science-2018:P.E.5
science-2018:P.E.5 Nuclear energy
Learning goal: Describe how alpha, beta, positron and gamma emission each change a nucleus, predict the product of a given decay or identify the decay from its reactants and products, explain why nuclear reactions release far more energy than chemical reactions, and use mass-energy equivalence, E = mc², to calculate the energy released in fission and fusion from given masses.
Nuclei and their notation. A nucleus is described by its atomic number Z (the number of protons, which fixes the element) and its mass number A (protons plus neutrons). Uranium-238 has Z = 92 and A = 238, so it has 146 neutrons. Isotopes of one element share Z and differ in A. Some nuclei are unstable and spontaneously change into other nuclei; this is radioactive decay. In every nuclear change two quantities are conserved and let you predict the products: the total mass number and the total charge. Write the reactants and products with A as a superscript and Z as a subscript, and make both columns balance.
Four kinds of emission, four different changes. Alpha emission ejects an alpha particle, which is a helium-4 nucleus (two protons, two neutrons). The nucleus loses 4 from A and 2 from Z, so it becomes the element two places earlier in the periodic table: uranium-238 becomes thorium-234. Beta-minus emission converts a neutron into a proton and ejects a high-speed electron (and an antineutrino). A is unchanged and Z rises by one: carbon-14 becomes nitrogen-14. Positron emission converts a proton into a neutron and ejects a positron, the electron's positively charged antiparticle (and a neutrino). A is unchanged and Z falls by one: sodium-22 becomes neon-22. Gamma emission is a photon of very high-frequency electromagnetic radiation released when a nucleus drops from an excited state to a lower-energy state; neither A nor Z changes, only the energy of the nucleus. The three particle emissions differ in mass, charge and penetrating power: alpha particles are massive and doubly charged and are stopped by paper or skin; beta particles are stopped by a few millimeters of metal; gamma rays need thick lead or concrete. The identity of the new element can be predicted for alpha and beta decays by the rules above. Many heavy nuclei decay through a decay series: a chain of alpha and beta steps ending at a stable nucleus, which is why uranium ore contains a whole family of daughter isotopes. Each step has its own half-life, the time for half of a large sample of that nuclide to decay, which can range from fractions of a second to billions of years.
Fission and fusion. In fission a heavy nucleus such as uranium-235 absorbs a neutron and splits into two medium-sized nuclei plus two or three neutrons. Those neutrons can cause further fissions, and a chain reaction results; in a nuclear power plant the chain reaction is held at a steady rate by control rods that absorb neutrons and by a moderator that slows the neutrons to the speeds at which uranium-235 absorbs them best. In fusion two light nuclei, such as the hydrogen isotopes deuterium (hydrogen-2) and tritium (hydrogen-3), join to form a heavier nucleus, releasing energy. Fusion of hydrogen nuclei is also the Sun's energy source, although in the Sun it runs through a different chain of reactions (the proton-proton chain, which turns four hydrogen nuclei into one helium nucleus). Fusion requires enormous temperatures and pressures because the positively charged nuclei must be forced close enough for the strong nuclear force to bind them. Nuclear reactions are accompanied by energy changes far larger than chemical reactions: a chemical bond change involves a few electronvolts per atom, while a single nuclear reaction releases millions of electronvolts (MeV) per nucleus, roughly a million times more per particle.
Mass-energy equivalence. During any nuclear process, the energy transferred out of the system is directly proportional to the decrease in the system's mass: E = mc², where c = 3.00 × 10⁸ m/s. Because c² is so large (9.00 × 10¹⁶ m²/s²), a very small loss of mass is accompanied by the release of a very large amount of energy. In the energy-releasing reactions studied here (radioactive decay, the fission of heavy nuclei and the fusion of light ones) the products have less total mass than the nuclei that went in; the difference, called the mass defect, shows up as released energy, mostly as kinetic energy of the products and as gamma radiation. In general the energy released is Q = (mass of reactants − mass of products) × c²: a positive Q means energy comes out, while a reaction whose products are more massive than its reactants (fusing nuclei heavier than iron, for example) has a negative Q and needs an energy input. Nuclear masses are usually given in atomic mass units (u), with 1 u = 1.6605 × 10⁻²⁷ kg, and a convenient conversion is that 1 u of mass corresponds to 931.5 MeV of energy, where 1 MeV = 1.602 × 10⁻¹³ J.
Worked model: deuterium-tritium fusion. ²H + ³H → ⁴He + n. Suppose the problem gives these masses: ²H 2.014102 u, ³H 3.016049 u, ⁴He 4.002603 u, neutron 1.008665 u. Reactants: 2.014102 + 3.016049 = 5.030151 u. Products: 4.002603 + 1.008665 = 5.011268 u. Mass defect: 5.030151 − 5.011268 = 0.018883 u. Energy released: 0.018883 × 931.5 = 17.6 MeV. In SI units: Δm = 0.018883 × 1.6605 × 10⁻²⁷ = 3.136 × 10⁻²⁹ kg, and E = (3.136 × 10⁻²⁹)(3.00 × 10⁸)² = 2.82 × 10⁻¹² J, which is the same 17.6 MeV. Both routes must agree; if they do not, a mass was copied wrongly. Check the balance too: A: 2 + 3 = 4 + 1; Z: 1 + 1 = 2 + 0.
Misconception: the nucleons disappear. They do not. In the D-T reaction five nucleons go in and five come out. What decreases is the mass of the system, because the products are more tightly bound; binding energy shows up as missing mass. The nuclei have not been "turned into energy" wholesale; a fraction of a percent of their mass has.
Electricity from fission: weighing the argument. A fission plant uses the thermal energy of the reactor core to boil water, drive a turbine and turn a generator, the same final steps as a coal-fired plant but with no combustion. Advantages include a very large energy output from a small mass of fuel and no carbon dioxide emitted during generation. Disadvantages include radioactive waste that must be stored securely for very long times because of long-lived isotopes in its decay series, the consequences of a cooling failure, the mining and enrichment of fuel and the cost of decommissioning. Fusion would avoid most of the waste problem but is not yet a practical source of electricity. A sound argument for or against a plant in a particular place relates these scientific principles, chain reaction control, half-lives of the waste, thermal energy needing cooling water, to the local conditions, and presents evidence rather than slogans.
Investigation: a model of random decay. Nothing radioactive is handled. Materials: 100 dice, a tray, graph paper. (With 100 coins instead, remove every coin showing heads; the chance of "decay" per toss is then 1/2, so the half-life is one toss rather than about four rolls.) Steps: roll all the dice; remove every die showing a six (it has "decayed"); count the survivors; roll the survivors again; repeat for 15 rolls and plot survivors against roll number. Compare with a second group's data. A result that supports the model: the curve falls by roughly the same fraction each roll (with random scatter that grows as the dice run out) rather than by the same number, and the number of rolls needed to halve the sample is about the same whether you start from 100 or from 50 (the model's half-life is ln 2 ÷ ln(6/5) ≈ 3.8 rolls). What the model cannot show: which particles are emitted, the energy released, the real half-lives (seconds to billions of years) or the way a decay series steps through several elements; it shows only that decay is random for each nucleus and predictable for a large sample.
Worked example 1 — predict an alpha-decay product. Radium-226 (Z = 88) emits an alpha particle. A: 226 − 4 = 222; Z: 88 − 2 = 86. Element 86 is radon, so the product is radon-222: ²²⁶Ra → ²²²Rn + ⁴He. Check: mass numbers 226 = 222 + 4; charges 88 = 86 + 2.
Worked example 2 — identify the process. Phosphorus-32 (Z = 15) becomes sulfur-32 (Z = 16). The mass number is unchanged and Z rose by one, so a neutron became a proton and an electron was emitted: beta-minus decay, ³²P → ³²S + e⁻ (+ antineutrino). Had Z fallen by one with A unchanged, it would be positron emission (or electron capture, which changes A and Z in the same way and is told apart by the particles observed); had A fallen by 4 and Z by 2, alpha emission; had nothing changed but the energy, gamma emission.
Worked example 3 — fusion energy from the mass defect. As in the lesson: D-T fusion, mass defect 0.018883 u, energy 17.6 MeV = 2.82 × 10⁻¹² J per reaction. To put the scale in view: one mole of such reactions (6.02 × 10²³) would release 1.7 × 10¹² J, while burning one mole of a typical fuel molecule releases of the order of 10⁶ J, roughly a million times less.
Worked example 4 — fission energy from given masses. ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n, with the problem giving ²³⁵U = 235.043930 u, ¹⁴¹Ba = 140.914411 u, ⁹²Kr = 91.926156 u, n = 1.008665 u. Reactants: 235.043930 + 1.008665 = 236.052595 u. Products: 140.914411 + 91.926156 + 3(1.008665) = 235.866562 u. Mass defect: 0.186033 u. Energy: 0.186033 × 931.5 = 173 MeV, about ten times the D-T fusion energy per reaction, from a nucleus about fifty times heavier. The two fragments are themselves radioactive, and their later decays release more energy, so the total per uranium-235 fission is about 200 MeV (sources give about 200 to 215 MeV, depending on whether the energy carried off by neutrinos is counted). Balance check: A: 235 + 1 = 141 + 92 + 3; Z: 92 + 0 = 56 + 36 + 0.
Representation task: a decay chart. Make a four-row table for alpha, beta-minus, positron and gamma emission with columns for the particle emitted, its charge, the change in A, the change in Z, an example reaction written with superscripts and subscripts, and one sentence on penetrating power. Below it, write out and balance two decays given only the parent: bismuth-210 (Z = 83) by beta-minus emission, and polonium-210 (Z = 84) by alpha emission, naming each product element from a periodic table.
Independent performance: the energy of a nuclear reaction and a position on a power plant. Part 1: using masses provided by your teacher, calculate the mass defect and the energy released for one fusion and one fission reaction, in both MeV and joules, showing the balance of A and Z, and compare each with the energy of a chemical reaction of your choice per atom. Part 2: write a one-page evidence-based recommendation for or against building a fission power plant in a hypothetical river town. Use this invented site description as the local conditions: a town of 40 000 people on a large river whose flow can supply cooling water all year; the proposed site is 6 km upstream of the town, on ground above the highest recorded flood; the region has no record of damaging earthquakes; the nearest existing power station burns coal. Success criteria: the argument explains how a controlled chain reaction produces thermal energy and how that becomes electrical energy; it uses the ideas of half-life and decay series to explain why waste storage is a long-term issue; it identifies one design feature (for example, cooling arrangements or containment) that would make the plant more suitable for the location; and it names at least one advantage and one disadvantage supported by physics rather than opinion.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Polonium-210 (Z = 84) undergoes alpha decay. Write the balanced equation and name the product.
Alpha decay removes 4 from A and 2 from Z: A = 210 − 4 = 206, Z = 84 − 2 = 82. Element 82 is lead, so ²¹⁰Po → ²⁰⁶Pb + ⁴He. Both mass number (210 = 206 + 4) and charge (84 = 82 + 2) balance, which is how the product is predicted.
Potassium-40 (Z = 19) can decay to calcium-40 (Z = 20). Identify the decay process and explain your reasoning.
A is unchanged (40 → 40) and Z rises by one (19 → 20), so a neutron in the nucleus has become a proton, which is beta-minus decay: an electron (and an antineutrino) is emitted. Alpha decay would lower A by 4; positron emission would lower Z; gamma emission would change neither. Only beta-minus fits the change.
What is the mass defect of a nuclear reaction, and how is it related to the energy released?
In an energy-releasing nuclear reaction, the mass defect is the amount by which the total mass of the products is less than the total mass of the reactants. By mass-energy equivalence, E = mc², that missing mass appears as energy released, mostly as kinetic energy of the products and gamma radiation (a reaction whose products have more mass than its reactants absorbs energy instead). Because c² = 9.00 × 10¹⁶ m²/s², a very small mass loss corresponds to a very large energy; 1 u of mass corresponds to 931.5 MeV.
A fusion reaction has reactants of total mass 6.000000 u and products of total mass 5.980000 u (values given). Calculate the energy released in MeV and in joules.
Mass defect = 6.000000 − 5.980000 = 0.020000 u. Energy = 0.020000 × 931.5 MeV/u = 18.63 MeV. In joules: 18.63 MeV × 1.602 × 10⁻¹³ J/MeV = 2.98 × 10⁻¹² J. Alternatively Δm = 0.020000 × 1.6605 × 10⁻²⁷ kg = 3.32 × 10⁻²⁹ kg and E = (3.32 × 10⁻²⁹)(3.00 × 10⁸)² = 2.99 × 10⁻¹² J; the two routes agree to rounding.
Why do nuclear reactions release so much more energy than chemical reactions, and roughly how much more per particle?
Chemical reactions rearrange electrons, changing energies of a few electronvolts per atom, while nuclear reactions rearrange nucleons bound by the strong nuclear force, changing energies of millions of electronvolts per nucleus. The D-T fusion reaction releases 17.6 MeV and the fission of one uranium-235 nucleus about 200 MeV once the later radioactive decays of the fragments are counted (the reaction in Worked example 4, which stops at the first products, gives 173 MeV), roughly a million times the energy of a chemical bond change. The mass defect makes this visible: a measurable fraction of a percent of the mass is converted to energy in a nuclear reaction, an amount far too small to detect in a chemical one.
Describe how a fission power plant turns nuclear energy into electrical energy and name one advantage and one disadvantage supported by physics.
Uranium-235 nuclei absorb neutrons and split, releasing energy and more neutrons; control rods absorb neutrons so the chain reaction runs at a steady rate. The kinetic energy of the fission fragments becomes thermal energy in the fuel, which heats water to make steam that drives a turbine connected to a generator, where electromagnetic induction produces the electrical energy. Advantage: a very small mass of fuel gives a very large energy output with no combustion products. Disadvantage: fission products include isotopes with long half-lives and long decay series, so the waste must be isolated for a very long time, and the core must be cooled continuously.
In the dice model of decay, why does removing the sixes each roll model radioactive decay well, and what does the model fail to show?
Each die has the same fixed chance (1 in 6) of 'decaying' on each roll regardless of its history, and which dice decay is unpredictable, just as each nucleus has a fixed probability of decaying per unit time and no memory. As a result about the same fraction of the sample decays per roll on average, so the predicted halving interval (about 3.8 rolls) is the same whatever the starting number, which is the meaning of half-life; any single run scatters around that prediction, and the scatter is larger when few dice are left. The model cannot show the particles emitted, the energy released, the actual time scale, or the way a decay series passes through several elements.