rapidly and uncontrollably if k is well above 1, as in a nuclear weapon. Critical mass is the minimum amount of fissile material (such as uranium-235 or plutonium-239) needed to sustain k = 1, since a larger mass has a larger volume-to-surface-area ratio, reducing the fraction of neutrons that escape before causing further fissions. Reactors stay safely subcritical to supercritical-by-design using control rods that absorb neutrons, letting operators fine-tune k to hold the reaction at a steady, controllable power level.

Worked Example: Energy Released in Fission of Uranium-235

A single U-235 nucleus undergoes fission, and the total mass of the fission products (plus neutrons) is measured to be 0.215 atomic mass units (u) less than the starting U-235 plus the neutron that triggered it. Using E = mc² with 1 u = 1.6605×10⁻²⁷ kg: Δm = 0.215 u = 0.215 × 1.6605×10⁻²⁷ kg = 3.570×10⁻²⁸ kg E = Δmc² = 3.570×10⁻²⁸ × (3×10⁸)² = 3.570×10⁻²⁸ × 9×10¹⁶ ≈ 3.21×10⁻¹¹ J Converting to the more commonly quoted unit of MeV (1 MeV = 1.602×10⁻¹³ J): E ≈ 3.21×10⁻¹¹ / 1.602×10⁻¹³ ≈ 200 MeV — matching the textbook figure for U-235 fission. Scaling up to a realistic mass: one mole of U-235 fully fissioning (6.022×10²³ nuclei) releases 200 MeV × 1.602×10⁻¹³ J/MeV × 6.022×10²³ ≈ 1.93×10¹³ J, roughly 19.3 terajoules from just 235 grams of fuel — illustrating why nuclear fuel has an energy density roughly a million times greater than chemical fuels like coal or oil.

Fission vs. Fusion Energy Yield Compared

Although a single fission event (200 MeV) releases roughly ten times more energy than a single fusion event (17.6 MeV), fusion fuel is vastly lighter per atom involved (hydrogen isotopes vs. uranium), so fusion actually releases more energy per kilogram of fuel than fission does. Both nuclear processes dwarf chemical combustion by a factor of millions per reaction, which is the fundamental reason nuclear fuel — whether fission or fusion — needs only a tiny mass to match the energy output of a huge mass of fossil fuel.

Critical Mass and Chain Reaction Basics

A fission chain reaction is sustained when, on average, each fission event triggers at least one more fission event — described by the neutron multiplication factor k. When k = 1 exactly, the reaction is 'critical' and proceeds at a steady rate (as in a nuclear power reactor, deliberately controlled to hold k at 1). When k > 1, the reaction is 'supercritical' and the rate grows exponentially (as in a weapon, or an uncontrolled reactor excursion). When k < 1, the reaction is 'subcritical' and dies out. The critical mass — the minimum amount of fissile material needed to sustain k = 1 — depends on the material's shape, density, purity, and whether a neutron-reflecting material surrounds it, which is why reactor and weapon designs manipulate geometry and reflectors as carefully as they manipulate the quantity of fissile material itself.

Half-Life: A Quick Worked Example

A radioactive isotope has a half-life of 8 days. Starting with 80 g, how much remains after 24 days? Since 24 days is exactly 3 half-lives (24/8 = 3), the remaining amount is found by halving three times: 80 → 40 → 20 → 10 g. Using the general formula N = N₀ × (1/2)^(t/T½): N = 80 × (1/2)^3 = 80 × 0.125 = 10 g, confirming the step-by-step result. This simple doubling/halving pattern only works cleanly when the elapsed time is a whole-number multiple of the half-life; for any other elapsed time, the exponential formula is needed directly rather than relying on simple halving steps.