Chapter 13 · Nuclei

13.3 Nuclear Energy: Fission and Fusion

A nuclear reaction releases energy $Q=(m_\text{initial}-m_\text{final})c^2$. Splitting a uranium nucleus gives about $200$ MeV and, through a chain reaction, powers reactors; joining hydrogen nuclei powers the Sun. Both move nucleons up the binding-energy curve.

In December 1938 the chemists Otto Hahn and Fritz Strassmann in Berlin bombarded uranium with neutrons and found, to their astonishment, barium among the products — an element with roughly half the mass of uranium. Lise Meitner and Otto Frisch worked out over the Christmas holidays what had happened: the uranium nucleus had split in two, releasing about $200$ MeV, and they called the process fission. Within four years the first nuclear reactor was running in Chicago. Today about one-tenth of the world's electricity comes from fission reactors, in countries from France (about two-thirds of its electricity) to India, whose reactors are being developed to run eventually on thorium.

The opposite process, fusion — joining light nuclei — is what makes the Sun and every other star shine. On Earth it has been achieved in an uncontrolled way in thermonuclear weapons, and in experiments such as the JET tokamak in the United Kingdom, the National Ignition Facility in the USA and the international ITER project under construction in France, scientists are trying to tame it. In Section 13.2 we saw why both processes release energy: they move nucleons up the binding-energy-per-nucleon curve. In this section we calculate how much energy, see how a chain reaction of fissions can be started and controlled, and follow the fusion reactions in the Sun.

Nuclear Reactions and Their $Q$-value

In a nuclear reaction, nuclei (and particles such as neutrons, protons and alpha particles) collide or break apart and new nuclei are formed. We write a reaction like a chemical equation, for example $${}^{2}_{1}\text{H}+{}^{3}_{1}\text{H}\to{}^{4}_{2}\text{He}+{}^{1}_{0}n.$$ Two quantities must balance on both sides: the total charge (the sum of the lower numbers, $1+1=2+0$) and the total number of nucleons (the upper numbers, $2+3=4+1$). The total mass does not balance exactly — and the difference is the energy released.

📘 Definition 13.6 — $Q$-value of a nuclear reaction
The $Q$-value of a nuclear reaction is the energy released, equal to the decrease in total rest-mass energy: $$Q=\left(m_\text{initial}-m_\text{final}\right)c^2=\left(\sum m_\text{reactants}-\sum m_\text{products}\right)\times931.5\ \text{MeV/u}.$$ If $Q\gt 0$ the reaction is exothermic: rest-mass energy is converted into kinetic energy of the products (and gamma rays). If $Q\lt 0$ it is endothermic: it can happen only if the reactants bring in at least that much kinetic energy. Atomic masses may be used when the number of electrons balances on the two sides (it does whenever the charge balances and no free electrons appear in the equation).
📐 Result 13.6 — $Q$ from binding energies

Because protons and neutrons are conserved in a nuclear reaction (only their grouping changes), the masses of the free nucleons cancel and $$Q=\sum E_b(\text{products})-\sum E_b(\text{reactants}).$$ Energy is released when the products are more tightly bound than the reactants.

✏️ Example 1 — The deuterium–tritium reaction
(a) Find the $Q$-value of ${}^{2}_{1}\text{H}+{}^{3}_{1}\text{H}\to{}^{4}_{2}\text{He}+n$, given the atomic masses ${}^{2}\text{H}$ $2.014102$ u, ${}^{3}\text{H}$ $3.016049$ u, ${}^{4}\text{He}$ $4.002603$ u and $m_n=1.008665$ u. (b) The reactants are nearly at rest. Use conservation of momentum to find how the energy is shared between the helium nucleus and the neutron.
Masses Initial: $2.014102+3.016049=5.030151$ u. Final: $4.002603+1.008665=5.011268$ u. Decrease $0.018883$ u (electrons balance: one on each side, in the atoms).
(a) Q $Q=0.018883\times931.5\approx17.6$ MeV, exothermic.
(b) Sharing Total momentum is zero, so $p_\text{He}=p_n=p$, and $K=\dfrac{p^2}{2m}$: kinetic energies are in the inverse ratio of the masses, $K_n:K_\text{He}=4:1$. The neutron gets $\tfrac45\times17.6\approx14.1$ MeV and the helium nucleus $\approx3.5$ MeV.
Check Same answer from binding energies (Result 13.6): $28.30-2.22-8.48=17.6$ MeV. ✓ Fusion reactors must capture those $14$ MeV neutrons in a surrounding "blanket" to collect the energy.

Nuclear Fission

📘 Definition 13.7 — Nuclear fission
Nuclear fission is the splitting of a heavy nucleus into two nuclei of intermediate mass (the fission fragments), usually with the emission of two or three neutrons and the release of about $200$ MeV. Uranium-235 undergoes fission when it absorbs a slow (thermal) neutron: $${}^{235}_{92}\text{U}+{}^{1}_{0}n\to{}^{236}_{92}\text{U}^{*}\to\text{two fragments}+(2\text{ or }3)\,{}^{1}_{0}n+\text{energy}.$$

The intermediate nucleus ${}^{236}\text{U}^{*}$ is formed in a highly excited state; it wobbles like a liquid drop, stretches, and the electric repulsion between its two halves drives them apart. Many different pairs of fragments are possible — barium and krypton, xenon and strontium, and dozens of others — for example $${}^{235}_{92}\text{U}+{}^{1}_{0}n\to{}^{141}_{56}\text{Ba}+{}^{92}_{36}\text{Kr}+3\,{}^{1}_{0}n,$$ $${}^{235}_{92}\text{U}+{}^{1}_{0}n\to{}^{140}_{54}\text{Xe}+{}^{94}_{38}\text{Sr}+2\,{}^{1}_{0}n.$$ Most of the energy appears as kinetic energy of the two fragments (about $170$ MeV), which fly apart and heat the surrounding material; the rest comes out as prompt neutrons and gamma rays and, later, from the decay of the radioactive fragments. Because heavy nuclei have more neutrons per proton than medium ones, the fragments are neutron-rich and spare neutrons are released — which is what makes a chain reaction possible.

✏️ Example 2 — Energy of one fission
Find the energy released in ${}^{235}_{92}\text{U}+n\to{}^{141}_{56}\text{Ba}+{}^{92}_{36}\text{Kr}+3n$. Atomic masses: ${}^{235}\text{U}$ $235.043930$ u, ${}^{141}\text{Ba}$ $140.914411$ u, ${}^{92}\text{Kr}$ $91.926173$ u; $m_n=1.008665$ u.
Balance Charge: $92+0=56+36+0$ ✓. Nucleons: $235+1=141+92+3$ ✓. Electrons balance ($92=56+36$).
Masses Initial: $235.043930+1.008665=236.052595$ u. Final: $140.914411+91.926173+3(1.008665)$ $=235.866579$ u.
Q $\Delta m=0.186016$ u, so $Q=0.186016\times931.5\approx173$ MeV.
Comment The later radioactive decays of the barium and krypton fragments add roughly another $20$–$30$ MeV, bringing the total to about $200$ MeV per fission — consistent with the estimate from the binding-energy curve (Section 13.2, Example 5).
✏️ Example 3 — Fuel for a power station
Take $200$ MeV per fission. (a) How much energy is released by the complete fission of $1.0$ kg of uranium-235? (b) A power station produces $1000$ MW of electrical power with an efficiency of $33\%$. How many fissions per second does it need, and how much uranium-235 does it use per day? ($N_A=6.022\times10^{23}\ \text{mol}^{-1}$, $1\ \text{MeV}=1.6\times10^{-13}$ J.)
(a) Nuclei $N=\dfrac{1000\ \text{g}}{235\ \text{g/mol}}\times6.022\times10^{23}\approx2.56\times10^{24}$.
(a) Energy $E=N\times200\times1.6\times10^{-13}\ \text{J}\approx8.2\times10^{13}$ J — the same as burning about $2500$ tonnes of coal.
(b) Fission rate Heat needed: $\dfrac{1000\ \text{MW}}{0.33}\approx3000$ MW $=3.0\times10^9$ J/s. Each fission gives $3.2\times10^{-11}$ J, so the rate is $\dfrac{3.0\times10^9}{3.2\times10^{-11}}\approx9.4\times10^{19}$ fissions per second.
(b) Fuel Per day: $(9.4\times10^{19})(86400)\approx8.1\times10^{24}$ nuclei $=\dfrac{8.1\times10^{24}}{6.022\times10^{23}}\times235$ g $\approx3.2$ kg of uranium-235 per day.

Chain Reactions and Nuclear Reactors

Each fission is started by one neutron and releases on average about $2.5$. If at least one of these goes on to cause another fission, the process keeps itself going: a chain reaction. Enrico Fermi's team built the first controlled one, a pile of graphite and uranium under a sports stadium in Chicago, on 2 December 1942.

📘 Definition 13.8 — Chain reaction and multiplication factor
In a chain reaction, neutrons released by one generation of fissions cause the next generation of fissions. The multiplication factor $k$ is $$k=\frac{\text{number of fissions (or neutrons) in one generation}}{\text{number in the previous generation}}.$$ After $n$ generations, starting from $N_0$, the number is $N_n=N_0k^n$. The reaction is
  • subcritical if $k\lt 1$ — it dies away;
  • critical if $k=1$ — it runs at a steady rate (a working reactor);
  • supercritical if $k\gt 1$ — it grows exponentially (uncontrolled in a bomb).

Not every neutron produces a fission. Some escape through the surface of the fuel; others are captured by uranium-238 or other materials without causing fission. The fraction lost through the surface depends on the ratio of surface area to volume, which falls as the lump grows. So for a given shape there is a critical size (and corresponding critical mass) below which $k\lt 1$ however the fuel is arranged.

A second difficulty is that the neutrons from fission are fast (about $2$ MeV), while uranium-235 is far more likely to absorb slow neutrons (about $0.025$ eV, "thermal" neutrons). A nuclear reactor is designed to keep $k$ exactly at $1$ using slowed-down neutrons. Its main parts are:

Fuel — uranium enriched to about $3$–$5\%$ uranium-235 (natural uranium is only $0.7\%$ uranium-235), in sealed rods. Moderator — ordinary water, heavy water ($\text{D}_2\text{O}$) or graphite, which slows the neutrons by elastic collisions. A neutron loses most energy per collision with a nucleus of similar mass (as in a head-on collision of equal balls), which is why light nuclei such as hydrogen, deuterium and carbon are used. Control rods — made of cadmium or boron, which absorb neutrons strongly; pushing them in lowers $k$, pulling them out raises it. Coolant — water, heavy water, carbon dioxide or liquid sodium, which carries the heat to a steam generator that drives a turbine. Shielding — thick concrete and steel to absorb neutrons and gamma rays.

Why can $k$ be held at exactly $1$ by mechanical rods? About $0.7\%$ of fission neutrons are emitted not at once but seconds later, by the decay of some fragments. These delayed neutrons stretch the effective generation time to about $0.1$ s, slow enough for the control system to respond.

✏️ Example 4 — How fast does a chain reaction grow?
(a) In a lump of pure uranium-235 with $k=2$, how many generations are needed to go from one fission to $10^{24}$ fissions (about the number of nuclei in $0.4$ kg)? With a generation time of $10^{-8}$ s, how long does this take? (b) In a reactor, a control rod is withdrawn slightly so that $k=1.001$; the effective generation time is $0.1$ s. By what factor does the power grow in one minute?
(a) Generations $2^n=10^{24}\Rightarrow n=\dfrac{24}{\log_{10}2}=\dfrac{24}{0.301}\approx80$ generations.
(a) Time $80\times10^{-8}\ \text{s}\approx0.8\ \mu\text{s}$ — less than a millionth of a second: a supercritical chain reaction is explosive.
(b) Reactor One minute is $60/0.1=600$ generations: $1.001^{600}\approx1.8$. The power nearly doubles in a minute — a slow, controllable rise.
Lesson The difference between a reactor and a bomb is not the fuel's energy but how far $k$ exceeds $1$ and how long a generation lasts. Reactor fuel (a few percent uranium-235) cannot explode like a bomb.
Interactive 1 — A Chain Reaction, Generation by Generation

Each flash is a fission. Every fission releases $2$ or $3$ neutrons (average $2.5$); each neutron causes a new fission with probability $p$ — the rest escape or are absorbed by the control rods. So $k=2.5p$. Choose how far the control rods are pushed in (which sets $p$) and watch successive generations. The graph shows the number of fissions per generation on a logarithmic scale, with the expected $N_0k^n$ as a dashed line. The random choices come from a fixed seed, so a run can be repeated with Restart.

0.40
multiplication factor $k$—
generation—
fissions this generation—
state—

Nuclear Fusion and the Energy of the Sun

📘 Definition 13.9 — Nuclear fusion
Nuclear fusion is the joining of two light nuclei to form a heavier nucleus, with the release of energy (because the product lies higher on the steep left-hand side of the binding-energy curve). Examples: $${}^{2}_{1}\text{H}+{}^{2}_{1}\text{H}\to{}^{3}_{2}\text{He}+n+3.27\ \text{MeV},$$ $${}^{2}_{1}\text{H}+{}^{3}_{1}\text{H}\to{}^{4}_{2}\text{He}+n+17.6\ \text{MeV}.$$ Fusion that happens because the nuclei are moving fast in a very hot gas (a plasma) is called thermonuclear fusion.

The Coulomb barrier. Two nuclei can fuse only if they come within the range of the nuclear force, a few femtometres. Both are positively charged, so they repel each other, and they must approach with enough kinetic energy to climb this electric "hill". In a gas, the energy comes from temperature: the average kinetic energy of a particle is $\tfrac32k_BT$, where $k_B=1.38\times10^{-23}$ J/K is the Boltzmann constant (Class 11 Physics, Chapter 12 — Kinetic Theory). Fusion therefore needs temperatures of millions of kelvin — hence "thermonuclear". It happens in practice at lower temperatures than the simple estimate below suggests, because some particles move much faster than average and because quantum mechanics lets nuclei "tunnel" through the top of the barrier.

✏️ Example 5 — How hot must deuterium be?
Treat two deuterons as spheres of radius $R=R_0A^{1/3}$ ($R_0=1.2$ fm). (a) Find the electric potential energy when they just touch. (b) If two deuterons collide head-on with equal speeds, each needs half of this energy. Estimate the temperature at which this is the average kinetic energy of a deuteron. (Use $\dfrac{ke^2}{1\ \text{fm}}=1.44$ MeV, $k_B=1.38\times10^{-23}$ J/K, $1\ \text{eV}=1.6\times10^{-19}$ J.)
(a) Touching $R=1.2\times2^{1/3}\approx1.51$ fm, so the centres are $d=2R\approx3.02$ fm apart: $U=\dfrac{ke^2}{d}=\dfrac{1.44\ \text{MeV fm}}{3.02\ \text{fm}}\approx0.48$ MeV.
(b) Each deuteron $K=\tfrac12U\approx0.24$ MeV $=0.24\times10^6\times1.6\times10^{-19}\approx3.8\times10^{-14}$ J.
(b) Temperature $\tfrac32k_BT=K\Rightarrow T=\dfrac{2K}{3k_B}$ $=\dfrac{2(3.8\times10^{-14})}{3(1.38\times10^{-23})}$ $\approx1.8\times10^9$ K.
Reality check Fusion reactors work at about $1.5\times10^8$ K, and the Sun's core is only about $1.5\times10^7$ K. The fast "tail" of the speed distribution and quantum tunnelling make fusion possible far below this estimate — but the estimate shows why fusion needs extreme temperatures.

Fusion in the Sun. In the Sun's core, at about $1.5\times10^7$ K and a density over $100$ times that of water, hydrogen is turned into helium mainly by the proton–proton chain: $$\begin{aligned}{}^{1}_{1}\text{H}+{}^{1}_{1}\text{H}&\to{}^{2}_{1}\text{H}+e^{+}+\nu\quad(\text{twice}),\\ {}^{2}_{1}\text{H}+{}^{1}_{1}\text{H}&\to{}^{3}_{2}\text{He}+\gamma\quad(\text{twice}),\\ {}^{3}_{2}\text{He}+{}^{3}_{2}\text{He}&\to{}^{4}_{2}\text{He}+2\,{}^{1}_{1}\text{H}.\end{aligned}$$ Here $e^{+}$ is a positron (which soon annihilates with an electron) and $\nu$ a neutrino, an almost massless neutral particle that escapes the Sun freely. The net effect is $$4\,{}^{1}_{1}\text{H}+2e^{-}\to{}^{4}_{2}\text{He}+2\nu+26.7\ \text{MeV}.$$ The very first step is extremely slow (a given proton waits, on average, billions of years), which is why the Sun burns steadily for ten billion years instead of exploding. Heavier stars also burn helium into carbon and oxygen, and so on up to the iron peak.

✏️ Example 6 — The Sun's hydrogen budget
(a) Using atomic masses (${}^{1}\text{H}$ $1.007825$ u, ${}^{4}\text{He}$ $4.002603$ u), show that the net reaction of the proton–proton chain releases $26.7$ MeV. (b) The Sun's luminosity is $3.8\times10^{26}$ W. How many helium nuclei does it make per second, and how much hydrogen does it consume per second? (c) What fraction of the hydrogen's mass is converted to energy?
(a) Q Four hydrogen atoms contain the $4$ protons and $4$ electrons; one helium atom contains $2$ electrons, and the other $2$ electrons annihilate with the $2$ positrons. So with atomic masses, $Q=[4(1.007825)-4.002603]\times931.5$ $=0.028697\times931.5\approx26.7$ MeV. (About $0.5$ MeV of this is carried off by the neutrinos.)
(b) Rate $26.7$ MeV $=4.27\times10^{-12}$ J, so the Sun makes $\dfrac{3.8\times10^{26}}{4.27\times10^{-12}}\approx8.9\times10^{37}$ helium nuclei per second.
(b) Hydrogen Four protons each: $4\times8.9\times10^{37}\times1.67\times10^{-27}$ kg $\approx5.9\times10^{11}$ kg/s — about $600$ million tonnes of hydrogen every second.
(c) Fraction $\dfrac{26.7}{4(1.007825)(931.5)}\approx0.0071$, i.e. $0.71\%$. Check: $0.71\%$ of $5.9\times10^{11}$ kg/s is $4.2\times10^9$ kg/s, the mass-loss rate found in Section 13.2. ✓
Interactive 2 — Bookkeeping the Proton–Proton Chain

Follow four protons through the Sun's proton–proton chain. At each step the widget computes the energy released from the atomic masses ($Q=\Delta m\times931.5$ MeV) and adds it to the running total; the bar shows how much of the original mass has been converted. Press Play to run the steps in order, or pick one to study it.

this step releases—
running total—
mass converted—
per kg of hydrogen—

Fusion on Earth. The easiest reaction to start is deuterium–tritium (Example 1). Deuterium is plentiful — about one hydrogen atom in $6400$ in seawater is deuterium — and tritium can be made inside the reactor from lithium. The difficulty is to hold a plasma at $10^8$ K long enough, and dense enough, for more energy to come out than goes in. Tokamaks such as JET and ITER hold the plasma in a doughnut-shaped magnetic field (charged particles spiral along field lines — Section 4.1); inertial confinement machines crush a tiny fuel pellet with powerful lasers. In December 2022 the National Ignition Facility for the first time released more fusion energy than the laser energy delivered to the pellet — a milestone, though still far from a power station.

⚠️ Common mistakes with fission and fusion

1. Forgetting to balance the equation. Before computing $Q$, check that both charge and nucleon number balance, and count all the neutrons (in fission, $1$ in and $2$ or $3$ out).

2. Using $A$ in place of the mass. $Q$ comes from the tiny differences in the decimals: $235+1=141+92+3$ exactly, so using mass numbers gives $Q=0$.

3. Thinking fission needs fast neutrons. Uranium-235 fissions most readily with slow neutrons; that is why reactors have a moderator. A moderator slows neutrons; control rods absorb them. They are different jobs.

4. "More energy per reaction means more energy per kilogram." One fission (about $200$ MeV) releases more than one D–T fusion ($17.6$ MeV), but per nucleon (per kilogram of fuel) fusion releases about four times more.

5. Confusing critical with dangerous. "Critical" means $k=1$: a steady chain reaction. Every working reactor is critical.

🚀 Beyond the board syllabus (JEE Advanced) — radioactivity and the law of radioactive decay

Syllabus note: radioactivity and the decay law are in the JEE Advanced syllabus only — not in the CBSE, NEET or JEE Main syllabus. The NCERT book keeps only a qualitative mention that unstable nuclei emit α, β or γ radiation.

Radioactive decay. Some nuclei are unstable and change spontaneously into other nuclei, emitting radiation. In alpha decay a heavy nucleus emits a helium-4 nucleus: ${}^{A}_{Z}\text{X}\to{}^{A-4}_{Z-2}\text{Y}+{}^{4}_{2}\text{He}$; for example ${}^{238}_{92}\text{U}\to{}^{234}_{90}\text{Th}+{}^{4}_{2}\text{He}$, with $Q=[238.050788-234.043601$ $-\,4.002603]\times931.5$ $\approx4.27$ MeV (atomic masses), most of it ($\approx4.2$ MeV) carried by the light alpha particle. In beta-minus decay a neutron in the nucleus turns into a proton, an electron and an antineutrino: ${}^{A}_{Z}\text{X}\to{}^{A}_{Z+1}\text{Y}+e^-+\bar\nu$. In beta-plus decay a proton turns into a neutron, a positron and a neutrino: ${}^{A}_{Z}\text{X}\to{}^{A}_{Z-1}\text{Y}+e^++\nu$ (the source of the positrons in PET scans). In gamma decay a nucleus left in an excited state drops to a lower state by emitting a high-energy photon; $Z$ and $A$ do not change.

The decay law. Each unstable nucleus has the same, fixed probability $\lambda$ per unit time of decaying (the decay constant), whatever its age. So the number of decays per second — the activity — is proportional to the number $N$ of undecayed nuclei, and (uses integration) $$R=-\frac{dN}{dt}=\lambda N\quad\Rightarrow\quad N=N_0e^{-\lambda t},\qquad R=R_0e^{-\lambda t}.$$ The half-life $T_{1/2}$ is the time for half the nuclei to decay, and the mean life $\tau$ is the average lifetime: $$T_{1/2}=\frac{\ln2}{\lambda}=\frac{0.693}{\lambda},\qquad \tau=\frac1\lambda=1.44\,T_{1/2}.$$ After $n$ half-lives, $N=N_0/2^n$. Activity is measured in becquerel ($1\ \text{Bq}=1$ decay per second; an older unit is the curie, $1\ \text{Ci}=3.7\times10^{10}$ Bq). Example: the activity of a sample falls from $800$ Bq to $100$ Bq in $15$ days. Since $800/100=8=2^3$, three half-lives have passed, so $T_{1/2}=5.0$ days and $\lambda=0.693/5.0\approx0.139$ per day. Radiocarbon dating uses exactly this law with carbon-14 ($T_{1/2}\approx5730$ years).

Exercises

I
Nuclear reactions and $Q$-values. Atomic masses (u): ${}^{1}\text{H}$ $1.007825$, $n$ $1.008665$, ${}^{2}\text{H}$ $2.014102$, ${}^{3}\text{H}$ $3.016049$, ${}^{3}\text{He}$ $3.016029$, ${}^{4}\text{He}$ $4.002603$, ${}^{6}\text{Li}$ $6.015122$; $1\ \text{u}=931.5\ \text{MeV}/c^2$.
  • 1 Identify X: ${}^{235}_{92}\text{U}+n\to{}^{144}_{56}\text{Ba}+\text{X}+3n$.
  • 2 Identify Y in the reaction used to make tritium for fusion reactors: ${}^{6}_{3}\text{Li}+n\to{}^{4}_{2}\text{He}+\text{Y}$.
  • 3 Find the $Q$-value of ${}^{2}\text{H}+{}^{2}\text{H}\to{}^{3}\text{He}+n$.
  • 4 Find the $Q$-value of ${}^{2}\text{H}+{}^{2}\text{H}\to{}^{3}\text{H}+{}^{1}\text{H}$.
  • 5 Find the $Q$-value of the reaction in part 2.
  • 6 Can a helium-4 nucleus split by itself into two deuterons? Justify with its $Q$-value.
  • 7 Using binding energies of $2.22$ MeV (${}^{2}\text{H}$), $8.48$ MeV (${}^{3}\text{H}$) and $28.30$ MeV (${}^{4}\text{He}$), find the $Q$-value of the D–T reaction, and explain why the neutron does not appear in the calculation.
1 Charge: $92=56+Z$, so $Z=36$. Nucleons: $236=144+A+3$, so $A=89$: ${}^{89}_{36}\text{Kr}$.
2 $Z=3-2=1$, $A=7-4=3$: ${}^{3}_{1}\text{H}$, tritium.
3 $\Delta m=2(2.014102)-(3.016029+1.008665)$ $=0.003510$ u; $Q\approx3.27$ MeV.
4 $\Delta m=4.028204-(3.016049+1.007825)$ $=0.004330$ u; $Q\approx4.03$ MeV.
5 $\Delta m=(6.015122+1.008665)$ $-\,(4.002603+3.016049)$ $=0.005135$ u; $Q\approx4.78$ MeV — exothermic, so the tritium-breeding blanket also adds heat.
6 $Q=[4.002603-2(2.014102)]\times931.5\approx-23.8$ MeV. $Q\lt 0$: it cannot happen spontaneously; $23.8$ MeV would have to be supplied.
7 $Q=28.30-(2.22+8.48)=17.6$ MeV. A free neutron is not bound to anything, so its binding energy is zero.
II
Fission and chain reactions. Take $200$ MeV per fission of uranium-235, $1\ \text{MeV}=1.6\times10^{-13}$ J, $N_A=6.022\times10^{23}\ \text{mol}^{-1}$.
  • 1 Express $200$ MeV in joules.
  • 2 How many fissions per second produce a power of $1.0$ W?
  • 3 How much uranium-235 does a reactor producing $500$ MW of heat consume in one year ($3.156\times10^7$ s)?
  • 4 What mass of coal ($3.3\times10^7$ J/kg) releases as much energy as the fission of $1.0$ g of uranium-235 ($8.2\times10^{10}$ J)?
  • 5 A chain reaction with $k=0.80$ starts with $1000$ neutrons. How many are there in the tenth generation?
  • 6 With $k=1.2$, how many generations does it take for the number of fissions per generation to grow a thousandfold?
  • 7 State the purpose of (a) the moderator, (b) the control rods and (c) the coolant in a nuclear reactor.
  • 8 In a head-on elastic collision with a stationary nucleus of mass number $A$, a neutron keeps a fraction $\left(\dfrac{A-1}{A+1}\right)^2$ of its kinetic energy. Find this fraction for hydrogen and for carbon-12, and explain why moderators are made of light nuclei.
1 $200\times1.6\times10^{-13}=3.2\times10^{-11}$ J.
2 $\dfrac{1.0}{3.2\times10^{-11}}\approx3.1\times10^{10}$ fissions per second.
3 Rate $=\dfrac{5.0\times10^8}{3.2\times10^{-11}}\approx1.56\times10^{19}$ s⁻¹; in a year $4.9\times10^{26}$ nuclei $=\dfrac{4.9\times10^{26}}{6.022\times10^{23}}\times235$ g $\approx1.9\times10^2$ kg.
4 $\dfrac{8.2\times10^{10}}{3.3\times10^7}\approx2.5\times10^3$ kg — two and a half tonnes of coal for one gram of uranium-235.
5 $1000\times0.80^{10}\approx107$ — the reaction is dying away (subcritical).
6 $1.2^n=1000\Rightarrow n=\dfrac{\log1000}{\log1.2}=\dfrac{3}{0.0792}\approx38$ generations.
7 (a) Slows fast neutrons to thermal speeds, at which uranium-235 fissions readily. (b) Absorb neutrons to adjust $k$ and keep it at $1$ (or shut the reactor down). (c) Carries the heat from the core to the steam generator, and keeps the core from overheating.
8 Hydrogen ($A=1$): $0$ — the neutron stops dead, giving all its energy to the proton. Carbon-12: $(11/13)^2\approx0.72$, so it loses $28\%$. The lighter the nucleus, the more energy each collision removes, so fewer collisions (and less chance of capture) are needed.
III
Fusion. Use $\dfrac{ke^2}{1\ \text{fm}}=1.44$ MeV, $k_B=1.38\times10^{-23}$ J/K, $1\ \text{MeV}=1.6\times10^{-13}$ J, $1\ \text{u}=1.66\times10^{-27}$ kg.
  • 1 Two protons are treated as spheres of radius $1.2$ fm. Find their electric potential energy when they just touch.
  • 2 At what temperature is the average kinetic energy $\tfrac32k_BT$ of a particle equal to $0.30$ MeV?
  • 3 Find the average kinetic energy of a particle at the Sun's central temperature, $1.5\times10^7$ K, in joules and in keV. Compare it with your answer to part 1.
  • 4 The D–T reaction releases $17.6$ MeV from reactants of total mass $5.030$ u. Find the energy released per kilogram of fuel.
  • 5 Compare your answer to part 4 with the energy from the fission of $1.0$ kg of uranium-235 ($8.2\times10^{13}$ J).
  • 6 Estimate. The Sun ($2.0\times10^{30}$ kg) is $70\%$ hydrogen by mass, and about one-tenth of that hydrogen (the part in the core) can be burnt. At $5.9\times10^{11}$ kg of hydrogen per second, how long can the Sun shine?
  • 7 Give two reasons why controlled fusion is much harder to achieve than controlled fission.
1 Centres $2.4$ fm apart: $U=1.44/2.4=0.60$ MeV.
2 $T=\dfrac{2K}{3k_B}$ $=\dfrac{2(0.30\times1.6\times10^{-13})}{3(1.38\times10^{-23})}$ $\approx2.3\times10^9$ K.
3 $\tfrac32(1.38\times10^{-23})(1.5\times10^7)\approx3.1\times10^{-16}$ J $\approx1.9$ keV — about $300$ times smaller than the $0.6$ MeV barrier. Only the fastest protons, helped by quantum tunnelling, ever fuse.
4 $\dfrac{17.6\times1.6\times10^{-13}}{5.030\times1.66\times10^{-27}}\approx3.4\times10^{14}$ J/kg.
5 $\dfrac{3.4\times10^{14}}{8.2\times10^{13}}\approx4$: per kilogram, D–T fusion gives about four times as much energy as fission.
6 Burnable hydrogen $=0.1\times0.7\times2.0\times10^{30}=1.4\times10^{29}$ kg; time $=\dfrac{1.4\times10^{29}}{5.9\times10^{11}}\approx2.4\times10^{17}$ s $\approx7.5\times10^9$ years — consistent with a total lifetime of about ten billion years.
7 (i) The fuel must be heated to about $10^8$ K to overcome the Coulomb barrier, and no material wall can touch such a plasma, so it must be held by magnetic fields or inertia. (ii) It must be kept hot and dense for long enough that the energy released exceeds the energy put in; fission, by contrast, is started by slow neutrons at ordinary temperatures.
IV
Applications and longer problems.
  • 1 Modelling. A region needs $2.0$ GW of electrical power, generated at $33\%$ efficiency. Compare the mass of fuel needed per year by (a) uranium-235 fission ($8.2\times10^{13}$ J/kg) and (b) D–T fusion ($3.37\times10^{14}$ J/kg).
  • 2 In the D–T reaction the neutron carries $14.1$ MeV. Find its speed ($m_n=1.67\times10^{-27}$ kg).
  • 3 A uranium nucleus splits into ${}^{141}\text{Ba}$ and ${}^{92}\text{Kr}$, which share $170$ MeV of kinetic energy. Ignoring the neutrons and the initial momentum, find the kinetic energy of each fragment.
  • 4 In fission the number of nucleons is unchanged. Explain where the released energy comes from.
  • 5 How much mass is converted to energy each day in a reactor producing $3000$ MW of heat?
1 Heat per year $=\dfrac{2.0\times10^9}{0.33}\times3.156\times10^7\approx1.9\times10^{17}$ J. (a) $\dfrac{1.9\times10^{17}}{8.2\times10^{13}}\approx2.3\times10^3$ kg of uranium-235. (b) $\dfrac{1.9\times10^{17}}{3.37\times10^{14}}\approx5.6\times10^2$ kg of D–T fuel — compared with roughly six million tonnes of coal.
2 $v=\sqrt{\dfrac{2K}{m}}=\sqrt{\dfrac{2(14.1)(1.6\times10^{-13})}{1.67\times10^{-27}}}$ $\approx5.2\times10^7$ m/s (about $0.17c$, so the non-relativistic formula is still reasonable).
3 Equal and opposite momenta, so $K\propto1/m$: $K_\text{Kr}=170\times\dfrac{141}{233}\approx103$ MeV, $K_\text{Ba}=170\times\dfrac{92}{233}\approx67$ MeV. The lighter fragment moves faster.
4 The nucleons are regrouped into more tightly bound nuclei ($E_{bn}$ rises from about $7.6$ to $8.5$ MeV). The total binding energy increases, so the total rest mass decreases, and the mass difference appears as kinetic energy: $Q=\Delta m\,c^2$.
5 $\Delta m=\dfrac{Pt}{c^2}$ $=\dfrac{(3.0\times10^9)(86400)}{9.0\times10^{16}}$ $\approx2.9\times10^{-3}$ kg $\approx2.9$ g per day.
🎯 Exam corner — nuclear energy: fission and fusion BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Calculate the energy released in the fusion reaction ${}^{2}_{1}\text{H}+{}^{2}_{1}\text{H}\to{}^{3}_{2}\text{He}+{}^{1}_{0}n$. Given: $m({}^{2}\text{H})=2.014102$ u, $m({}^{3}\text{He})=3.016029$ u, $m_n=1.008665$ u, $1\ \text{u}=931.5\ \text{MeV}/c^2$. Is the reaction exothermic or endothermic?

Masses Reactants: $2\times2.014102=4.028204$ u. Products: $3.016029+1.008665=4.024694$ u.
Q $\Delta m=0.003510$ u; $Q=0.003510\times931.5\approx3.27$ MeV.
Answer $Q\approx3.27$ MeV $\gt 0$: exothermic. Exam tip: state "mass of reactants $\gt$ mass of products" in words — it earns the reasoning mark.

E2. Board · 2 marks Distinguish between nuclear fission and nuclear fusion. Why does fusion require a very high temperature?

Distinction Fission: a heavy nucleus splits into two medium nuclei (started by a slow neutron; $\approx200$ MeV per event). Fusion: two light nuclei join into a heavier one (a few to about $20$ MeV per event, but more per nucleon). Both raise $E_b/A$.
Temperature The nuclei are both positive and must overcome their Coulomb repulsion to come within the few-femtometre range of the nuclear force; only at temperatures of $10^7$–$10^8$ K do enough of them move fast enough. Exam tip: use the words "Coulomb barrier" and "range of the nuclear force".

E3. JEE Main A research reactor produces $1.6$ MW of heat. If each fission of uranium-235 releases $200$ MeV, the number of fissions per second is (A) $5.0\times10^{16}$ (B) $8.0\times10^{18}$ (C) $5.0\times10^{19}$ (D) $3.2\times10^{-11}$.

Concept $200$ MeV $=3.2\times10^{-11}$ J; rate $=\dfrac{1.6\times10^6}{3.2\times10^{-11}}=5.0\times10^{16}$ s⁻¹.
Answer (A). Exam tip: (D) is the energy per fission in joules, not a rate — check the units of every option.

E4. NEET In a nuclear reactor using uranium-235, the function of the moderator is to (A) absorb neutrons and stop the chain reaction (B) slow down fast neutrons (C) cool the reactor core (D) speed up the neutrons so that they cause fission.

Concept Uranium-235 fissions most readily with slow (thermal) neutrons; fission neutrons are fast and must be slowed by collisions with light nuclei.
Answer (B). Exam tip: (A) describes control rods and (C) the coolant.

E5. JEE Advanced (Numerical; radioactivity — beyond NCERT, see the 🚀 box.) A radioactive sample has a half-life of $2.0$ h and an initial activity of $6.4\times10^4$ Bq. (a) Find its activity after $6.0$ h, in units of $10^3$ Bq. (b) Estimate the number of radioactive nuclei present initially.

(a) $6.0$ h is three half-lives: $R=\dfrac{6.4\times10^4}{2^3}=8.0\times10^3$ Bq, so the answer is 8.
(b) $\lambda=\dfrac{0.693}{7200\ \text{s}}\approx9.6\times10^{-5}\ \text{s}^{-1}$; $N_0=\dfrac{R_0}{\lambda}=\dfrac{6.4\times10^4}{9.6\times10^{-5}}\approx6.6\times10^8$ nuclei.
Note Quantitative radioactive decay is in the JEE Advanced syllabus only — not in CBSE, NEET or JEE Main. Exam tip: convert the half-life to seconds before using $R=\lambda N$ with activity in Bq.

E6. Board · Case study A pressurised-water reactor uses uranium enriched to $4\%$ uranium-235. Ordinary water flows through the core, and rods of boron carbide can be raised or lowered between the fuel rods. (i) (2 marks) Define the multiplication factor $k$, and state its value when the reactor runs at constant power. What happens to the power if $k$ becomes $1.002$? (ii) (1 mark) The water does two jobs. Name them. (iii) (1 mark) The power starts to rise. What should the operators do with the boron carbide rods, and why?

(i) $k=$ (number of fissions in one generation)/(number in the previous generation). Constant power: $k=1$ (critical). With $k=1.002$ the power grows steadily (exponentially) — supercritical.
(ii) Moderator (slows neutrons) and coolant (carries heat to the steam generator).
(iii) Lower them further into the core: boron absorbs neutrons, reducing $k$ back to $1$. Exam tip: moderator → slows, control rods → absorb, coolant → removes heat. Keep the three apart.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How much fusion energy is locked up in the deuterium of one litre of seawater, and could the oceans supply the world's energy?

Why it matters. Fusion research programmes in Europe, China, India, Japan, Korea, Russia and the USA — partners in the ITER project — argue that fusion fuel is effectively inexhaustible. This task tests that claim with a back-of-the-envelope calculation of the kind researchers use to judge whether an idea is worth pursuing.

What to do. Assume that all the deuterium can eventually be burnt through a sequence of D–D and follow-on reactions whose net effect is $6\,{}^{2}\text{H}\to2\,{}^{4}\text{He}+2\,{}^{1}\text{H}+2n+43.2$ MeV.

  1. Check the $43.2$ MeV from the atomic masses ${}^{2}\text{H}$ $2.014102$ u, ${}^{4}\text{He}$ $4.002603$ u, ${}^{1}\text{H}$ $1.007825$ u and $m_n=1.008665$ u. What is the energy per deuteron?
  2. One litre of seawater is about $1.0$ kg of water ($18$ g/mol). Find the number of hydrogen atoms, and the number of deuterium atoms if $1.56\times10^{-4}$ of all hydrogen atoms are deuterium.
  3. Find the fusion energy per litre, and compare it with petrol ($3.4\times10^7$ J per litre).
  4. The world uses roughly $6\times10^{20}$ J of primary energy per year (an approximate, rounded figure). How many litres of seawater per year would supply it? The oceans hold about $1.3\times10^{21}$ litres. Comment.

Tools: Calculator or spreadsheet.

Hand in: Your calculations, the petrol comparison, and a paragraph on why — despite these numbers — fusion power is not yet available.

Going further. Most fusion reactor designs use D–T rather than D–D. Find out why, and what limits the supply of lithium needed to breed tritium.

Instructor note: $\Delta m=0.046426$ u → $43.2$ MeV, i.e. $7.2$ MeV per deuteron. One litre: $3.35\times10^{25}$ molecules, $6.7\times10^{25}$ H atoms, $1.04\times10^{22}$ deuterons; energy $\approx1.04\times10^{22}\times7.2\times1.6\times10^{-13}\approx1.2\times10^{10}$ J — about $350$ litres of petrol. World demand: $6\times10^{20}/1.2\times10^{10}\approx5\times10^{10}$ litres per year, about $4\times10^{-11}$ of the ocean per year — effectively inexhaustible. The obstacles are engineering (confinement, materials damaged by $14$ MeV neutrons, tritium breeding, cost), not fuel supply.

🧪 Self-check quiz

0/5 correct

Q1. A nuclear reaction has a positive $Q$-value. This means that

Q2. On average, how many neutrons are released in one fission of uranium-235?

Q3. A nuclear reactor running at constant power has a multiplication factor

Q4. The Sun's energy comes mainly from

Q5. Which pair correctly matches a reactor component with its job?