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E.4

Fission

Theme E · Nuclear and quantum physics · SL and HL

E.3 established that splitting a heavy nucleus releases energy. This topic is about what it takes to do that a hundred billion billion times a second, in a controlled way, in a building — and about the problems that come with succeeding.

🎯What you need to be able to do

  • Describe induced fission, and write a balanced fission equation.
  • Explain the chain reaction in terms of the multiplication factor \(k\), and what critical mass means.
  • Explain why the neutrons must be slowed, and what the moderator does about it.
  • State the job of each part of a reactor: fuel, moderator, control rods, coolant, heat exchanger, containment.
  • Explain why uranium must be enriched, and what that has to do with weapons.
  • Explain why fission products are radioactive, and discuss the waste that follows.
  • Evaluate nuclear power against the alternatives, with both halves of the argument.
  • Outline nuclear fusion as a power source, the conditions it needs, and why it is not yet working.

⚛️Induced fission

A uranium-235 nucleus that absorbs a neutron becomes uranium-236, and uranium-236 is so unstable that it deforms and splits within about \( 10^{-14} \) s. That is induced fission: not something the nucleus does on its own, but something a neutron makes it do.

\[ ^{235}_{\ 92}\text{U} + ^{1}_{0}\text{n} \rightarrow\ ^{141}_{\ 56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n} \]

The split is not always into barium and krypton — a nucleus this size can come apart in hundreds of different ways, and the fragments are usually unequal, with nucleon numbers around 95 and 140. What is always true is that two or three neutrons come out, and that the two balances hold: \( 235 + 1 = 141 + 92 + 3 \), and \( 92 + 0 = 56 + 36 + 0 \).

A slow neutron approaches a uranium-235 nucleus, which absorbs it and deforms, then splits into a barium-141 nucleus and a krypton-92 nucleus flying apart in opposite directions, repelled electrostatically, together with three more neutrons. The equation is written out underneath with both balances shown: nucleon number 235 plus 1 equals 141 plus 92 plus 3, and proton number 92 plus 0 equals 56 plus 36. A bar chart shows where the energy goes: 168 MeV as kinetic energy of the fragments, 14 MeV as gamma rays, 8 MeV as beta particles, 5 MeV as kinetic energy of the neutrons and 12 MeV carried away by antineutrinos, of a total 207 MeV, of which 195 MeV becomes heat. A note explains that working the mass difference for this exact split gives 173 MeV rather than the 200 MeV usually quoted, because the fragments are themselves radioactive and their later decays release the rest.
Over 80% of the energy is simply the kinetic energy of the two fragments, flying apart because each is highly positively charged and they are suddenly next to each other. It is electrostatic repulsion that does the work.

Two details in that figure repay attention. First, the neutron that starts it must be slow — the next section explains why. Second, if you work out the energy released from the mass difference for this particular split you get 173 MeV, not the 200 MeV usually quoted. Both figures are right: 173 MeV is the prompt release for one channel, and the fission fragments are themselves radioactive, so their subsequent beta and gamma decays give up the rest over the following hours and years.

✏️Worked example 1 — completing a fission equation

A uranium-235 nucleus absorbs a neutron and splits into strontium-94 and a second fragment, releasing two neutrons. Identify the second fragment. Then estimate the energy released, given that uranium-235 has a binding energy per nucleon of 7.59 MeV and both fragments have about 8.5 MeV per nucleon.

Balance the nucleon numbers. The left-hand side has \( 235 + 1 = 236 \). The right has \( 94 + A + 2 \), so \( A = 140 \).

Balance the proton numbers. Strontium is \( Z = 38 \). The left has \( 92 + 0 = 92 \), so \( Z = 92 - 38 = 54 \), which is xenon.

\[ ^{235}_{\ 92}\text{U} + ^{1}_{0}\text{n} \rightarrow\ ^{94}_{38}\text{Sr} + ^{140}_{\ 54}\text{Xe} + 2\,^{1}_{0}\text{n} \]

Energy, from the binding-energy curve. Total binding energy before, for the 236 nucleons in the compound nucleus:

\[ 236 \times 7.59 = 1791\ \text{MeV} \]

And after, for the two fragments (the free neutrons have no binding energy):

\[ (94 + 140) \times 8.5 = 234 \times 8.5 = 1989\ \text{MeV} \]
\[ E = 1989 - 1791 = 198 \approx 2 \times 10^{2}\ \text{MeV} \]
Sanity check. Around 200 MeV is the number to expect for any fission of a uranium-sized nucleus, so this agrees. Note the direction: binding energy increased, and the increase is what was released. Getting that sign the wrong way round is the standard error here — if your answer comes out negative, you have subtracted the wrong way.

🔗The chain reaction, and one number that decides everything

Each fission produces two or three neutrons, so each fission can in principle cause more than one more. Whether it actually does is the whole question, and it is captured in a single number: the multiplication factor \(k\), the average number of neutrons from one fission that go on to cause another.

Three panels comparing successive generations of fissions. Subcritical, with k less than one, starts with eight fissions and falls to four, then two, then one, so the reaction dies out because too many neutrons escape or are absorbed without fissioning. Critical, with k equal to one, keeps four fissions in every generation, giving the steady output of a power reactor because exactly one neutron per fission goes on to cause the next. Supercritical, with k greater than one, grows from one to two to four to eight, so the rate grows every generation, which is how a reactor is started up and how a weapon works. A note explains that a fission makes two or three neutrons so k could in principle be three, but in practice most are lost by escaping the surface, being absorbed by uranium-238, or being absorbed by a control rod.
A power reactor is a chain reaction held deliberately at \( k = 1 \), minute after minute, for months. Start-up means nudging \(k\) fractionally above 1 and then bringing it back; shutdown means driving it well below.

A fission makes two or three neutrons, so why is \(k\) not simply 2 or 3? Because most neutrons never cause a fission. Some escape through the surface of the fuel. Some are absorbed by uranium-238, which is over 99% of natural uranium and which mostly captures neutrons without fissioning. Some are absorbed by the moderator, the coolant, or the structure. And in a working reactor, some are absorbed deliberately, by the control rods.

⚖️Critical mass

Because escape through the surface is one of the loss routes, the size of the lump of fuel matters directly. Fissions happen throughout the volume, which goes as \( r^3 \); escapes happen through the surface, which goes as \( r^2 \). So the fraction lost scales as

\[ \frac{\text{surface}}{\text{volume}} = \frac{4\pi r^2}{\tfrac{4}{3}\pi r^3} = \frac{3}{r} \]

Double the radius and you halve the fraction escaping. Above some radius, enough neutrons stay in that the chain sustains itself, and the mass at that radius is the critical mass.

Two spheres of fissile material compared. The small one, labelled too small, has nine neutrons escaping through its surface and only three absorbed inside, so most neutrons reach the surface before they hit anything and k is less than one. The larger one, labelled big enough, has only four escaping and fourteen absorbed inside, so the chain sustains and k is at least one. A panel explains the geometry: fissions happen throughout the volume, which goes as r cubed, while escapes happen through the surface, which goes as r squared, so the escaping fraction is proportional to area over volume, which is three over r, and above some radius enough neutrons stay in. Figures follow: a bare sphere of pure uranium-235 has a critical mass of about 52 kilograms, a radius of 8.7 centimetres at a density of 18700 kilograms per cubic metre, under 20 kilograms if surrounded by a neutron reflector, and no single critical lump at all when spread through a moderator in a reactor core.
A bare sphere of pure uranium-235 goes critical at about 52 kg — a ball 17 cm across. Surround it with a neutron reflector, which bounces escaping neutrons back in, and the figure drops below 20 kg.

🎸Why the neutrons have to be slowed down

Here is the awkwardness at the centre of reactor design. Fission produces neutrons with kinetic energies around 2 MeV — fast neutrons. But uranium-235 is hundreds of times more likely to absorb a slow one. A neutron in thermal equilibrium with matter at room temperature has an energy of about \( kT = 0.025 \) eV, and at that energy U-235 fissions readily; at 2 MeV it mostly does not.

On the left, a log-log graph of the fission cross-section of uranium-235 in barns against neutron energy in electronvolts. The cross-section falls steeply from 585 barns at the thermal energy of 0.025 electronvolts, following an inverse-speed law, through a shaded region between 1 and ten thousand electronvolts described as a forest of resonances, and down to about 1 barn at the 2 mega-electronvolt energy of a neutron straight out of a fission. On the right, a bar chart of the maximum fraction of kinetic energy a neutron can lose in one head-on collision: 100 per cent with hydrogen in ordinary water, 88.9 per cent with deuterium in heavy water, 28.4 per cent with carbon as graphite, and only 1.7 per cent with uranium itself, so the moderator must be water or graphite. A note explains that a neutron bouncing off a heavy nucleus keeps almost all its speed, like a marble hitting a bowling ball, while off a nucleus of its own mass it can stop dead.
The vertical scale spans three decades: at thermal energies U-235 presents a target some five hundred times larger than it does to a fast neutron. Slowing the neutrons down is not an optimisation, it is the difference between working and not working.

So the neutrons must be slowed, from 2 MeV to 0.025 eV — a factor of about \( 10^{8} \) in energy. That is the moderator's job, and it does it by elastic collisions. The physics of those collisions decides what the moderator can be made of: in a head-on elastic collision, a light particle transfers most of its kinetic energy only to a target of similar mass. A neutron bouncing off a uranium nucleus loses under 2% of its energy; off a hydrogen nucleus, of almost exactly its own mass, it can lose all of it.

Hence water or graphite. Around twenty collisions with hydrogen will thermalise a fission neutron; carbon needs about a hundred. In a pressurised water reactor the moderator and the coolant are the same water, which turns out to be a useful safety property: lose the water and the chain reaction stops, because there is nothing left to slow the neutrons.

The trap: a moderator and a control rod do opposite things. The moderator slows neutrons and keeps them in play, making fission more likely. Control rods absorb neutrons and take them out of play, making fission less likely. Water and graphite moderate; boron and cadmium absorb. Writing that the moderator “controls the reaction by absorbing neutrons” is one of the most commonly seen errors on this topic, and it loses the mark outright.

🏭Inside a reactor

A pressurised water reactor is the commonest design in the world, and every part of it has exactly one job. Exam questions almost always ask for the job, not the name.

A schematic of a pressurised water reactor. Inside a containment building of thick steel and concrete sits the reactor vessel, holding fuel rods in water with control rods withdrawn above them on dashed insertion paths. Water leaves the vessel hot at 325 degrees Celsius, passes through a heat exchanger, also called the steam generator, and is pumped back cooler: this primary loop is sealed and radioactive and held at 155 bar so the water cannot boil. On the secondary side, steam leaves the heat exchanger, drives a turbine coupled to a generator producing about a gigawatt of electricity for the grid, then passes to a condenser where it turns back to water and is pumped to the heat exchanger again; this secondary loop is clean water and steam and is never radioactive. Cooling water from a river or tower passes through the condenser and leaves warm, carrying the waste heat. A table lists the job of each part: fuel, moderator, control rods, coolant, heat exchanger and containment.
The primary water is both coolant and moderator, and it never leaves containment. From the heat exchanger onwards this is an ordinary steam power station, subject to the same Carnot limit as any other — about a third of the heat becomes electricity.
fueluranium enriched to 3–5% U-235, sealed in rods
moderatorslows fast neutrons so U-235 will absorb them
control rodsboron or cadmium: absorb neutrons, and so set \(k\)
coolantcarries the heat to the exchanger — and stops the core melting
heat exchangerheat crosses from primary to secondary; the water does not
containmentthe last barrier if every other one fails

The pressure is worth a sentence of its own. The primary water is held at about 155 times atmospheric pressure so that it stays liquid at 325 °C. Water that hot at ordinary pressure would flash to steam, and steam is a far worse coolant — and a far worse moderator.

⚛️Enrichment, and what it has to do with weapons

Natural uranium is 99.27% uranium-238 and only 0.72% uranium-235. Uranium-238 does not fission with slow neutrons; it absorbs them. So natural uranium in a PWR cannot sustain a chain reaction, and the fuel must be enriched to a few per cent U-235.

Enrichment is difficult in a specific way: U-235 and U-238 are the same element, so no chemical process can separate them. It has to be done on the mass difference alone — about 1% — by gas centrifuges running in long cascades.

On the left, a bar chart of uranium-235 content by grade: natural uranium as mined is 0.72 per cent and will not sustain a chain in a pressurised water reactor; PWR reactor fuel is about 4 per cent, enriched typically to between 3 and 5; research reactor fuel is 20 per cent, the legal line for low enriched uranium; and weapons grade is 90 per cent. A note says enrichment separates two isotopes of one element, so no chemistry can do it. On the right, a logarithmic comparison of the energy released by one kilogram of different fuels: firewood at 16 megajoules, coal at 30, petrol at 46, and uranium-235 fissioned at 82 terajoules, which is about 2700 tonnes of coal. Below, a note that a one gigawatt station burns about 1.2 tonnes of uranium-235 a year while a coal station of the same output burns 8700 tonnes a day, and a warning that the same enrichment plant that makes 4 per cent fuel can, run for longer, make 90 per cent weapons material.
The same cascade that enriches to 4% will, run for longer, enrich to 90%. That single fact is why civil nuclear power carries a proliferation question that no other generating technology does, and why enrichment plants are inspected.

✏️Worked example 2 — how much fuel a power station actually uses

A nuclear power station delivers 1.0 GW of electrical power at an overall efficiency of 33%. Taking each fission to release 200 MeV, find (a) the number of fissions per second and (b) the mass of uranium-235 consumed per day. Take \( 1\ \text{eV} = 1.60 \times 10^{-19} \) J and \( N_A = 6.02 \times 10^{23}\ \text{mol}^{-1} \).

(a) Work back to thermal power first. The 1.0 GW is what comes out; the reactor must produce three times as much heat:

\[ P_{\text{thermal}} = \frac{1.0 \times 10^{9}}{0.33} = 3.0 \times 10^{9}\ \text{W} \]

One fission gives \( 200 \times 10^{6} \times 1.60 \times 10^{-19} = 3.2 \times 10^{-11} \) J, so

\[ \text{rate} = \frac{3.0 \times 10^{9}}{3.2 \times 10^{-11}} = 9.5 \times 10^{19}\ \text{fissions per second} \]

(b) Turn a number of nuclei into a mass. In one day there are \( 9.5 \times 10^{19} \times 86400 = 8.2 \times 10^{24} \) fissions, and each consumes one U-235 nucleus. One mole is 235 g and contains \( N_A \) nuclei, so

\[ m = \frac{8.2 \times 10^{24}}{6.02 \times 10^{23}} \times 235 = 13.6 \times 235 \approx 3.2 \times 10^{3}\ \text{g} = 3.2\ \text{kg per day} \]
Sanity check. That is about 1.2 tonnes of U-235 a year. A coal station of the same output burns roughly 8700 tonnes a day — a ratio of about two million, which is the ratio of the energy per event in the two cases. The two independent routes agree, so the arithmetic is sound. Note the two places this calculation is usually lost: forgetting to divide by the efficiency, and forgetting that the fuel is only a few per cent U-235, so the mass of fuel loaded is far greater than the mass of U-235 burnt.

☢️The waste, and why there is any

Fission fragments are radioactive, and the reason is worth deriving rather than remembering. A nucleus with 92 protons needs about 1.6 neutrons per proton to sit in the stability band of E.3. A fragment with 50 protons needs only about 1.4. But the fragment inherits its neutrons from the uranium, so it arrives with far too many for its new proton number — it lands well above the band. Every fission fragment is therefore a beta-minus emitter, usually through a chain of several decays.

A log-log graph of the activity of spent nuclear fuel, relative to the uranium ore it was mined from, against time since it left the reactor. The activity starts around a million times the ore level after one year and falls steadily, crossing the dashed line marking the ore level after roughly a hundred thousand years. Beneath the time axis a bar shows where the fuel sits during that time: cooling ponds for the first years, dry storage on site for decades, then deep geological disposal in stable rock hundreds of metres down. A panel to the right sets out both sides of the argument: for nuclear power, no carbon dioxide while running, enormous energy density, plentiful and widely spread fuel, output independent of weather or time of day, and small land area per gigawatt; against it, waste that stays dangerous for thousands of years, very high and slow build cost, accidents that are rare but severe, links to weapons material, and the fact that uranium is finite too.
The activity falls by six orders of magnitude, but slowly at the end: the long tail belongs to the actinides, not the fission products. Reprocessing, which separates those actinides out, is what would shorten the tail.

The practical consequence is a fuel assembly that comes out of the reactor generating several kilowatts of heat purely from its own decay, and that stays hazardous long after any institution now in existence can be expected to look after it. It is handled in three stages: years in cooling ponds, where water absorbs both the heat and the radiation; decades in dry casks on site; and then, in principle, deep geological disposal in stable rock. The qualification matters — as of now only a handful of countries have a repository actually built.

The trap: “nuclear power produces no greenhouse gases” needs a qualifier. Fission itself produces none, and that is a real and large advantage over burning anything. But mining, enrichment, construction and decommissioning all use energy, and today that energy is largely fossil. The honest statement is that nuclear generation emits far less carbon dioxide per kilowatt-hour than any fossil fuel — comparable to wind — not that it emits none. An evaluation question wants the qualified version.

⚖️Weighing it up

“Discuss the advantages and disadvantages” is a standard question, and it is marked on whether both sides are actually argued rather than listed. The strongest points on each side:

forno CO2 at the point of generation, unlike any fossil fuel
forenergy density around \( 10^{6} \) times that of coal, so tiny fuel volumes
foroutput does not depend on the weather or the time of day
againstwaste stays hazardous for far longer than any institution has lasted
againstaccidents are rare but can render land unusable for decades
againstbuild costs are very high and construction takes a decade or more
againstthe fuel cycle is entangled with weapons material

Two things are worth saying plainly because they are often got wrong. First, a reactor cannot detonate like a bomb. A weapon needs a critical mass of nearly pure U-235 assembled in microseconds; reactor fuel is a few per cent U-235 dispersed through a moderator, and no rearrangement of it forms a supercritical lump. Reactor accidents are steam explosions, chemical fires and melting, which are quite bad enough, but they are not nuclear detonations. Second, uranium is finite too — nuclear fission is a low-carbon source, but it is not a renewable one.

☀️Fusion: the other way up the curve

The binding-energy curve has two downhill directions, and fission uses only one of them. Fusing light nuclei climbs the same curve from the left, and climbs it far more steeply: fusing deuterium with tritium releases 17.6 MeV from five nucleons, or 3.5 MeV per nucleon, against fission's 0.85.

\[ ^{2}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow\ ^{4}_{2}\text{He} + ^{1}_{0}\text{n} + 17.6\ \text{MeV} \]
On the left, the deuterium-tritium reaction: a deuterium nucleus and a tritium nucleus combine to give a helium-4 nucleus, described as harmless, and a fast neutron carrying most of the energy out, releasing 17.6 mega-electronvolts, or 3.5 MeV per nucleon. On the right, what it costs to make that happen: both nuclei are positive so they repel hard right up until they touch; they must arrive at enormous speed, meaning a temperature of about a hundred million kelvin; at that temperature matter is a plasma and no solid container can hold it; so it is held by magnetic fields in a tokamak, a doughnut of plasma touching nothing; and it must be held long enough and dense enough, with the product of density and confinement time as the criterion. A comparison table sets fission against fusion on energy per nucleon, fuel, main waste, runaway risk and status.
Fusion’s appeal is the fuel and the waste: deuterium comes out of ordinary water, and the products are helium and a neutron. Its difficulty is entirely in the conditions — and those are not an engineering detail, they follow from the Coulomb barrier.

Both nuclei are positively charged, so they repel until they are close enough for the strong force to take over — about \( 10^{-15} \) m. Getting there needs kinetic energies corresponding to temperatures around \( 10^{8} \) K. At those temperatures matter is a fully ionised plasma, and no material container can hold it: contact would both vaporise the wall and quench the plasma. So it is confined by magnetic fields instead, in a torus — a tokamak — where the charged particles spiral along field lines and never touch anything (the physics of D.3).

The remaining condition is that the plasma must be held long enough and densely enough for the reactions to release more energy than went into heating it. Density and confinement time trade off against each other, and no device has yet run at sustained net gain. Which is the honest status of fusion power: the physics is settled, the engineering is not.

Stars, of course, solve the confinement problem for free. A star has \( 10^{30} \) kg of overlying material pressing inwards under gravity, so it can hold a plasma at fusion conditions for billions of years without any magnets at all — which is where E.5 picks the story up.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A uranium-235 nucleus absorbs a neutron and splits, giving barium-144, three neutrons and one other fragment. Identify it, and write the complete equation.
Nucleon number: \( 235 + 1 = 236 \) on the left, and \( 144 + A + 3 \) on the right, so \( A = 89 \).
Proton number: \( 92 + 0 = 92 \) on the left. Barium is \( Z = 56 \), so \( Z = 92 - 56 = 36 \), which is krypton. \[ ^{235}_{\ 92}\text{U} + ^{1}_{0}\text{n} \rightarrow\ ^{144}_{\ 56}\text{Ba} + ^{89}_{36}\text{Kr} + 3\,^{1}_{0}\text{n} \] Check both sums: \( 144 + 89 + 3 = 236 \) and \( 56 + 36 = 92 \).
2. Explain why the neutrons produced by fission must be slowed down before they can sustain a chain reaction, and name a suitable material for doing it.
Fission produces neutrons with energies around 2 MeV, but uranium-235 has a fission cross-section of only about 1 barn at that energy, against 585 barns at thermal energies of about 0.025 eV. A fast neutron is therefore several hundred times less likely to cause the next fission, and the chain would die out. The moderator slows them by repeated elastic collisions. It must be a light nucleus, because a neutron transfers most of its kinetic energy only to a target of similar mass — water (hydrogen) or graphite (carbon) are the usual choices. A neutron loses up to 100% of its energy in a head-on collision with hydrogen but under 2% with uranium.
3. Define critical mass, and explain why a critical mass exists at all.
The critical mass is the minimum mass of fissile material for which a chain reaction is self-sustaining, that is, for which \( k = 1 \). It exists because neutrons are produced throughout the volume of the material but are lost through its surface. Volume goes as \( r^3 \) and surface area as \( r^2 \), so the fraction escaping goes as \( 3/r \) — the larger the lump, the smaller the proportion lost. Below a certain radius so many escape that \( k < 1 \) and the reaction dies out. For a bare sphere of pure U-235 the critical mass is about 52 kg; a neutron reflector around it, bouncing escaping neutrons back, reduces that to under 20 kg.
4. State the function of the moderator and of the control rods in a reactor, and explain why they must be made of different materials.
The moderator slows fast neutrons to thermal speeds so that U-235 will absorb them: it keeps neutrons in play. It must be a light nucleus that scatters rather than absorbs — water or graphite.
The control rods absorb neutrons and remove them from the chain, so raising and lowering them sets \(k\) and therefore the power. They must be made of a strong neutron absorber — boron or cadmium.
The two jobs are opposite, which is why the materials must be different: a moderator that absorbed neutrons would stop the reaction it exists to sustain.
5. A compound nucleus of 236 nucleons with a binding energy per nucleon of 7.6 MeV splits into two fragments with 234 nucleons between them, at 8.4 MeV per nucleon. Estimate the energy released.
Binding energy before: \( 236 \times 7.6 = 1794 \) MeV.
Binding energy after: \( 234 \times 8.4 = 1966 \) MeV (the two free neutrons have no binding energy).
Energy released: \( 1966 - 1794 = 172 \approx 1.7 \times 10^{2} \) MeV.
The binding energy has increased, and that increase is what is released — the products are more tightly bound than the parent, which is exactly what moving up the binding-energy curve means.
6. A reactor delivers 500 MW of electrical power at an efficiency of 35%. Each fission releases 200 MeV. Find the number of fissions per second and the mass of U-235 used per year.
Thermal power: \( 500 \times 10^{6} / 0.35 = 1.43 \times 10^{9} \) W.
Energy per fission: \( 200 \times 10^{6} \times 1.60 \times 10^{-19} = 3.2 \times 10^{-11} \) J.
Rate: \( (1.43 \times 10^{9})/(3.2 \times 10^{-11}) = 4.5 \times 10^{19} \) fissions per second.
In a year (\( 3.16 \times 10^{7} \) s) that is \( 1.4 \times 10^{27} \) fissions, so \[ m = \frac{1.4 \times 10^{27}}{6.02 \times 10^{23}} \times 235 \approx 5.5 \times 10^{5}\ \text{g} = 5.5 \times 10^{2}\ \text{kg} \] About half a tonne of U-235 a year. Watch the efficiency: forgetting to divide by 0.35 understates the fuel by a factor of three.
7. Explain why the fragments produced by fission are radioactive, and state which type of decay they undergo.
Stability requires roughly 1.6 neutrons per proton at \( Z = 92 \) but only about 1.4 at \( Z \approx 50 \). The fragments inherit the uranium’s neutron-rich composition, so relative to their own, much smaller, proton number they have far too many neutrons and sit well above the band of stability on an N–Z plot. They therefore decay by beta-minus emission, converting a neutron into a proton and moving diagonally back towards the band — usually through a chain of several such decays before reaching a stable nuclide.
8. A student claims that a nuclear power station could explode like a nuclear weapon. Explain why this is not possible.
A weapon requires a critical mass of nearly pure fissile material — typically over 90% U-235 — brought together in microseconds so that the chain reaction multiplies before the assembly blows itself apart. Reactor fuel is only 3–5% U-235 and is dispersed in rods through a moderator, so it can never form a supercritical bare mass however it is rearranged; and the reaction cannot run away because it depends on moderated neutrons, so losing the coolant (which is also the moderator) shuts the chain reaction down. Reactor accidents are steam explosions, hydrogen fires and core melting, driven by decay heat rather than by a runaway chain — serious, but not a nuclear detonation.
9. State two conditions that must be met for nuclear fusion to occur in a reactor, and explain why each is necessary.
Very high temperature, of order \( 10^{8} \) K: both nuclei are positively charged, so they repel electrostatically, and they must arrive with enough kinetic energy to get within about \( 10^{-15} \) m, where the strong nuclear force can take over.
Sufficient density, maintained for long enough: fusion is a two-body collision process, so the reaction rate depends on how many nuclei are packed together and how long they are held there. The product of density and confinement time must exceed a threshold, or the energy released never exceeds the energy spent heating the plasma.
A third point worth adding: at those temperatures the fuel is a plasma that no material can contain, so it must be confined by magnetic fields — in a tokamak — or, in a star, by gravity.
10. Compare the energy released per nucleon in the fusion reaction \( ^{2}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow\ ^{4}_{2}\text{He} + ^{1}_{0}\text{n} \), which releases 17.6 MeV, with that of a fission releasing 200 MeV from 236 nucleons. Comment on the result.
Fusion: five nucleons are involved (2 + 3), so \( 17.6/5 = 3.5 \) MeV per nucleon.
Fission: \( 200/236 = 0.85 \) MeV per nucleon.
Fusion releases about four times as much energy per nucleon. This follows directly from the binding-energy-per-nucleon curve: the climb from hydrogen to helium is enormously steeper than the shallow descent from uranium down towards the middle of the curve. So although one fission event releases far more energy than one fusion event, fusion is much the better use of a given mass of fuel — and its fuel is deuterium from ordinary water rather than mined and enriched uranium.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — nuclear binding energy and fission
  • The Physics Hypertextbook — nuclear energy