---
title: "Nuclear Energy: Fission, Fusion, E=mc2"
book: "High School Physics"
subject: physics
language: en
chapter: 33
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/33-nuclear-energy-fission-fusion-e-mc2
---

# Chapter 33 — Nuclear Energy: Fission, Fusion, E=mc2

A fuel pellet the size of a pencil eraser covers a household’s electricity for a year; the same service from coal takes three tonnes. Overhead, the Sun pays for daylight by vanishing — four million tonnes of itself a second. This chapter opens the account book behind both bargains: mass itself is [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy), and nuclei ([Chapter 19](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#ch-g11-nucleus-radioactivity)) are where it is cashed.

## 33.1 Mass is energy

**Theorem 33.1 (Mass–energy equivalence).**

A body of mass $m$, merely by existing, holds the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy)

$$
E = m c^2, \qquad c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s}.
$$

Whenever a system loses mass $\Delta m$, it releases the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) $\Delta m\, c^2$ — whatever the mechanism.

**Proof.** *Admitted at this level.* ∎

**Remark 33.2 (Einstein, 1905).**

This is Einstein’s relation, admitted here; where it comes from is the business of the special-relativity chapter that closes this year ([Chapter 35](https://one-course.com/books/physics/2/en/chapter/35-special-relativity-time-dilation#ch-g12-special-relativity)). Note the rate: $c^2 = 9.00 \times 10^{16}\,\mathrm{J}/\mathrm{kg}$ — a rounding error of mass is a fortune of [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy).

**Definition 33.3 (Electron-volt and atomic mass unit).**

Two [units](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) tailored to the [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder). The *electron-volt*, the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) an electron gains crossing $1\,\mathrm{V}$ ([Chapter 12](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#ch-g11-circuits-and-power)): $1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$, $1\,\mathrm{MeV} = 1.60 \times 10^{-13}\,\mathrm{J}$ — chemical bonds trade a few $\mathrm{eV}$, nuclear reactions [megaelectron-volts](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage). The *unified atomic mass unit*, one twelfth of the mass of a carbon-12 [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder): $1\,\mathrm{u} = 1.660\,54 \times 10^{-27}\,\mathrm{kg}$, of [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) equivalent $931.5\,\mathrm{MeV}$ — $1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$.

**Example 33.4 (The sleeping fortune).**

One [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of anything, fully converted, is $9.00 \times 10^{16}\,\mathrm{J}$ — three years of output of a large ($1\,\mathrm{GW}$) [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) station. Nobody noticed before 1905 because ordinary physics barely touches mass: burning a [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of coal ($3.0 \times 10^{7}\,\mathrm{J}$) lightens it by a third of a microgram. Nuclei, we shall see, move one part in a thousand: still small, but a million times chemistry.

## 33.2 The mass defect

**Definition 33.5 (Mass defect and binding energy).**

Weigh a [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) ${}^{A}_{Z}\mathrm{X}$ (mass $m$), then weigh its $Z$ protons and $A - Z$ neutrons separately: the parts are *heavier* than the whole. The difference

$$
\Delta m = Z\,m_p + (A - Z)\,m_n - m > 0,
\qquad m_p = 1.007\,28\,\mathrm{u},\; m_n = 1.008\,66\,\mathrm{u},
$$

is the *mass defect*; the *binding energy* $E_b = \Delta m\,c^2$ is the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) needed to pull the [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) apart into free [nucleons](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) — equally, the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) released when it is assembled. A bound [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) sits *below* its parts.

![An energy ladder: the bound nucleus lies E_b below its separated nucleons — assembling it releases E_b, dismantling costs it.](https://one-course.com/images/onecourse/chapters/physics-2/g12-nuclear-energy/fig-7ee271e97b73.svg)

*An [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) ladder: the bound [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) lies $E_b$ below its separated [nucleons](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) — assembling it releases $E_b$, dismantling costs it.*

**Example 33.6 (Helium-4).**

The ${}^{4}_{2}\mathrm{He}$ [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) has $m = 4.001\,51\,\mathrm{u}$; its parts: $2 \times 1.00728 + 2 \times 1.00866 = 4.031\,88\,\mathrm{u}$. So $\Delta m = 0.030\,37\,\mathrm{u}$ and $E_b = 0.03037 \times 931.5 \approx
28.3\,\mathrm{MeV}$: the [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) weighs $0.75\%$ less than its parts — the “one part in a thousand” of [Example 33.4](#ex-g12-nuclear-energy-fortune).

**Method 33.7 (Energy balance of a nuclear reaction).**

1. Balance the equation: total $A$ and $Z$ conserved ( [Chapter 19](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#ch-g11-nucleus-radioactivity) ).
2. Add the masses before, then after, in $\mathrm{u}$ — keep all five decimals.
3. $\Delta m = m_{\text{before}} - m_{\text{after}}$ , then $E = \Delta m\,(\text{in u}) \times 931.5\,\mathrm{MeV}$ , carried off as [kinetic energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-kinetic) of the products; negative means the reaction must be paid.

## 33.3 The curve that runs the universe

**Definition 33.8 (Binding energy per nucleon).**

Cohesion is compared fairly by the *binding energy per nucleon*, $E_b/A$, the average cost of extracting one [nucleon](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder): $28.3/4 \approx 7.1\,\mathrm{MeV}$ for helium-4.

**Proposition 33.9 (The iron peak, and two roads down).**

Plotted against $A$, the [binding energy per nucleon](#def-g12-nuclear-energy-pernucleon) climbs steeply through the light nuclei, peaks near iron, ${}^{56}_{26}\mathrm{Fe}$, at about $8.8\,\mathrm{MeV}$ per [nucleon](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder), then slides gently to about $7.6\,\mathrm{MeV}$ at uranium: iron is the most tightly bound [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) in nature. A reaction releases [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) exactly when its products sit *higher* on the curve; two opposite strategies therefore both pay, and both walk toward iron — *splitting* the heaviest nuclei, and *merging* the lightest.

**Proof.** *Admitted at this level.* ∎

**Remark 33.10 (Why a peak).**

The tug-of-war of [Chapter 13](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#ch-g11-fundamental-interactions): strong glue binds only touching neighbours, Coulomb repulsion spans the whole [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) — small nuclei are mostly surface, big ones mostly repulsion, iron the compromise. The honest accounting lives in the university volumes.

![Binding energy per nucleon: steep climb, iron peak, long slide — fusion climbs from the left, fission from the right, both releasing energy.](https://one-course.com/images/onecourse/chapters/physics-2/g12-nuclear-energy/fig-2bb05f424ed6.svg)

*[Binding energy per nucleon](#def-g12-nuclear-energy-pernucleon): steep climb, [iron peak](#prop-g12-nuclear-energy-ironpeak), long slide — [fusion](#def-g12-nuclear-energy-fusion) climbs from the left, [fission](#def-g12-nuclear-energy-fission) from the right, both releasing [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy).*

## 33.4 Fission and the chain reaction

**Definition 33.11 (Fission).**

*Fission* is the splitting of a heavy [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) into two mid-sized fragments. Uranium-235 is *fissile*: a slow neutron, absorbed, leaves the [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) so agitated that it tears in two, spitting out two or three fresh neutrons — ${}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \to \text{two fragments} +
2\text{--}3\;{}^{1}_{0}\mathrm{n}$, the pair varying from fission to fission.

**Example 33.12 (One fission, weighed).**

One frequent route, by [Method 33.7](#met-g12-nuclear-energy-balance):

$$
{}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \longrightarrow
{}^{94}_{38}\mathrm{Sr} + {}^{140}_{54}\mathrm{Xe} + 2\,{}^{1}_{0}\mathrm{n}.
$$

Before: $1.00866 + 234.99350 = 236.002\,16\,\mathrm{u}$. After: $93.89450 + 139.89200 + 2 \times 1.00866 = 235.803\,82\,\mathrm{u}$. Hence $\Delta m = 0.198\,34\,\mathrm{u}$ and $E \approx 185\,\mathrm{MeV}$; counting the later decays of the radioactive fragments ([Chapter 32](https://one-course.com/books/physics/2/en/chapter/32-radioactive-decay#ch-g12-radioactive-decay)), each [fission](#def-g12-nuclear-energy-fission) is worth about $200\,\mathrm{MeV}$ — a few $\mathrm{eV}$ buys one [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) of coal.

**Definition 33.13 (Chain reaction and critical mass).**

Each [fission](#def-g12-nuclear-energy-fission) is lit by one neutron and frees two or three: in a large enough lump they strike other nuclei and the reaction feeds itself — a *chain reaction*. In a small lump most neutrons escape through the surface first; the minimum quantity sustaining the chain is the *critical mass* — below it the chain fizzles, above it every generation multiplies: the principle of the bomb.

![A chain reaction: one neutron splits one nucleus, the three neutrons freed split three more — a reactor allows exactly one to carry on.](https://one-course.com/images/onecourse/chapters/physics-2/g12-nuclear-energy/fig-64904b846f63.svg)

*A [chain reaction](#def-g12-nuclear-energy-chain): one neutron splits one [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder), the three neutrons freed split three more — a reactor allows exactly one to carry on.*

**Definition 33.14 (The reactor, tamed).**

A [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) reactor holds the chain at exactly one neutron per [fission](#def-g12-nuclear-energy-fission). The fuel: uranium pellets modestly enriched in uranium-235. The *moderator* (ordinary water, usually) slows the fast [fission](#def-g12-nuclear-energy-fission) neutrons by collisions — slow neutrons are far better at triggering the next [fission](#def-g12-nuclear-energy-fission); the *control rods* (boron, cadmium) devour neutrons — pushed in, the chain dies; drawn out, it quickens. The fragments’ agitation becomes heat, then steam, then electricity ([Chapter 9](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#ch-g10-energy-conservation)).

**Example 33.15 (A peppercorn against a truck).**

One gram of uranium-235 holds $N = 1.0 \times 10^{-3} / (235 \times 1.660\,54 \times 10^{-27}) \approx 2.56 \times 10^{21}$ nuclei; at $200\,\mathrm{MeV} = 3.2 \times 10^{-11}\,\mathrm{J}$ each, fissioning them all yields about $8.2 \times 10^{10}\,\mathrm{J}$ — the [chemical energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-forms) of $8.2 \times 10^{10} /
3.0 \times 10^{7} \approx 2.7 \times 10^{3}\,\mathrm{kg}$ of coal: a peppercorn against three tonnes.

## 33.5 Fusion: the Sun’s own fire

**Definition 33.16 (Fusion).**

*Fusion* is the merging of two light nuclei into a heavier, more tightly bound one. The most accessible reaction weds deuterium and tritium, the heavy [isotopes](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation) of hydrogen ([Chapter 19](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#ch-g11-nucleus-radioactivity)):

$$
{}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \longrightarrow
{}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}.
$$

**Example 33.17 (D–T, weighed).**

Before: $2.01355 + 3.01550 = 5.029\,05\,\mathrm{u}$; after: $4.00151 + 1.00866 = 5.010\,17\,\mathrm{u}$; so $\Delta m = 0.018\,88\,\mathrm{u}$ and $E \approx 17.6\,\mathrm{MeV}$, four fifths of it on the neutron — less than a [fission](#def-g12-nuclear-energy-fission), but from five [nucleons](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) instead of $236$: about $3.4 \times 10^{14}\,\mathrm{J}$ per [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of fuel, four times [fission](#def-g12-nuclear-energy-fission). Helium-4 ([Example 33.6](#ex-g12-nuclear-energy-helium)) is once again the ash.

**Definition 33.18 (The Coulomb barrier).**

Two nuclei are both positive: to touch, they must first climb the hill of their electric repulsion ([Chapter 14](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#ch-g11-electric-gravitational-fields)) — the *Coulomb barrier*. Only near $1 \times 10^{8}\,\mathrm{K}$ do collisions carry the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) to cross it; the fuel is then a *plasma* of bare nuclei and electrons. [Fission](#def-g12-nuclear-energy-fission) pays no such toll: its trigger, the neutron, is neutral, and [works](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) cold.

**Example 33.19 (The Sun runs on it).**

The Sun’s core, at $1.5 \times 10^{7}\,\mathrm{K}$, [fuses](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-joule) ordinary hydrogen step by step — the *proton–proton chain*, of net effect $4\,{}^{1}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + 2\,{}^{0}_{+1}\mathrm{e}
+ \text{radiation}$, about $26\,\mathrm{MeV}$ per helium: $0.7\%$ of the hydrogen’s mass leaves as [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy). The sunlight whose [spectrum](https://one-course.com/books/physics/2/en/chapter/21-sound-and-acoustics#def-g12-sound-acoustics-timbre) we read in [Chapter 2](https://one-course.com/books/physics/2/en/chapter/2-light-spectra-and-the-message-of-light#ch-g10-light-spectra) is this mass, arriving eight minutes late; every star shines by walking the curve toward iron.

**Remark 33.20 (ITER: bottling a star).**

No solid can hold a [plasma](#def-g12-nuclear-energy-barrier) at $1.5 \times 10^{8}\,\mathrm{K}$; a *tokamak* cages it in a [magnetic field](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-bfield) ([Chapter 15](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#ch-g11-magnetic-fields)), a doughnut of [field lines](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-lines) the charged particles spiral along without touching the wall; ITER, built to release ten times its heating [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power), is the current step. Break the confinement and the [plasma](#def-g12-nuclear-energy-barrier) cools and stops in seconds: no [chain reaction](#def-g12-nuclear-energy-chain), no [critical mass](#def-g12-nuclear-energy-chain) — the hard part is not stopping the fire but keeping it lit.

**Example 33.21 (The energy ladder, per kilogram).**

[Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) per [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of fuel: coal, $3.0 \times 10^{7}\,\mathrm{J}$; uranium-235 by [fission](#def-g12-nuclear-energy-fission), $8.2 \times 10^{13}\,\mathrm{J}$; deuterium–tritium by [fusion](#def-g12-nuclear-energy-fusion), $3.4 \times 10^{14}\,\mathrm{J}$; total conversion ($E = mc^2$), $9.0 \times 10^{16}\,\mathrm{J}$. Six orders of magnitude separate chemistry from the [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) — the reason a reactor is refuelled by the truck and a coal plant by the trainload.

## 33.6 Exercises

**Exercise 33.1 ★.**

Convert: (a) $1\,\mathrm{eV}$ and $1\,\mathrm{MeV}$ into [joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work); (b) the $200\,\mathrm{MeV}$ of one [fission](#def-g12-nuclear-energy-fission) into [joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work); (c) check from $1\,\mathrm{u} = 1.660\,54 \times 10^{-27}\,\mathrm{kg}$ that its [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) equivalent is close to $931.5\,\mathrm{MeV}$.

**Solution of Exercise 33.1.**

(a) $1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$; $1\,\mathrm{MeV} = 1.60 \times 10^{-13}\,\mathrm{J}$. (b) $200 \times 1.60 \times 10^{-13} = 3.2 \times 10^{-11}\,\mathrm{J}$. (c) $E = 1.660\,54 \times 10^{-27} \times 9.00 \times 10^{16} = 1.49 \times 10^{-10}\,\mathrm{J}
= 1.49 \times 10^{-10}/1.60 \times 10^{-13} \approx 9.3 \times 10^{2}\,\mathrm{MeV}$ — the $931.5\,\mathrm{MeV}$ of the course, up to our rounded $c$.

**Exercise 33.2 ★.**

A sugar cube has mass $6.0\,\mathrm{g}$: compute its full $E = mc^2$ value. A household draws $10\,\mathrm{kW}\,\mathrm{h} = 3.6 \times 10^{7}\,\mathrm{J}$ a day — for how many years could the cube, fully converted, [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) it?

**Solution of Exercise 33.2.**

$E = 6.0 \times 10^{-3} \times 9.00 \times 10^{16} = 5.4 \times 10^{14}\,\mathrm{J}$; $5.4 \times 10^{14}/3.6 \times 10^{7} = 1.5 \times 10^{7}$ days $\approx 4.1 \times 10^{4}$ years — forty millennia on one sugar cube.

**Exercise 33.3 ★.**

The carbon-12 [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) has mass $11.996\,71\,\mathrm{u}$: compute its [mass defect](#def-g12-nuclear-energy-defect), its [binding energy](#def-g12-nuclear-energy-defect) in $\mathrm{MeV}$, and its [binding energy per nucleon](#def-g12-nuclear-energy-pernucleon) ($m_p = 1.007\,28\,\mathrm{u}$, $m_n = 1.008\,66\,\mathrm{u}$).

**Solution of Exercise 33.3.**

$\Delta m = 6 \times 1.00728 + 6 \times 1.00866 - 11.99671 =
0.098\,93\,\mathrm{u}$; $E_b = 0.09893 \times 931.5 \approx 92.2\,\mathrm{MeV}$; $E_b/A \approx 7.68\,\mathrm{MeV}$ per [nucleon](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder).

**Exercise 33.4 ★.**

Complete each [fission](#def-g12-nuclear-energy-fission), giving $A$, $Z$ and the element ($Z = 36$: krypton, $Z = 52$: tellurium): (a) ${}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \to
{}^{144}_{56}\mathrm{Ba} + {}^{A}_{Z}\mathrm{X} + 3\,{}^{1}_{0}\mathrm{n}$; (b) ${}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \to
{}^{97}_{40}\mathrm{Zr} + {}^{A}_{Z}\mathrm{Y} + 2\,{}^{1}_{0}\mathrm{n}$.

**Solution of Exercise 33.4.**

(a) $A = 236 - 144 - 3 = 89$, $Z = 92 - 56 = 36$: ${}^{89}_{36}\mathrm{Kr}$, krypton. (b) $A = 236 - 97 - 2 = 137$, $Z = 92 - 40 = 52$: ${}^{137}_{52}\mathrm{Te}$, tellurium.

**Exercise 33.5 ★.**

Read the curve of [Proposition 33.9](#prop-g12-nuclear-energy-ironpeak): estimate $E_b/A$ for helium-4, carbon-12, iron-56 and uranium-235. Which is most tightly bound? Which could release [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) by [fission](#def-g12-nuclear-energy-fission), and which by [fusion](#def-g12-nuclear-energy-fusion)?

**Solution of Exercise 33.5.**

About $7.1$, $7.7$, $8.8$ and $7.6\,\mathrm{MeV}$ per [nucleon](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder); iron-56 is the most bound. Uranium can release [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) by [fission](#def-g12-nuclear-energy-fission), helium-4 and carbon-12 by [fusion](#def-g12-nuclear-energy-fusion) — both moves climb toward iron; iron itself has nowhere to go.

**Exercise 33.6 ★★.**

The deuteron ${}^{2}_{1}\mathrm{H}$ has mass $2.013\,55\,\mathrm{u}$: compute its [binding energy](#def-g12-nuclear-energy-defect) in $\mathrm{MeV}$. A chemical bond holds a few $\mathrm{eV}$ — by what factor does the nuclear “bond” beat it?

**Solution of Exercise 33.6.**

$\Delta m = 1.00728 + 1.00866 - 2.01355 = 0.002\,39\,\mathrm{u}$, so $E_b \approx 2.23\,\mathrm{MeV}$ — about a million times the few $\mathrm{eV}$ of a chemical bond.

**Exercise 33.7 ★★.**

Weigh the [fission](#def-g12-nuclear-energy-fission) ${}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \to
{}^{144}_{56}\mathrm{Ba} + {}^{89}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n}$: masses $234.993\,50\,\mathrm{u}$ (U), $143.892\,20\,\mathrm{u}$ (Ba), $88.897\,90\,\mathrm{u}$ (Kr), $1.008\,66\,\mathrm{u}$ (n). [Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) released, in $\mathrm{MeV}$ and in [joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work)?

**Solution of Exercise 33.7.**

Before: $236.002\,16\,\mathrm{u}$; after: $143.89220 + 88.89790 +
3 \times 1.00866 = 235.816\,08\,\mathrm{u}$; $\Delta m = 0.186\,08\,\mathrm{u}$, so $E \approx 173\,\mathrm{MeV} = 173 \times 1.60 \times 10^{-13} \approx
2.8 \times 10^{-11}\,\mathrm{J}$.

**Exercise 33.8 ★★.**

Another [fusion](#def-g12-nuclear-energy-fusion): ${}^{2}_{1}\mathrm{H} + {}^{2}_{1}\mathrm{H} \to
{}^{3}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}$, with $m({}^{3}_{2}\mathrm{He}) = 3.014\,93\,\mathrm{u}$. Compute the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) released. Why so much less than D–T? (Consider where each product sits on the curve.)

**Solution of Exercise 33.8.**

$\Delta m = 2 \times 2.01355 - 3.01493 - 1.00866 = 0.003\,51\,\mathrm{u}$: $E \approx 3.27\,\mathrm{MeV}$. D–T makes helium-4, anomalously bound ($7.1\,\mathrm{MeV}$ per [nucleon](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder)); helium-3 sits far lower on the curve, so the step releases much less.

**Exercise 33.9 ★★.**

Explain, a few lines each: (a) why [fission](#def-g12-nuclear-energy-fission) needs no heating while [fusion](#def-g12-nuclear-energy-fusion) demands about $1 \times 10^{8}\,\mathrm{K}$; (b) how the Sun manages at “only” $1.5 \times 10^{7}\,\mathrm{K}$ (the crowd’s size, its fastest members); (c) why a [fusion](#def-g12-nuclear-energy-fusion) plant cannot run away like a [chain reaction](#def-g12-nuclear-energy-chain).

**Solution of Exercise 33.9.**

(a) The neutron is neutral: no [Coulomb barrier](#def-g12-nuclear-energy-barrier), [fission](#def-g12-nuclear-energy-fission) fires cold. Fusing nuclei are both positive and must climb the barrier, so only $\sim 1 \times 10^{8}\,\mathrm{K}$ agitation gives collisions violent enough. (b) The core holds an astronomical number of protons; the rare fastest ones in the crowd do the fusing, and the Sun has billions of years to spare — a slow burn is exactly what a star wants. (c) There is no neutron avalanche: each [fusion](#def-g12-nuclear-energy-fusion) is bought by its own collision, and any loss of confinement cools the [plasma](#def-g12-nuclear-energy-barrier) and puts the fire out.

**Exercise 33.10 ★★.**

The Sun’s net reaction turns four protons ($m_p = 1.007\,28\,\mathrm{u}$) into one helium-4 [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) ($4.001\,51\,\mathrm{u}$): compute the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) per helium made, in $\mathrm{MeV}$, and the fraction of the initial mass converted.

**Solution of Exercise 33.10.**

$\Delta m = 4 \times 1.00728 - 4.00151 = 0.027\,61\,\mathrm{u}$, so $E \approx 25.7\,\mathrm{MeV}$; fraction $0.02761/4.02912 \approx 0.7\%$ of the mass.

**Exercise 33.11 ★★.**

In a water-moderated reactor: (a) what does the [moderator](#def-g12-nuclear-energy-reactor) do, and why does the chain need it? (b) the [control rods](#def-g12-nuclear-energy-reactor)? (c) the cooling water *is* the [moderator](#def-g12-nuclear-energy-reactor): why does losing it tend to choke the chain rather than speed it up?

**Solution of Exercise 33.11.**

(a) It slows [fission](#def-g12-nuclear-energy-fission) neutrons by collisions; slow neutrons are far more likely to [fission](#def-g12-nuclear-energy-fission) uranium-235, so the chain needs them. (b) They absorb neutrons, holding the chain at one neutron per [fission](#def-g12-nuclear-energy-fission) — pushed in, it dies. (c) No water, no moderation: the neutrons stay fast, miss, and the chain chokes — the design fails toward “off”.

**Exercise 33.12 ★★★.**

An uncontrolled chain doubles at each generation. From one [fission](#def-g12-nuclear-energy-fission), about how many generations until the $2.56 \times 10^{21}$ nuclei of one gram of uranium-235 have split ($2^{10} \approx 10^{3}$)? At roughly $10\,\mathrm{ns}$ per generation, how long is that — and why, then, is criticality guarded so carefully?

**Solution of Exercise 33.12.**

[Fissions](#def-g12-nuclear-energy-fission) double each generation, so after $n$ generations about $2^n$ have split; $2^{n} = 2.56 \times 10^{21} \approx 2 \times (10^{3})^{7}
\approx 2^{71}$ gives $n \approx 71$. Time: $71 \times 10\,\mathrm{ns}
\approx 0.7\,\text{µ}\mathrm{s}$ — a gram of fuel, three tonnes of coal’s worth, in under a microsecond: criticality is a cliff, not a slope.

**Exercise 33.13 ★★★.**

Could we mine [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) by fissioning iron? Weigh ${}^{56}_{26}\mathrm{Fe} \to 2\,{}^{28}_{14}\mathrm{Si}$, with $m(\mathrm{Fe}) = 55.920\,66\,\mathrm{u}$, $m(\mathrm{Si}) = 27.969\,24\,\mathrm{u}$. Compute $\Delta m$ and conclude, in terms of the curve, why iron is nuclear ash: fuel for nothing.

**Solution of Exercise 33.13.**

$\Delta m = 55.92066 - 2 \times 27.96924 = -0.017\,82\,\mathrm{u}$: $E \approx -16.6\,\mathrm{MeV}$ — the split *absorbs* [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy). Iron sits at the peak: fissioning or fusing it both go downhill in binding, both cost [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy). It is the ash of every nuclear fire.

**Exercise 33.14 ★★★.**

From $m({}^{56}_{26}\mathrm{Fe}) = 55.920\,66\,\mathrm{u}$ and $m({}^{235}_{92}\mathrm{U}) = 234.993\,50\,\mathrm{u}$, compute $E_b/A$ for both ($m_p = 1.007\,28\,\mathrm{u}$, $m_n = 1.008\,66\,\mathrm{u}$). [Fission](#def-g12-nuclear-energy-fission) fragments sit near $8.5\,\mathrm{MeV}$ per [nucleon](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder): *estimate* from the two levels the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) of one [fission](#def-g12-nuclear-energy-fission), and compare with [Example 33.12](#ex-g12-nuclear-energy-fissionbalance).

**Solution of Exercise 33.14.**

Fe: $\Delta m = 26 \times 1.00728 + 30 \times 1.00866 - 55.92066 =
0.528\,42\,\mathrm{u}$, $E_b \approx 492\,\mathrm{MeV}$, $E_b/A \approx
8.79\,\mathrm{MeV}$. U: $\Delta m = 92 \times 1.00728 + 143 \times
1.00866 - 234.99350 = 1.914\,64\,\mathrm{u}$, $E_b \approx 1784\,\mathrm{MeV}$, $E_b/A \approx 7.59\,\mathrm{MeV}$. Repacking $235$ [nucleons](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) at $8.5\,\mathrm{MeV}$ instead of $7.59\,\mathrm{MeV}$ gains $235 \times 0.9
\approx 2 \times 10^{2}\,\mathrm{MeV}$ — the $185$–$200$ of the worked example.

**Exercise 33.15 ★★★.**

ITER aims at $500\,\mathrm{MW}$ of [fusion](#def-g12-nuclear-energy-fusion) [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power). At $17.6\,\mathrm{MeV}$ per D–T reaction ($5.029\,\mathrm{u}$ of fuel consumed), find the reactions per second, then the fuel burned per day. A $500\,\mathrm{MW}$ coal furnace burns about $1.4 \times 10^{6}\,\mathrm{kg}$ a day: compare.

**Solution of Exercise 33.15.**

$5.0 \times 10^{8}/2.82 \times 10^{-12} \approx 1.8 \times 10^{20}$ reactions per second; mass rate $1.8 \times 10^{20} \times 5.029 \times 1.660\,54 \times 10^{-27}
\approx 1.5 \times 10^{-6}\,\mathrm{kg}/\mathrm{s}$, i.e. about $0.13\,\mathrm{kg}$ per day — against $1.4 \times 10^{6}$ [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of coal: ten million to one.

## 33.7 Problem: Powering a City

**Problem 33.1.**

Weekend problem — powering a city: one steady gigawatt for half a million people, bought four ways — uranium pellets, trainloads of coal, a pool of seawater, the Sun’s own substance — with the same kilogram of vanished mass hiding under all four

A city of $500\,000$ inhabitants draws a steady electric [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) $P = 1.0\,\mathrm{GW}$; its plants, whatever the fuel, turn heat into electricity with [efficiency](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) $33\%$ ([Chapter 9](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#ch-g10-energy-conservation)). Data: one uranium-235 [fission](#def-g12-nuclear-energy-fission), $200\,\mathrm{MeV}$ in all; coal, $3.0 \times 10^{7}\,\mathrm{J}/\mathrm{kg}$; one D–T [fusion](#def-g12-nuclear-energy-fusion), $17.6\,\mathrm{MeV}$ from $5.029\,\mathrm{u}$ of fuel; $1\,\mathrm{u} = 1.660\,54 \times 10^{-27}\,\mathrm{kg}$; $1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$; one year $= 3.16 \times 10^{7}\,\mathrm{s}$; the Sun: $3.9 \times 10^{26}\,\mathrm{W}$, $2.0 \times 10^{30}\,\mathrm{kg}$; a cubic [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of seawater holds about $33\,\mathrm{g}$ of deuterium.

**Part I — The [fission](#def-g12-nuclear-energy-fission) plant.**

1. What thermal [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) must the reactor deliver?
2. Express the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) of one [fission](#def-g12-nuclear-energy-fission) in [joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) .
3. How many [fissions](#def-g12-nuclear-energy-fission) per second run the city?
4. The mass of one uranium-235 [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) ( $A = 235$ ), then the uranium-235 consumed per second?
5. Deduce the uranium-235 consumed in one year, in [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) .
6. The fuel is enriched to $4.0\%$ uranium-235: what mass does a year take, and does it fit on one truck?

**Part II — The coal ledger.**

7. Compute the [thermal energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-forms) the city consumes in one year.
8. What mass of coal delivers it?
9. Compute the ratio (coal mass)/(uranium-235 mass) for one year, and check it against the ladder of [Example 33.21](#ex-g12-nuclear-energy-ladder) .
10. A freight train hauls $3.0 \times 10^{6}\,\mathrm{kg}$ : how many trains per year, and per day? One sentence: picture the two supply lines.

**Part III — The [fusion](#def-g12-nuclear-energy-fusion) dream.**

11. Express the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) of one D–T [fusion](#def-g12-nuclear-energy-fusion) in [joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) .
12. Compute the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) released per [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of D–T fuel; compare with the ladder.
13. What mass of D–T fuel covers the city’s year? What mass of that is deuterium ( $2.014$ of the $5.029\,\mathrm{u}$ )?
14. What volume of seawater contains that deuterium? Compare with an Olympic pool, about $2.5 \times 10^{3}\,\mathrm{m}^{3}$ . (Tritium is bred from lithium.)
15. Why does this plant carry no [critical mass](#def-g12-nuclear-energy-chain) and no [chain reaction](#def-g12-nuclear-energy-chain) to fear? What, then, is the hard part?

**Part IV — The star that pays in mass.**

16. From $E = mc^2$ , what mass does the Sun convert each second?
17. What mass per second, and per year, must *any* $3.0\,\mathrm{GW}$ -thermal source convert? Check that all three fuels surrender this same mass.
18. What total mass has the Sun radiated away in its $4.5 \times 10^{9}$ years, and what fraction of the Sun is that?
19. [Fusion](#def-g12-nuclear-energy-fusion) can tap about $0.07\%$ of the Sun’s mass (the burnable share of its core): estimate its total lifetime as a star, and what is left.
20. Close the audit in three sentences: one year of the city in tonnes of coal, [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of uranium-235 and [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of D–T fuel — and the [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of mass that, in every scenario, became the city’s light.

**Solution of Problem 33.1.**

**1.** $P_{\text{th}} = 1.0 \times 10^{9}/0.33 \approx 3.0\,\mathrm{GW}$.

**2.** $200 \times 1.60 \times 10^{-13} = 3.2 \times 10^{-11}\,\mathrm{J}$.

**3.** $3.0 \times 10^{9}/3.2 \times 10^{-11} \approx 9.4 \times 10^{19}$ [fissions](#def-g12-nuclear-energy-fission) per second.

**4.** $m_{\mathrm U} = 235 \times 1.660\,54 \times 10^{-27} =
3.90 \times 10^{-25}\,\mathrm{kg}$; $9.4 \times 10^{19} \times 3.90 \times 10^{-25} \approx
3.7 \times 10^{-5}\,\mathrm{kg}/\mathrm{s}$ — some forty micrograms a second.

**5.** $3.7 \times 10^{-5} \times 3.16 \times 10^{7} \approx 1.2 \times 10^{3}\,\mathrm{kg}$: about $1.2$ tonnes of uranium-235 a year.

**6.** $1.2 \times 10^{3}/0.040 \approx 2.9 \times 10^{4}\,\mathrm{kg}$ — some $29$ tonnes of fuel: one heavy truck, once a year.

**7.** $3.0 \times 10^{9} \times 3.16 \times 10^{7} \approx 9.5 \times 10^{16}\,\mathrm{J}$.

**8.** $9.5 \times 10^{16}/3.0 \times 10^{7} \approx 3.2 \times 10^{9}\,\mathrm{kg}$: $3.2$ million tonnes of coal.

**9.** $3.2 \times 10^{9}/1.2 \times 10^{3} \approx 2.7 \times 10^{6}$ — the $8.2 \times 10^{13}/3.0 \times 10^{7}$ of the ladder, as it must be.

**10.** $3.2 \times 10^{9}/3.0 \times 10^{6} \approx 1060$ trains a year, roughly three a day: a coal train at dawn, noon and dusk for ever, against one truck each spring.

**11.** $17.6 \times 1.60 \times 10^{-13} = 2.82 \times 10^{-12}\,\mathrm{J}$.

**12.** One pair weighs $5.029 \times 1.660\,54 \times 10^{-27} =
8.35 \times 10^{-27}\,\mathrm{kg}$: $2.82 \times 10^{-12}/8.35 \times 10^{-27} \approx
3.4 \times 10^{14}\,\mathrm{J}/\mathrm{kg}$ — the ladder’s [fusion](#def-g12-nuclear-energy-fusion) rung.

**13.** $9.5 \times 10^{16}/3.4 \times 10^{14} \approx 2.8 \times 10^{2}\,\mathrm{kg}$ of D–T; deuterium share $280 \times 2.014/5.029 \approx 1.1 \times 10^{2}\,\mathrm{kg}$.

**14.** $1.1 \times 10^{2}/3.3 \times 10^{-2} \approx 3.4 \times 10^{3}\,\mathrm{m}^{3}$ of seawater — about a pool and a half for the whole city’s year.

**15.** Nothing multiplies: no neutron lights the next [fusion](#def-g12-nuclear-energy-fusion), and losing confinement cools the [plasma](#def-g12-nuclear-energy-barrier) and stops it within seconds. The hard part is the opposite — holding $1.5 \times 10^{8}\,\mathrm{K}$ together long enough to burn.

**16.** $3.9 \times 10^{26}/9.00 \times 10^{16} \approx 4.3 \times 10^{9}\,\mathrm{kg}/\mathrm{s}$: four million tonnes a second.

**17.** $3.0 \times 10^{9}/9.00 \times 10^{16} \approx 3.3 \times 10^{-8}\,\mathrm{kg}/\mathrm{s}$, i.e. $3.3 \times 10^{-8} \times 3.16 \times 10^{7} \approx 1.0\,\mathrm{kg}$ a year — and yes: $1.2 \times 10^{3}\,\mathrm{kg} \times 0.00085 \approx 1$, $3.2 \times 10^{9}
\times 3.3 \times 10^{-10} \approx 1$, $280 \times 0.00375 \approx 1$. Every fuel surrenders the same [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit); they differ only in how much cargo must carry it.

**18.** $4.3 \times 10^{9} \times 3.16 \times 10^{7} \times 4.5 \times 10^{9}
\approx 6.1 \times 10^{26}\,\mathrm{kg}$ — only $6.1 \times 10^{26}/2.0 \times 10^{30} \approx
3 \times 10^{-4}$, three parts in ten thousand.

**19.** Convertible mass $0.0007 \times 2.0 \times 10^{30} =
1.4 \times 10^{27}\,\mathrm{kg}$, worth $1.4 \times 10^{27} \times 9.00 \times 10^{16} \approx
1.3 \times 10^{44}\,\mathrm{J}$; at $3.9 \times 10^{26}\,\mathrm{W}$ that lasts $1.3 \times 10^{44}/
3.9 \times 10^{26} \approx 3.2 \times 10^{17}\,\mathrm{s} \approx 1.0 \times 10^{10}$ years — ten billion, of which some five billion remain.

**20.** One year of the city: $3.2$ million tonnes of coal, or $1.2 \times 10^{3}$ [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of uranium-235, or $2.8 \times 10^{2}$ [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of D–T drawn from a pool and a half of seawater. Under every ledger the same entry: one [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of mass, gone. $E = mc^2$ does not care which fuel carries it — only how big a truck it takes.