Corrected exercises: nuclear physics and radioactivity (203-SN3-RE)

Aluminum-27: $R=1.20\times 27^{1/3}=1.20\times 3=3.60$ fm, a rare case where the calculation is exact. Uranium-238: $238^{1/3}=6.198$, so $R=7.44$ fm. The heaviest nucleus is therefore only twice as large as the aluminum nucleus, although it contains nine times as many nucleons. That is where the cube root comes from: the volume of the nucleus is proportional to the NUMBER of nucleons, $V\propto A$, which means that the nucleons are packed in contact with one another, like marbles in a bag, without ever being compressed further. It is the signature of a short-range, saturated nuclear force: each nucleon interacts only with its immediate neighbours, not with the whole nucleus.

Mass: $m=A\times 1.66054\times 10^{-27}$ kg. Volume: $V=\dfrac{4}{3}\pi R^{3}=\dfrac{4}{3}\pi r_{0}^{3}A$. The factor $A$ appears IN THE NUMERATOR AND IN THE DENOMINATOR and cancels: $\rho=\dfrac{1.66054\times 10^{-27}}{\frac{4}{3}\pi(1.20\times 10^{-15})^{3}}=\dfrac{1.66054\times 10^{-27}}{7.238\times 10^{-45}}=2.29\times 10^{17}$ kg/m$^{3}$. All nuclei, from hydrogen to uranium, therefore have the same density, to within a few percent. This is the direct consequence of the $A^{1/3}$ law, and the converse is true too: it was by measuring this constant density through electron scattering that the law was established.

Ratio of the radii for aluminum: $\dfrac{1.0\times 10^{-10}}{3.60\times 10^{-15}}=2.8\times 10^{4}$. The atom is thus about thirty thousand times wider than its nucleus. Ratio of the volumes: the cube, that is $2.1\times 10^{13}$, more than twenty trillion. In other words, if the nucleus were the size of a $1$ cm marble, the nearest electron would be $280$ metres away, and there would be nothing in between. Ordinary matter is essentially empty space, its mass is concentrated in a hundred-thousandth of its size, and its apparent solidity comes not from space being filled but from the repulsion between electron clouds.

$m=\rho V=2.29\times 10^{17}\times 5.0\times 10^{-6}=1.1\times 10^{12}$ kg, about a billion tonnes. That is the order of magnitude of the mass of a mountain, or three thousand times that of the largest ocean liner ever built, in the volume of a teaspoon. This figure is no game: it is exactly the density of a neutron star, which is nothing other than an atomic nucleus twenty kilometres in radius held together by gravity.

$B=[Z\,m(^{1}\mathrm{H})+N\,m_{n}-m(^{A}_{Z}\mathrm{X})]c^{2}$. The proof comes down to counting electrons. The right-hand side contains $Z$ hydrogen atoms, so $Z$ protons AND $Z$ electrons; the subtracted term is the complete atom, which also contains $Z$ electrons. The two packets of $Z$ electrons cancel exactly, leaving only what we want, the comparison between the free nucleons and the bound nucleus. That is why tables publish atomic masses: they are what a mass spectrometer actually measures, and they can be used as they stand. The only approximation is the binding energy of the electrons themselves, a few tens of eV, six orders of magnitude below what we are calculating.

Helium-4: $2(1.007825)+2(1.008665)-4.002602=0.030378$ u, so $B=0.030378\times 931.494=28.30$ MeV and $B/A=7.074$ MeV. Iron-56: $26(1.007825)+30(1.008665)-55.934936=0.528464$ u, so $B=492.3$ MeV and $B/A=8.790$ MeV. Uranium-238: $92(1.007825)+146(1.008665)-238.050788=1.934202$ u, so $B=1801.7$ MeV and $B/A=7.570$ MeV. The three points fall on the curve: helium low on the left despite its reputation as a very tightly bound nucleus, iron at the peak, uranium back down.

Because stability is not measured by the total binding energy, which grows trivially with the size of the nucleus, but by the binding energy PER NUCLEON, which tells how firmly each nucleon is held. Iron-56 is at $8.790$ MeV per nucleon, uranium-238 at $7.570$: each nucleon of uranium is less firmly held than each nucleon of iron, so an exothermic reaction toward fragments closer to the peak is possible. The physical cause is the Coulomb repulsion, which grows as $Z^{2}$ and acts over long range across the whole nucleus, whereas the attractive, saturated nuclear force grows only as $A$. Past iron, the repulsion gradually wins, and this is the whole energetics of fission.

Each hydrogen atom contains an electron of $0.000549$ u. With bare protons, the student subtracts $26\times 0.000549=0.014274$ u from the first term, without correcting the subtracted term, which still contains the $26$ electrons of iron. The result for $B$ is therefore too small by $0.014274\times 931.494=13.30$ MeV, giving $479.0$ MeV instead of $492.3$: an error of $2.7$%. It is insidious because it leaves a plausible result of the right order of magnitude, which will go unnoticed on rereading. The only safeguard is to ask yourself, before any calculation and every time, whether the table gives atomic or nuclear masses; in practice it almost always gives atomic masses.

Conservation of the mass number $A$ and conservation of the charge $Z$: the sum of the $A$ values and the sum of the $Z$ values are the same on both sides. These two lines are enough, there is nothing else to memorize. $^{238}_{92}\mathrm{U}\to\,^{234}_{90}\mathrm{Th}+^{4}_{2}\mathrm{He}$: $238=234+4$ and $92=90+2$. $^{226}_{88}\mathrm{Ra}\to\,^{222}_{86}\mathrm{Rn}+^{4}_{2}\mathrm{He}$: radium indeed gives radon, which incidentally explains why this gas appears in the basements of granitic regions.

$\beta^{-}$ of carbon-14: $^{14}_{6}\mathrm{C}\to\,^{14}_{7}\mathrm{N}+e^{-}+\bar{\nu}_{e}$. A neutron of the nucleus has turned into a proton, $Z$ increases by one, $A$ does not change, and it is an ANTINEUTRINO that accompanies the electron. $\beta^{+}$ of sodium-22: $^{22}_{11}\mathrm{Na}\to\,^{22}_{10}\mathrm{Ne}+e^{+}+\nu_{e}$. A proton becomes a neutron, $Z$ decreases, and this time it is a neutrino. The rule that avoids the mistake: the light particle emitted with the electron is an ANTIparticle, the one emitted with the positron is a particle, so that the lepton number is conserved as well.

$^{40}_{19}\mathrm{K}\to\,^{40}_{18}\mathrm{Ar}+e^{+}+\nu_{e}$, $\beta^{+}$ decay, which also occurs by electron capture, the nucleus swallowing an electron from its electron cloud. And $^{40}_{19}\mathrm{K}\to\,^{40}_{20}\mathrm{Ca}+e^{-}+\bar{\nu}_{e}$, $\beta^{-}$ decay. A nucleus that can decay in BOTH DIRECTIONS sits exactly on the ridge between the two sides of the valley of stability: its neighbours on the left and on the right, argon-40 and calcium-40, are both more tightly bound than it is, and it has no way of choosing. This dual mode is also what makes potassium-40 usable for dating rocks, the branch toward argon producing a gas that stays trapped in the crystal.

$^{60}_{27}\mathrm{Co}\to\,^{60}_{28}\mathrm{Ni}^{*}+e^{-}+\bar{\nu}_{e}$, then $^{60}_{28}\mathrm{Ni}^{*}\to\,^{60}_{28}\mathrm{Ni}+\gamma$ twice in succession. $\gamma$ emission changes neither $A$ nor $Z$: the nucleus remains rigorously the same element and the same isotope, and merely loses excitation energy, exactly like an atom falling back to its ground state by emitting light, except that the energies are a million times larger. It is therefore not a TRANSMUTATION, whereas $\alpha$ and $\beta$ decays are. In practice, $\gamma$ emission almost always accompanies another decay: it is its remainder, not the event itself.

Because the energy spectrum of $\beta$ electrons is CONTINUOUS, whereas the decay of a well-defined nucleus into another well-defined nucleus must release a fixed energy. If the electron were alone, it should always carry away the same energy, and experiment showed instead a whole range, from zero up to a maximum. In 1930, Pauli preferred to postulate an invisible particle sharing the energy with the electron rather than give up conservation of energy, which Bohr was ready to sacrifice. This particle had to be neutral, of almost zero mass, and to interact only through the weak interaction, which is to say practically undetectable: it took until 1956 and a nuclear reactor, which supplies a colossal flux of antineutrinos, to observe it directly. It is the best example in the chapter of a prediction forced by a conservation law.

$Q=[238.050788-234.043601-4.002602]\times 931.494=0.004585\times 931.494=4.271$ MeV, tabulated value $4.27$ MeV. Check of the electron count: on the left, the uranium atom carries $92$; on the right, thorium carries $90$ and helium $2$, that is $92$. Both sides have the same number, they cancel, and atomic masses are used as they stand. Note how delicate the calculation is: a difference of $0.0046$ between numbers of six significant figures. Rounding the masses to four decimals would completely destroy the result.

$Q=[14.003242-14.003074]\times 931.494=0.000168\times 931.494=0.1565$ MeV, that is $156$ keV, the tabulated value. Count: on the left, the carbon atom carries $6$ electrons; on the right, the nitrogen atom carries $7$, but the emitted $\beta$ electron is precisely the seventh. The total is $7$ on both sides, and the books balance without any correction term. This is a happy accident of $\beta^{-}$ decay, and a misleading one: it makes you think the rule is general.

It is not. Count for $\beta^{+}$: on the left, the sodium atom carries $11$ electrons; on the right, the neon atom carries only $10$, and a POSITRON must also be created. In terms of mass, there is therefore one electron too many on the left and one positron more on the right, a deficit of two electron masses: $Q=[m_{\text{Na}}-m_{\text{Ne}}-2m_{e}]c^{2}$. Numerically: $[21.994437-21.991385]\times 931.494=2.843$ MeV, minus $1.022$ MeV, so $Q=1.821$ MeV. The tabulated value is $1.820$ MeV. This $1.022$ MeV is the toll of the chapter, and forgetting it is the most common mistake on tests.

Conservation of momentum, the initial nucleus being at rest: $p_{\alpha}=p_{\text{Th}}$, so $K=\dfrac{p^{2}}{2m}$ is shared in inverse proportion to the masses. $K_{\alpha}=Q\dfrac{m_{\text{Th}}}{m_{\text{Th}}+m_{\alpha}}=4.271\times\dfrac{234}{238}=4.199$ MeV and $K_{\text{Th}}=4.271\times\dfrac{4}{238}=0.0718$ MeV, that is $71.8$ keV. The $\alpha$ particle carries away $98.3$% of the energy, which is exactly why the energy of $\alpha$ particles is tabulated as if it were $Q$. The nonrelativistic approximation is justified by comparing $K_{\alpha}$ with the rest energy of helium, $3727$ MeV: the ratio is $1.1\times 10^{-3}$, so $\gamma-1\approx 10^{-3}$, and the relativistic correction is of the order of a thousandth, far below the precision of the masses used.

This nucleus can NOT decay by $\beta^{+}$: it is short of $1.022$ MeV just to pay for creating the pair, and $Q$ would be negative. It is not stable for all that, because another route remains open to it, ELECTRON CAPTURE: the nucleus swallows an electron from its own $K$ shell, $p+e^{-}\to n+\nu_{e}$, which produces exactly the same daughter nucleus without having to create a positron. The threshold for electron capture is therefore zero, and it is the only mode open in the window between $0$ and $1.022$ MeV. It is detected experimentally by the characteristic X-ray emitted when an electron from a higher shell fills the vacancy.

$N=\dfrac{1.00\times 10^{-6}}{131}\times 6.022\times 10^{23}=4.60\times 10^{15}$ atoms. One microgram is already four million billion nuclei: this is why a dangerous radioactive source can be chemically undetectable. Decay constant: $\lambda=\dfrac{\ln 2}{T_{1/2}}=\dfrac{0.693}{8.02\times 86{,}400}=1.00\times 10^{-6}$ s$^{-1}$. Watch the unit of time: $\lambda$ must be in s$^{-1}$ for the activity to come out in becquerels, and converting the half-life to seconds is the first thing to do, before any other calculation.

$A=\lambda N=1.00\times 10^{-6}\times 4.60\times 10^{15}=4.60\times 10^{9}$ Bq, that is $4.60$ GBq or $\dfrac{4.60\times 10^{9}}{3.7\times 10^{7}}=124$ mCi. These two units still coexist in nuclear medicine departments, since doses are often prescribed in millicuries and measured in megabecquerels: being able to convert from one to the other is part of the chapter.

$24.1$ days is three half-lives give or take a few hours: $\dfrac{24.1}{8.02}=3.01$. The activity is therefore divided by $2^{3}=8$, which can be done in your head: about $0.58$ GBq. Check: $4.60\times 2^{-3.01}=0.573$ GBq. The mental estimate is worth making systematically, because it immediately catches an exponential error: whenever the duration is a whole multiple of the half-life, the answer must be a power of two.

$\dfrac{A}{A_{0}}=2^{-n}=0.0100$ gives $n=\dfrac{\ln 100}{\ln 2}=6.64$ half-lives. This is a number to remember, it comes up constantly: it takes about seven half-lives to fall to 1%, and ten to fall to a thousandth. In days: $t=6.64\times 8.02=53.3$ days. This is the time after which the waste of an iodine-131 department can be treated as ordinary waste, and it is fixed by this arithmetic and nothing else.

For equal masses, the numbers of atoms are comparable, up to the molar mass, but $\lambda$ is inversely proportional to the half-life: that of technetium-99m is $\dfrac{8.02\times 24}{6.0}=32$ times shorter, so its specific activity is about $32$ times GREATER. A short half-life means an intense activity for very little material. Choosing a medical isotope is a compromise between two opposing requirements: the half-life must be long enough for the product to survive its production, its transport and its uptake in the organ, and short enough for the patient to stop being irradiated quickly once the examination is over. A few hours is the optimum for diagnosis, which explains the dominance of technetium-99m; iodine-131, with its longer half-life and $\beta$ emission, is used mainly for therapy, where you WANT the irradiation to last.

$N_{\text{U}}(t)=N_{0}e^{-\lambda t}$, and each uranium atom that has disappeared has produced one lead atom, so $N_{\text{Pb}}(t)=N_{0}-N_{0}e^{-\lambda t}=N_{0}(1-e^{-\lambda t})$. The ratio is $\dfrac{N_{\text{Pb}}}{N_{\text{U}}}=\dfrac{1-e^{-\lambda t}}{e^{-\lambda t}}=e^{\lambda t}-1$. This is the form to remember: the daughter-to-parent ratio grows EXPONENTIALLY and without bound, whereas the fraction of parent remaining decreases. Inverting gives $t=\dfrac{1}{\lambda}\ln\left(1+\dfrac{N_{\text{Pb}}}{N_{\text{U}}}\right)$. Notice that $N_{0}$, the initial amount of uranium, has vanished: this is what makes the method usable, since nobody knows how much uranium the crystal contained when it formed.

$\lambda=\dfrac{\ln 2}{4.468\times 10^{9}}=1.551\times 10^{-10}$ yr$^{-1}$. Then $t=\dfrac{\ln 1.500}{1.551\times 10^{-10}}=\dfrac{0.4055}{1.551\times 10^{-10}}=2.61\times 10^{9}$ years, that is $2.61$ billion years. An order-of-magnitude check, to be made first: the ratio $0.500$ means that a third of the initial uranium has become lead, so two thirds remain, which is a little more than half, so the age must be a little less than one half-life. $2.61$ against $4.47$ billion: consistent.

Because the first step is by far the slowest, and a chain goes at the pace of its slowest link. The half-life of uranium-238 is four and a half billion years, those of all the intermediates range from a fraction of a second to the $245{,}000$ years of uranium-234, at least four orders of magnitude below. As soon as a uranium-238 atom finally decays, its descendants run through the whole chain in a time that is negligible on the geological scale and end up as lead-206. On the scale of a million years, the chain therefore behaves exactly like a direct transformation, and none of the thirteen intermediates needs to be followed.

$\lambda_{235}=\dfrac{\ln 2}{7.04\times 10^{8}}=9.845\times 10^{-10}$ yr$^{-1}$, so $\lambda_{235}t=9.845\times 10^{-10}\times 2.61\times 10^{9}=2.573$ and $\dfrac{N_{207}}{N_{235}}=e^{2.573}-1=12.1$. This second measurement is valuable because it is INDEPENDENT: two clocks with very different constants, one slow and the other six times faster, are running in the same crystal. If they give the same age, the dating is said to be concordant, and it becomes extremely unlikely that a disturbance has distorted both in exactly the same way. If they diverge, the gap itself tells us about the event that disturbed the crystal, and this is the principle of the concordia curve used by geochronologists.

First assumption: the crystal contained NO lead-206 when it formed. For zircon, it is solid for a chemical reason rather than a physical one: the zircon lattice readily accepts uranium in place of zirconium, whose ionic radius is close, and on the contrary rejects lead. A zircon is therefore born with practically no lead, and all the lead found in it is radiogenic. Second assumption: the system has remained CLOSED, with no uranium or lead entering or leaving since the formation. Here again zircon is exceptional: it is an extremely refractory and chemically inert mineral, which survives erosion, transport and even a metamorphic episode. These two properties, and not any superiority of the physics, make zircon the reference mineral of geochronology, and allowed the oldest known fragments of Earth's crust to be dated at $4.4$ billion years.

Secular equilibrium is the state in which the daughter is produced exactly as fast as it disappears, so that its quantity stays constant even though it is unstable. It is established on the condition that the half-life of the parent be VERY LONG compared with that of the daughter, $T_{1}\gg T_{2}$: the parent then feeds the daughter population at a practically constant rate over the lifetime of the daughter. Here the ratio is $4.47$ billion years against $1600$ years, a factor of three million: the condition is amply met.

The number of daughter atoms obeys $\dfrac{dN_{2}}{dt}=\lambda_{1}N_{1}-\lambda_{2}N_{2}$, the production by the parent minus the daughter's own disappearance. At equilibrium, $N_{2}$ no longer varies, so $\dfrac{dN_{2}}{dt}=0$, which immediately gives $\lambda_{1}N_{1}=\lambda_{2}N_{2}$, that is $A_{1}=A_{2}$: the activities are equal. Replacing $\lambda$ by $\dfrac{\ln 2}{T}$, the $\ln 2$ cancels and what remains is $\dfrac{N_{1}}{T_{1}}=\dfrac{N_{2}}{T_{2}}$, that is $\dfrac{N_{2}}{N_{1}}=\dfrac{T_{2}}{T_{1}}$. The result deserves to be read in words: the daughter is present in a quantity that is smaller the more short-lived it is, in the exact ratio of the half-lives.

$\dfrac{N_{\text{Ra}}}{N_{\text{U}}}=\dfrac{1600}{4.468\times 10^{9}}=3.58\times 10^{-7}$. To go to masses, we multiply by the ratio of the molar masses: $m_{\text{Ra}}=3.58\times 10^{-7}\times\dfrac{226}{238}\times 10^{6}$ g $=0.340$ g per tonne of uranium. About a third of a gram of radium per tonne: this is exactly the yield obtained by Marie Curie, who processed several tonnes of pitchblende residue to isolate a few decigrams of radium chloride. This figure, which nuclear physics gives in two lines, cost four years of chemical work in 1898, before it was known that a decay chain fixes its value.

One tonne of uranium-238 contains $N=\dfrac{10^{6}}{238}\times 6.022\times 10^{23}=2.53\times 10^{27}$ atoms. $\lambda_{\text{U}}=\dfrac{\ln 2}{4.468\times 10^{9}\times 3.156\times 10^{7}}=4.92\times 10^{-18}$ s$^{-1}$, in seconds. Activity: $A=1.24\times 10^{10}$ Bq, that is about $12$ GBq per tonne. Check through the radium: $N_{\text{Ra}}=3.58\times 10^{-7}\times 2.53\times 10^{27}=9.07\times 10^{20}$ atoms, $\lambda_{\text{Ra}}=\dfrac{\ln 2}{1600\times 3.156\times 10^{7}}=1.37\times 10^{-11}$ s$^{-1}$, hence $A=1.24\times 10^{10}$ Bq. The two coincide, as the theory demands. One tonne of natural uranium, a material that is handled routinely, therefore emits as much as the third of a gram of radium it contains.

The rise toward equilibrium follows $A_{2}(t)=A_{1}\left(1-e^{-\lambda_{2}t}\right)$, governed by the constant of the DAUGHTER and not that of the parent: students regularly swap the two. We want $1-e^{-\lambda_{2}t}=0.99$, that is $\lambda_{2}t=\ln 100=4.605$, so $t=\dfrac{4.605}{\ln 2}\times 3.82=6.64\times 3.82=25.4$ days. We find the same $6.64$ half-lives as for a decay to 1%, which is logical since it is the complement of the same exponential. One month after purification, a radium sample is therefore in equilibrium with its radon, and this is why a sealed radium source must be handled as a source of radioactive gas.

False, and the mistake is to treat an exponential as a straight line. After two half-lives, $\left(\tfrac{1}{2}\right)^{2}=25$% of the initial activity remains, not zero. Radioactive decay NEVER reaches zero in a finite time: it is a multiplicative process, each half-life divides what remains by two. Correct statement: after $16$ days a quarter of the activity remains, and it takes about seven half-lives, that is $53$ days, to get down to one hundredth.

The statement confuses two quantities. The TOTAL binding energy naturally grows with the size of the nucleus, a nucleus of $238$ nucleons simply having more bonds to account for than a nucleus of $56$. The right criterion is the binding energy PER NUCLEON: $7.570$ MeV for uranium against $8.790$ MeV for iron. Correct statement: iron-56 is the more tightly bound of the two, and it sits at the peak of the curve, which makes it the end point of all stellar nucleosynthesis and the point beyond which no fusion releases energy any more.

False. The $\beta^{-}$ electron is CREATED at the moment of the decay, when a neutron turns into a proton, $n\to p+e^{-}+\bar{\nu}_{e}$: it did not exist beforehand, neither in the nucleus nor anywhere else. The same goes for the positron of $\beta^{+}$ decay. The idea that an emitted particle must have pre-existed is a reflex of classical mechanics, and the weak interaction does not obey it: it converts particles into one another. Correct statement: the electron cloud takes part in only one specific mode, electron capture, where the nucleus on the contrary swallows an atomic electron.

The statement says the right thing for the wrong reason, and so gets wrong what matters. The low penetrating power of the $\alpha$ makes it harmless OUTSIDE the body, where the dead layer of the skin is enough to stop it. But it becomes by far the most dangerous in INTERNAL contamination, inhaled or ingested, precisely because it deposits all its energy within a few tens of micrometres, in living tissue: hence its quality factor of $20$ against $1$ for $\beta$ and $\gamma$. Correct statement: the danger of a radiation only makes sense together with an exposure scenario, external or internal, and inhaled radon is the typical example of a formidable $\alpha$ emitter.

False. The activity is $A=\lambda N$: for a given mass, it is inversely proportional to the half-life and also, more weakly, to the molar mass. One gram of technetium-99m, with a half-life of six hours, is billions of times more active than one gram of uranium-238, with a half-life of four and a half billion years. Correct statement: what distinguishes isotopes is not their mass but their $\lambda$, and the half-lives encountered range from the nanosecond to $10^{19}$ years, thirty-six orders of magnitude. No other quantity in the program has such a span.

Thermal power: $P_{\text{th}}=\dfrac{1000}{0.33}=3.03\times 10^{3}$ MW $=3.03\times 10^{9}$ W. Two thirds go to the condenser and the cooling tower, and this is not a design flaw but the Carnot limit applied to the temperatures of a water circuit. Energy of one fission: $200\times 10^{6}\times 1.602\times 10^{-19}=3.204\times 10^{-11}$ J. Number of fissions per second: $\dfrac{3.03\times 10^{9}}{3.204\times 10^{-11}}=9.46\times 10^{19}$ s$^{-1}$. Almost $10^{20}$ fissions per second, in a core that could fit in a room.

Number of fissions per day: $9.46\times 10^{19}\times 86{,}400=8.17\times 10^{24}$. Corresponding mass: $\dfrac{8.17\times 10^{24}}{6.022\times 10^{23}}\times 235=3.19\times 10^{3}$ g, that is $3.19$ kg per day. A little over three kilograms of uranium-235 fissioned per day to power a city: this is the figure that makes nuclear energy singular, and it must be handled honestly, since the mass of fuel actually loaded in the core is much larger, several tens of tonnes, because only a fraction of the uranium is fissile and undergoes fission before replacement.

$\dfrac{3.03\times 10^{9}}{3.0\times 10^{7}}=101$ kg/s, that is $101\times 86{,}400=8.73\times 10^{6}$ kg per day, about $8700$ tonnes of coal daily. The ratio with the $3.19$ kg of uranium is $2.7$ million. This is the order of magnitude that separates a chemical energy, a few electron-volts per atom, from a nuclear energy, two hundred million electron-volts per nucleus: a factor of the order of $10^{7}$ at the source, reduced here by the fact that not all the loaded uranium undergoes fission. One trainload of coal per day against one suitcase.

The thermal energy produced in one day is $3.03\times 10^{9}\times 86{,}400=2.62\times 10^{14}$ J. The corresponding mass is $m=\dfrac{E}{c^{2}}=\dfrac{2.62\times 10^{14}}{9.00\times 10^{16}}=2.91\times 10^{-3}$ kg, about $2.9$ grams. Relative to the $3.19$ kg consumed, this is $0.091$%, roughly one thousandth. This is the concrete reading of $E=mc^{2}$ in this chapter: fission does not make the fuel disappear, it rearranges its nucleons and converts only a thousandth of it into energy. The three kilograms of fissioned matter are still there afterward, as fission products, and that is exactly the waste problem.

If all fissions were instantaneous, the time between two generations of neutrons would be of the order of $10^{-4}$ s, even $10^{-8}$ s for prompt neutrons. A deviation of only $0.1$% in $k$ would then make the power grow by a factor of $e$ every fraction of a second: no control rod, no automatic system, no operator could react. What makes control possible is that about $0.65$% of fission neutrons are DELAYED neutrons, emitted a few seconds to a few minutes after the fission by the decay of certain products. The reactor is operated in a regime where criticality is reached ONLY with the contribution of these delayed neutrons, which brings the effective generation time to about ten seconds. The whole controllability of a reactor rests on this one percent of late neutrons, and a reactivity accident is precisely the crossing of the threshold beyond which the reactor becomes critical without them.

$\lambda=\dfrac{\ln 2}{3.82\times 86{,}400}=2.10\times 10^{-6}$ s$^{-1}$. From $A=\lambda N$ we get $N=\dfrac{400}{2.10\times 10^{-6}}=1.90\times 10^{8}$ atoms per cubic metre. Mass: $\dfrac{1.90\times 10^{8}}{6.022\times 10^{23}}\times 222=7.0\times 10^{-14}$ g per cubic metre, that is $70$ femtograms. The comment is the heart of the question: a concentration deemed worrying by every health agency corresponds to less than a tenth of a picogram per cubic metre. No chemical analysis in the world would detect that, and this is why radon is measured by its activity and never by its concentration. Radioactivity is, by a wide margin, the most sensitive analytical method available to physics.