Part 5 · Chapter 43

Energy from the Nucleus

People have drawn energy from atoms for millennia by burning wood and coal — rearranging the outer electrons of carbon and oxygen into tighter combinations. The last century opened a far deeper well: the nucleus. Pull an electron from an atom and it costs a few electron-volts; pull a nucleon from a nucleus and it costs a few million. That single factor is why a kilogram of uranium yields millions of times more energy than a kilogram of coal. In this chapter we follow that energy out of the nucleus along two opposite paths that meet at the same peak of the binding-energy curve: fission, in which a heavy nucleus splits and climbs toward iron, powering reactors and bombs; and fusion, in which light nuclei merge and climb toward iron from below, powering the Sun and every star. Along the way we will see why a chain reaction can be tamed into a steady glow, why a natural reactor once ran in Africa two billion years ago, and why building a fusion reactor on Earth remains one of the hardest engineering problems we have ever faced.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 100 min
i What you'll learn
  • Nuclear processes transform mass to energy about a million times more effectively per unit mass than chemical processes, because nucleons are bound by \(\sim\!\mathrm{MeV}\) energies, not \(\sim\!\mathrm{eV}\).
  • In fission, a thermal neutron splits a heavy nucleus into two middle-mass fragments plus neutrons, releasing \(Q \approx 200\,\mathrm{MeV}\) as the binding energy per nucleon climbs toward the iron peak; in symbols \(^{235}\mathrm{U}+n \to {}^{236}\mathrm{U} \to {}^{140}\mathrm{Xe}+{}^{94}\mathrm{Sr}+2n\).
  • The collective (liquid-drop) model treats fission as barrier tunneling; a nuclide fissions by thermal neutrons only if the excitation energy \(E_n\) exceeds (or nearly reaches) the barrier height \(E_b\).
  • Surplus neutrons enable a chain reaction, controlled in a reactor by the multiplication factor \(k\), a moderator, and neutron-absorbing control rods.
  • Fusion joins light nuclei that pierce the Coulomb barrier; thermonuclear temperatures express energy as \(K = kT\), and tunneling lets fusion proceed below the barrier peak.
  • The Sun burns hydrogen to helium by the proton–proton cycle, releasing \(Q = 26.7\,\mathrm{MeV}\) per helium nucleus formed.
  • Controlled fusion pursues the d-d and d-t reactions and must satisfy Lawson's criterion \(n\tau > 10^{20}\,\mathrm{s/m^3}\) at high temperature, via magnetic (tokamak) or inertial (laser) confinement.
Section 43-1

What Is Physics?

Can we get useful energy from nuclear sources, the way we have drawn it from atomic sources for thousands of years by burning wood and coal? The answer is yes — but the two are profoundly different. Burning coal rearranges the outer electrons of carbon and oxygen into more stable combinations; burning uranium in a reactor rearranges the nucleons of the uranium nucleus into more stable combinations. Electrons are held by the electromagnetic force and cost only a few electron-volts to remove; nucleons are held by the strong force and cost a few million electron-volts. That factor of a few million is exactly why a kilogram of uranium delivers a few million times more energy than a kilogram of coal. In both cases the release of energy is accompanied by a decrease in mass, \(Q = -\Delta m\,c^2\); the difference is simply that nuclear burning consumes a far larger fraction of the available mass.

Per-kilogram energy, ranked. A 50-m waterfall could light a 100-W bulb for about 5 s per kilogram of water; burning coal, about 8 hours; fission of the \(^{235}\mathrm{U}\) in enriched reactor fuel, about \(690\) years; complete fission of pure \(^{235}\mathrm{U}\) or complete fusion of deuterium, about \(3\times10^{4}\) years; and total matter–antimatter annihilation, about \(3\times10^{7}\) years — the ultimate limit, in which all the mass energy is released.
Section 43-2

Nuclear Fission: The Basic Process

After Chadwick discovered the neutron in 1932, Fermi realized that this uncharged particle — feeling no Coulomb repulsion — would be an ideal nuclear projectile, even when moving slowly. Such thermal neutrons, in thermal equilibrium with matter at room temperature, carry only about \(0.04\,\mathrm{eV}\) yet are remarkably effective. Bombarding uranium with them, Hahn and Strassmann found barium (\(Z=56\)) among the products — impossibly far from uranium (\(Z=92\)) unless the nucleus had split. Meitner and Frisch supplied the explanation and named the process fission.

A typical stepwise fission of uranium-235
\[ ^{235}\mathrm{U} + n \;\longrightarrow\; ^{236}\mathrm{U} \;\longrightarrow\; ^{140}\mathrm{Xe} + {}^{94}\mathrm{Sr} + 2n \]
The absorbed neutron forms a highly excited compound nucleus \(^{236}\mathrm{U}\), which then splits. The most probable fragment masses cluster near \(A \approx 95\) and \(A \approx 140\) (a still-unexplained "double-peaked" distribution). Throughout, protons and neutrons are separately conserved, so charge and nucleon number balance.

The primary fragments inherit uranium's high neutron-to-proton ratio (about \(1.6\)), but stable middle-mass nuclides need a ratio nearer \(1.3\!-\!1.4\). So the fragments are neutron rich: they promptly eject a couple of neutrons and then walk down a beta-decay chain, converting excess neutrons into protons until they reach stability.

Beta-decay chains of the primary fragments
\[ ^{140}\mathrm{Xe} \to {}^{140}\mathrm{Cs} \to {}^{140}\mathrm{Ba} \to {}^{140}\mathrm{La} \to {}^{140}\mathrm{Ce}, \qquad ^{94}\mathrm{Sr} \to {}^{94}\mathrm{Y} \to {}^{94}\mathrm{Zr} \]
As expected from beta decay, the mass numbers (140 and 94) stay fixed while \(Z\) rises by one at each step until a stable end product is reached.
📐
Estimating the energy release from the binding curve
Fission releases energy because the products are more tightly bound: the binding energy per nucleon \(\Delta E_{ben}\) rises from about \(7.6\,\mathrm{MeV}\) for a heavy nuclide (\(A\approx240\)) to about \(8.5\,\mathrm{MeV}\) for middle-mass nuclides (\(A\approx120\)).

Treating fission as splitting \(240\) nucleons into two well-bound halves,

Order-of-magnitude estimate of Q
\[ Q \approx \Big(8.5\,\tfrac{\mathrm{MeV}}{\text{nucleon}}\Big)(240) - \Big(7.6\,\tfrac{\mathrm{MeV}}{\text{nucleon}}\Big)(240) \approx 200\,\mathrm{MeV} \]
About \(200\,\mathrm{MeV}\) per fission — tens of millions of times the few eV released per atom in chemical burning. A precise mass-difference calculation (Example 1) confirms this estimate.
Section 43-3

A Model for Nuclear Fission

Bohr and Wheeler explained fission with the collective (liquid-drop) model of Chapter 42. A slow neutron captured by \(^{235}\mathrm{U}\) drops into the nuclear potential well, and its binding energy becomes excitation energy that sets the "drop" oscillating. If the oscillations grow large enough, the nucleus develops a neck, the two charged lobes repel, and the drop tears in two.

The fission barrier
\[ \text{Fission occurs if } \; E_n \gtrsim E_b \]
Plotted against a distortion parameter, the nucleus's potential energy rises to a peak — a barrier of height \(E_b\) — before falling to the separated fragments, whose energy lies below the start by \(Q\approx200\,\mathrm{MeV}\). Just as in alpha decay, the system can tunnel through, so the excitation energy \(E_n\) need not quite reach \(E_b\) for fission to occur.
Fissionability by thermal neutrons (Bohr–Wheeler test)
TargetNucleus fissioned\(E_n\) (MeV)\(E_b\) (MeV)Thermal fission?
\(^{235}\mathrm{U}\)\(^{236}\mathrm{U}\)6.55.2Yes
\(^{238}\mathrm{U}\)\(^{239}\mathrm{U}\)4.85.7No
\(^{239}\mathrm{Pu}\)\(^{240}\mathrm{Pu}\)6.44.8Yes
\(^{243}\mathrm{Am}\)\(^{244}\mathrm{Am}\)5.55.8No

The first and third rows are historically momentous: because \(E_n > E_b\) for \(^{235}\mathrm{U}\) and \(^{239}\mathrm{Pu}\), both fission on thermal neutrons — and both became the cores of the first atomic bombs. For \(^{238}\mathrm{U}\) and \(^{243}\mathrm{Am}\), \(E_n < E_b\), so a thermal neutron only leaves the nucleus to shed its excitation as a gamma ray; these nuclides fission only if struck by a sufficiently fast neutron (for \(^{238}\mathrm{U}\), at least about \(1.3\,\mathrm{MeV}\)).

Section 43-4

The Nuclear Reactor

For sustained power, each fission must trigger another — a chain reaction — since fission releases more neutrons than it consumes. Natural uranium is only \(0.7\%\) fissionable \(^{235}\mathrm{U}\); reactor fuel is enriched to a few percent. Three obstacles stand between that fuel and a working reactor.

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Three problems, three fixes
Leakage, neutron energy, and resonance capture each threaten the chain — and each has a remedy.

Leakage is a surface effect (\(\propto a^2\)) while production is a volume effect (\(\propto a^3\)), so a large enough core wins. Neutron energy: fission neutrons are fast (\(\sim 2\,\mathrm{MeV}\)) but fission best on thermal neutrons, so a light moderator (often water — its protons match the neutron mass and steal the most energy per elastic collision) slows them. Resonance capture: while slowing, neutrons pass through an energy band where \(^{238}\mathrm{U}\) greedily captures them, so fuel and moderator are kept in separate regions to make capture less likely.

Multiplication factor
\[ k = \frac{\text{neutrons at end of a generation}}{\text{neutrons at start of that generation}} \]
\(k = 1\) is exactly critical — steady power. Reactors are built slightly supercritical (\(k>1\)) and trimmed to \(k=1\) by inserting neutron-absorbing control rods (e.g. cadmium). A small fraction of delayed neutrons, emitted seconds after fission by decaying fragments, slows the reactor's response enough for mechanical control to keep up.

In a pressurized-water reactor (PWR), water serves as both moderator and heat-transfer fluid. A high-pressure primary loop carries heat from the core to a steam generator; the secondary loop raises steam that spins a turbine and electric generator. By the second law, only part of the thermal power becomes electricity — the rest is dumped to the environment.

Section 43-5

A Natural Nuclear Reactor

Fermi's 1942 pile was assumed to be the first fission reactor ever — but about two billion years ago, a uranium deposit at Oklo in Gabon, West Africa, apparently went critical on its own and ran for perhaps a few hundred thousand years. Two questions test the claim, and both are answered in the affirmative.

📐
Was there enough fuel, and where is the evidence?
Because \(^{235}\mathrm{U}\) (half-life \(7.0\times10^{8}\,\mathrm{y}\)) decays about \(6.4\) times faster than \(^{238}\mathrm{U}\) (\(4.5\times10^{9}\,\mathrm{y}\)), the past held proportionally more of it.

Two billion years ago the \(^{235}\mathrm{U}\) abundance was about \(3.8\%\) — essentially modern reactor-grade enrichment — so a natural reactor is plausible. The clincher is the fission-product fingerprint: the isotopic mix of neodymium in the Oklo ore matches the pattern produced by reactor fission (notably missing \(^{142}\mathrm{Nd}\), which is abundant in ordinary neodymium but not a fission product), not the natural pattern. A reactor really did run there.

Section 43-6

Thermonuclear Fusion: The Basic Process

The binding-energy curve also rises if two light nuclei merge into a heavier one — nuclear fusion. But the two positive nuclei must first pierce a Coulomb barrier: for two protons it is about \(400\,\mathrm{keV}\), and higher for more-charged nuclei. The way to get there in bulk matter is heat — raise the temperature until thermal motion alone supplies the energy. This is thermonuclear fusion.

Temperature expressed as energy
\[ K = kT \]
In fusion work, temperatures are quoted as the kinetic energy of the most probable particle. The Sun's central \(1.5\times10^{7}\,\mathrm{K}\) corresponds to only \(kT \approx 1.3\,\mathrm{keV}\) — far below the \(400\,\mathrm{keV}\) barrier. Fusion happens anyway for two reasons: the Maxwell distribution has a long high-energy tail, and quantum barrier tunneling lets nuclei fuse well below the barrier peak. The product of "enough fast particles" and "enough tunneling" peaks at a particular energy where fusion runs fastest.
Section 43-7

Thermonuclear Fusion in the Sun and Other Stars

The Sun has radiated at \(3.9\times10^{26}\,\mathrm{W}\) for billions of years. Chemical burning would last a thousand years; gravitational contraction, only a little longer. The true source is fusion: the Sun burns hydrogen ("fuel") into helium ("ashes") through the proton–proton (p-p) cycle.

Net result of the proton–proton cycle
\[ 4\,{}^{1}\mathrm{H} + 2e^- \;\longrightarrow\; {}^{4}\mathrm{He} + 2\nu + 6\gamma, \qquad Q = 26.7\,\mathrm{MeV} \]
The bottleneck is the very first step — two protons forming a deuteron — which happens only about once in \(10^{26}\) collisions. That slowness regulates the Sun's output and keeps it from exploding. Subsequent steps build \(^{3}\mathrm{He}\) and then \(^{4}\mathrm{He}\). Adding two electrons to each side makes both parentheses neutral atoms, so \(Q\) follows from atomic masses: \(Q = [\,4.002603 - 4(1.007825)\,]\,(931.5\,\mathrm{MeV/u}) = 26.7\,\mathrm{MeV}\).
📐
We are the children of the stars
When a star exhausts its hydrogen, its core contracts and heats; helium then fuses to carbon, and heavier elements form in hotter stages — but only up to the iron peak (\(A\approx56\)), beyond which fusion no longer releases energy.

Elements heavier than the peak are forged by neutron capture in supernova explosions, which scatter the new material into space. Later stars and planets condense from this enriched medium. Every atom heavier than hydrogen and helium in your body was manufactured inside a star that has since died.

Section 43-8

Controlled Thermonuclear Fusion

The first terrestrial fusion was the 1952 hydrogen bomb, triggered by a fission device to supply the needed heat and density. A controlled fusion reactor is far harder. The Sun's p-p cycle is hopelessly slow for Earth — it works only because of the Sun's enormous proton density — so terrestrial designs favor reactions among the heavy hydrogen isotopes.

The candidate fuel reactions (d-d and d-t)
\[ ^{2}\mathrm{H}+{}^{2}\mathrm{H} \to {}^{3}\mathrm{He}+n \;\;(3.27\,\mathrm{MeV}), \qquad ^{2}\mathrm{H}+{}^{2}\mathrm{H} \to {}^{3}\mathrm{H}+{}^{1}\mathrm{H} \;\;(4.03\,\mathrm{MeV}) \]
\[ ^{2}\mathrm{H}+{}^{3}\mathrm{H} \to {}^{4}\mathrm{He}+n \qquad (Q = 17.59\,\mathrm{MeV}) \]
Deuterium is just \(1\) part in \(6700\) of hydrogen but is essentially unlimited in seawater — the choice between "burning rocks" (uranium) and "burning water" (deuterium). The d-t reaction gives the largest \(Q\) and is the leading candidate.
Lawson's criterion
\[ n\tau > 10^{20}\,\mathrm{s/m^3} \]
A working d-t reactor needs a high enough particle density \(n\) held for a long enough confinement time \(\tau\) — a trade-off between many particles briefly or fewer particles longer — and a high plasma temperature (laboratory ion temperatures of \(35\,\mathrm{keV} \approx 4\times10^{8}\,\mathrm{K}\) have been reached). Meeting it is breakeven; a self-sustaining burn is ignition.
📐
Two ways to hold a star in a box
No solid wall survives a fusion plasma, so the hot, ionized gas must be confined without touching anything.

Magnetic confinement traps the charged plasma in a shaped field inside a doughnut-shaped chamber — a tokamak — heated by induced currents and particle beams. Inertial confinement instead zaps a tiny d-t pellet from all sides with synchronized laser pulses; the surface boils off, driving an inward shock that briefly compresses and heats the core while the fuel's own inertia holds it together. Neither approach has yet reached a practical, net-energy reactor, but both are pursued worldwide.

Worked Examples

Putting It to Work

1 Q value of a uranium-235 fission

Problem. Including the beta decay of the fragments, the net fission of Eq. 43-1 reduces to \(^{235}\mathrm{U} \to {}^{140}\mathrm{Ce} + {}^{94}\mathrm{Zr} + n\). With masses \(235.0439\), \(139.9054\), \(93.9063\), and \(1.00866\,\mathrm{u}\), find \(Q\).

Solution. The mass that vanishes is \(\Delta m = m_{\text{products}} - m_{\text{reactant}}\); then \(Q = -\Delta m\,c^2\).

Δm, then Q = −Δm·c²
\[ \Delta m = (139.9054 + 93.9063 + 1.00866) - 235.0439 = -0.22354\,\mathrm{u} \]
\[ Q = (0.22354\,\mathrm{u})(931.494\,\mathrm{MeV/u}) \approx 208\,\mathrm{MeV} \]

The precise \(208\,\mathrm{MeV}\) agrees beautifully with the \(\sim 200\,\mathrm{MeV}\) binding-curve estimate. In bulk fuel most of this becomes heat; only a few percent escapes as neutrino energy during the fragments' beta decay.

2 Energy from a kilogram of \(^{235}\mathrm{U}\)

Problem. How much energy is released by completely fissioning \(1.0\,\mathrm{kg}\) of pure \(^{235}\mathrm{U}\), taking \(Q = 200\,\mathrm{MeV}\) per fission? For how long could it power a \(100\,\mathrm{W}\) lamp?

Solution. Count atoms with \(N = (m/M)N_A\), multiply by \(Q\), then divide by the lamp power.

N atoms; E = N·Q; t = E/P
\[ N = \frac{1000\,\mathrm{g}}{235\,\mathrm{g/mol}}(6.02\times10^{23}) \approx 2.56\times10^{24} \]
\[ E = NQ = (2.56\times10^{24})(200\times10^{6}\,\mathrm{eV})(1.6\times10^{-19}\,\mathrm{J/eV}) \approx 8.2\times10^{13}\,\mathrm{J} \]
\[ t = \frac{E}{P} = \frac{8.2\times10^{13}\,\mathrm{J}}{100\,\mathrm{W}} \approx 8.2\times10^{11}\,\mathrm{s} \approx 2.6\times10^{4}\,\mathrm{y} \]

One kilogram of \(^{235}\mathrm{U}\) could run the lamp for about \(26{,}000\) years — matching Table 43-1 and dwarfing the \(8\) hours from a kilogram of coal.

3 Excitation energy from neutron capture

Problem. A thermal neutron is absorbed by \(^{235}\mathrm{U}\), forming \(^{236}\mathrm{U}\). With masses \(235.043922\), \(1.008665\), and \(236.045562\,\mathrm{u}\), find the excitation energy \(E_n\) deposited.

Solution. The mass lost in capture reappears as excitation energy: \(E_n = (m_{235} + m_n - m_{236})c^2\).

E_n = (m_U + m_n − m_U236) c²
\[ \Delta m = (235.043922 + 1.008665) - 236.045562 = 7.025\times10^{-3}\,\mathrm{u} \]
\[ E_n = (7.025\times10^{-3}\,\mathrm{u})(931.494\,\mathrm{MeV/u}) \approx 6.5\,\mathrm{MeV} \]

The \(6.5\,\mathrm{MeV}\) exceeds the \(5.2\,\mathrm{MeV}\) barrier of \(^{236}\mathrm{U}\), so a thermal neutron alone can drive fission — exactly why \(^{235}\mathrm{U}\) is reactor and weapons fuel, while \(^{238}\mathrm{U}\) (with \(E_n < E_b\)) is not.

4 Reactor: efficiency and fission rate

Problem. A PWR produces \(3400\,\mathrm{MW}\) of thermal power and \(1100\,\mathrm{MW}\) of electricity. (a) Find its efficiency. (b) At \(Q = 200\,\mathrm{MeV}\) per fission, find the fission rate \(R\).

Solution. Efficiency is output over input; the fission rate is the input power divided by the energy per fission.

eff = P_elec/P_therm; R = P/Q
\[ \text{eff} = \frac{1100\,\mathrm{MW}}{3400\,\mathrm{MW}} \approx 0.32 = 32\% \]
\[ R = \frac{P}{Q} = \frac{3.4\times10^{9}\,\mathrm{J/s}}{(200\times10^{6}\,\mathrm{eV})(1.6\times10^{-19}\,\mathrm{J/eV})} \approx 1.06\times10^{20}\,\mathrm{fissions/s} \]

The \(32\%\) efficiency is capped by the second law; the other \(2300\,\mathrm{MW}\) is dumped as waste heat. The core fissions roughly \(10^{20}\) nuclei per second — converting mass to energy at only about \(3.3\,\mathrm{g/day}\), the mass of a small coin.

5 Coulomb barrier and the temperature for fusion

Problem. Model a proton as a sphere of radius \(R = 1\,\mathrm{fm}\). (a) What kinetic energy \(K\) each must two protons have to just touch against their mutual repulsion? (b) What temperature gives this as the average energy?

Solution. Energy conservation gives \(2K = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{2R}\), so \(K = \dfrac{e^2}{16\pi\varepsilon_0 R}\); then set \(K_{\text{avg}} = \tfrac{3}{2}kT\).

K = e²/(16πε₀R); T = 2K/(3k)
\[ K = \frac{(1.60\times10^{-19})^2}{16\pi(8.85\times10^{-12})(1\times10^{-15})} \approx 5.75\times10^{-14}\,\mathrm{J} \approx 360\,\mathrm{keV} \]
\[ T = \frac{2K}{3k} = \frac{2(5.75\times10^{-14})}{3(1.38\times10^{-23})} \approx 3\times10^{9}\,\mathrm{K} \]

The barrier is a few hundred keV, and the temperature to surmount it on average is about \(3\times10^{9}\,\mathrm{K}\) — yet the Sun's core is only \(1.5\times10^{7}\,\mathrm{K}\). Fusion proceeds only because the fastest tail of protons, aided by tunneling, gets through.

6 Rate the Sun consumes hydrogen

Problem. The p-p cycle deposits \(26.2\,\mathrm{MeV}\) of thermal energy per four protons consumed. With the Sun radiating \(P = 3.9\times10^{26}\,\mathrm{W}\), find \(dm/dt\).

Solution. Write \(P = \dfrac{\Delta E}{\Delta m}\dfrac{dm}{dt}\) with \(\Delta E = 26.2\,\mathrm{MeV} = 4.20\times10^{-12}\,\mathrm{J}\) for \(\Delta m = 4(1.67\times10^{-27}\,\mathrm{kg})\).

dm/dt = (Δm/ΔE)·P
\[ \frac{dm}{dt} = \frac{4(1.67\times10^{-27}\,\mathrm{kg})}{4.20\times10^{-12}\,\mathrm{J}}\,(3.90\times10^{26}\,\mathrm{W}) \approx 6.2\times10^{11}\,\mathrm{kg/s} \]

The Sun burns about \(6\times10^{11}\,\mathrm{kg}\) of hydrogen every second — astonishing, yet trivial against its mass of \(2\times10^{30}\,\mathrm{kg}\), enough to keep shining for billions of years more.

Review

Chapter Summary

Nuclear vs. chemical

Nuclear processes convert mass to energy about a million times more effectively per kilogram, because binding is in MeV, not eV.

Fission

A thermal neutron splits \(^{235}\mathrm{U}\) into middle-mass fragments plus neutrons; \(Q \approx 200\,\mathrm{MeV}\) as \(\Delta E_{ben}\) rises toward iron.

Fission model

Liquid-drop picture: fission is barrier tunneling. Thermal fission needs \(E_n \gtrsim E_b\) (true for \(^{235}\mathrm{U}\), \(^{239}\mathrm{Pu}\)).

Reactor

Chain reaction tuned to \(k = 1\) via control rods; a moderator thermalizes neutrons; leakage and resonance capture are managed by size and geometry.

Oklo

A natural reactor ran \(\sim 2\) billion years ago when \(^{235}\mathrm{U}\) was \(\sim 3.8\%\); neodymium isotopes are the fingerprint.

Fusion basics

Light nuclei must pierce the Coulomb barrier (\(\sim 400\,\mathrm{keV}\) for two protons); a high-energy tail plus tunneling makes it possible (\(K = kT\)).

Stars

The Sun runs the p-p cycle (\(Q = 26.7\,\mathrm{MeV}\)); elements up to \(A\approx56\) form by fusion, heavier ones in supernovae.

Controlled fusion

d-d and d-t reactions; must satisfy Lawson's \(n\tau > 10^{20}\,\mathrm{s/m^3}\) at high \(T\), via tokamak or laser confinement.

Practice

Problems

Take \(c^2 = 931.494\,\mathrm{MeV/u}\), \(N_A = 6.02\times10^{23}\,\mathrm{mol^{-1}}\), \(k = 1.38\times10^{-23}\,\mathrm{J/K} = 8.62\times10^{-5}\,\mathrm{eV/K}\), \(\dfrac{e^2}{4\pi\varepsilon_0} = 1.44\,\mathrm{MeV\cdot fm}\), and \(1\,\mathrm{MeV} = 1.60\times10^{-13}\,\mathrm{J}\). For fission take \(Q = 200\,\mathrm{MeV}\) unless told otherwise.

  1. In the fission process \(^{235}\mathrm{U} + n \to {}^{132}\mathrm{Sn} + {}^{A}\!X + 3n\), find the mass number \(A\) and the atomic number \(Z\) of the second fragment \(X\).
  2. Do the initial fragments formed by fission have more protons than neutrons, more neutrons than protons, or about equal numbers? Explain using the nuclide chart's stability band.
  3. Calculate the energy released by completely fissioning \(1.00\,\mathrm{kg}\) of pure \(^{235}\mathrm{U}\), and state how long it could power a \(100\,\mathrm{W}\) lamp.
  4. At what rate must \(^{235}\mathrm{U}\) nuclei fission to generate \(1.0\,\mathrm{W}\) of power?
  5. Calculate \(Q\) for \(^{235}\mathrm{U} + n \to {}^{141}\mathrm{Cs} + {}^{93}\mathrm{Rb} + 2n\), using masses \(235.04392\), \(140.91963\), \(92.92157\), and \(1.00866\,\mathrm{u}\).
  6. The plutonium isotope \(^{239}\mathrm{Pu}\) releases on average \(180\,\mathrm{MeV}\) per fission. How much energy is released if all the atoms in \(1.00\,\mathrm{kg}\) of pure \(^{239}\mathrm{Pu}\) fission?
  7. The nuclide \(^{238}\mathrm{Np}\) requires \(4.2\,\mathrm{MeV}\) for fission, while removing a neutron from it costs \(5.0\,\mathrm{MeV}\). Is \(^{237}\mathrm{Np}\) fissionable by thermal neutrons? Explain.
  8. A neutron of mass \(m_n\) and energy \(K\) makes a head-on elastic collision with a stationary atom of mass \(m\). Show that \(\dfrac{\Delta K}{K} = \dfrac{4 m_n m}{(m+m_n)^2}\), and evaluate it for hydrogen, deuterium, carbon, and lead.
  9. Explain, using the multiplication factor \(k\), the roles of being supercritical by design, the control rods, and the delayed neutrons in steady reactor operation.
  10. A \(200\,\mathrm{MW}\) fission reactor consumes half its \(^{235}\mathrm{U}\) in \(3.00\,\mathrm{y}\). How much \(^{235}\mathrm{U}\) did it contain initially? (Assume all energy comes from \(^{235}\mathrm{U}\) fission.)
  11. How long ago was the \(^{235}\mathrm{U}/^{238}\mathrm{U}\) ratio in natural uranium equal to \(0.15\)? Use half-lives \(7.0\times10^{8}\,\mathrm{y}\) and \(4.5\times10^{9}\,\mathrm{y}\); today's ratio is \(0.0072\).
  12. Calculate the height of the Coulomb barrier for a head-on collision of two deuterons of effective radius \(2.1\,\mathrm{fm}\).
  13. Find \(Q\) for \(^{2}\mathrm{H}+{}^{2}\mathrm{H} \to {}^{3}\mathrm{He} + n\) using masses \(2.014102\), \(3.016029\), and \(1.008665\,\mathrm{u}\), and confirm the quoted \(3.27\,\mathrm{MeV}\).
  14. The Sun has mass \(2.0\times10^{30}\,\mathrm{kg}\) and radiates \(3.9\times10^{26}\,\mathrm{W}\). (a) At what rate is its mass decreasing? (b) What fraction of its original mass has it lost this way in \(4.5\times10^{9}\,\mathrm{y}\)?
  15. For a laser-fusion pellet whose compressed particle density is \(n \approx 4.8\times10^{31}\,\mathrm{m^{-3}}\), how long must that density be held to satisfy Lawson's criterion for breakeven?
Tip: three habits keep fission and fusion straight. First, hold the binding-energy curve in your mind as the single organizing picture: both fission (heavy → middle) and fusion (light → middle) release energy because they move products up toward the iron peak, so a positive \(Q\) always means "more tightly bound after." Second, every energy number is the same mass-difference calculation, \(Q = -\Delta m\,c^2\) with \(931.494\,\mathrm{MeV/u}\) — use atomic masses so the electrons cancel, and for power-plant questions chain it as energy per event → events per second → kilograms per day. Third, always separate whether a reaction can go from whether it goes fast enough: fission needs \(E_n\) to beat the barrier \(E_b\) (or tunnel it), fusion needs temperature to beat the Coulomb barrier (or tunnel it), and a reactor needs \(k = 1\) while fusion needs Lawson's \(n\tau\) — energy balance alone never tells you the rate.