Apply independently

Given

green $\lambda_\text{v} = 557.7\,\text{nm}$, red $\lambda_\text{p} = 630.0\,\text{nm}$, the energy levels of the figure, $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$ and $c = 2.998 \cdot 10^{8}\,\text{m/s}$

Find

the transitions corresponding to the colours and the wavelength $\lambda_3$ of the third transition

Principle

When an atom moves from an upper energy level to a lower one, it emits a photon whose energy is the difference of the levels. We calculate the photon energy of each colour and compare it with the differences of the levels. The third possible transition is from the second excited state directly to the ground state.

Reading the graph

From the figure we read the ground state $0$, the first excited state $1.97\,\text{eV}$ and the second excited state $4.19\,\text{eV}$. The differences of the levels are $4.19\,\text{eV} - 1.97\,\text{eV} = 2.22\,\text{eV}$, $1.97\,\text{eV} - 0 = 1.97\,\text{eV}$ and $4.19\,\text{eV} - 0 = 4.19\,\text{eV}$.

Formula

$E = hf = \dfrac{hc}{\lambda}$, so for the third transition $\lambda_3 = \dfrac{hc}{E_3}$

Conversions

$\lambda_\text{v} = 557.7 \cdot 10^{-9}\,\text{m}$ and $\lambda_\text{p} = 630.0 \cdot 10^{-9}\,\text{m}$. The electronvolts are converted by substituting $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$:

Substitution

Green:

so green light is produced in the transition from the second excited state to the lower one. Red: in the same way

so red light is produced in the transition from the first excited state to the ground state. Third transition:

Answer

Green light is produced in the transition from the second excited state to the lower one and red light from the first excited state to the ground state. The wavelength of the third transition is $\lambda_3 \approx 296\,\text{nm}$, three significant figures like the energy levels. It is ultraviolet radiation, which the eye cannot see.

Reasonableness

The greatest energy difference gives the shortest wavelength, and $296\,\text{nm}$ is shorter than the lower limit of visible light, about $380\,\text{nm}$. The energies of photons of visible light are $1.7$–$3.3\,\text{eV}$, and both colours of the aurora fall in this range, but $4.19\,\text{eV}$ does not.

Given

kinetic energy of the electron $E_\text{k} = 5\,\text{keV}$, energy of the second excited state $\Delta E = 4.19\,\text{eV}$ above the ground state

Find

the greatest possible number of exciting collisions $n$

Principle

In every exciting collision the electron gives the atom at least the difference of the levels in energy. The greatest number of excitations is obtained if the whole kinetic energy is used for them and for nothing else.

Formula

$n = \dfrac{E_\text{k}}{\Delta E}$

Conversions

A conversion to joules is not needed, because both energies are in electronvolts.

Substitution

Answer

One electron can excite at most about $1200$ oxygen atoms to the second excited state.

Reasonableness

The assumption was that all the energy is used for this excitation. In reality the electron also ionises atoms and molecules, excites nitrogen molecules and other levels, and transfers energy into the kinetic energy of particles, so there are clearly fewer excitations. In any case one fast electron produces hundreds of photons, which explains why auroras are bright.

Find

why red light is not produced lower down and why the spectrum consists of lines

Principle

The energy of an atom can have only certain values. An excited atom can give up its energy in two ways: by emitting a photon or in a collision in which the energy is transferred to another particle. Which happens first depends on the mean lifetime of the excited state and on how often the atom collides.

Model answer

Spectral lines. The energy of an oxygen atom can have only the values of the energy levels. When an atom moves from one level to another, the energy of the photon produced is the difference of the levels, and a definite wavelength follows from it. This is why the light of the gas contains only these wavelengths, that is, separate spectral lines, and not a continuous spectrum.

Altitude and colours. The first excited state, which produces the red light, decays by emitting a photon only after about $110\,\text{s}$ on average. At an altitude of $100$–$150\,\text{km}$ the air is so dense that an excited atom collides with other particles many times in a much shorter time. A collision releases the excited state: the energy is transferred to another particle, and no photon is produced. At altitudes above $200\,\text{km}$ the air is very thin and there are few collisions, so many atoms have time to emit a red photon.

The second excited state decays in only about $0.7\,\text{s}$, so many atoms emit a green photon before a collision even in denser air. Red light therefore requires thin air, and this is why it is seen in the upper part of the aurora.

Given

band gap of silicon $E = 1.12\,\text{eV}$, $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^{8}\,\text{m/s}$ and the wavelength of the claim $1500\,\text{nm}$

Find

the longest detectable wavelength $\lambda_\text{max}$ and an evaluation of the claim

Principle

One photon raises one electron to the conduction band if the energy of the photon is at least $1.12\,\text{eV}$. The energy of a photon decreases as the wavelength increases, so the longest detectable wavelength corresponds to exactly this energy. The intensity of the radiation tells the number of photons, not the energy of a single photon.

Formula

$E = \dfrac{hc}{\lambda}$, so $\lambda_\text{max} = \dfrac{hc}{E}$

Conversions

Substitution

For the radiation of the claim

Answer

The sensor detects wavelengths up to $\lambda_\text{max} \approx 1110\,\text{nm}$, so it detects the green and red light of the aurora. The claim is false: the energy $0.83\,\text{eV}$ of a $1500\,\text{nm}$ photon is smaller than $1.12\,\text{eV}$. More intense radiation brings more photons to the sensor, but none of them raises an electron to the conduction band, so the pixel detects nothing.

Reasonableness

The limit $1.1\,\mu\text{m}$ is in the near infrared, and silicon sensors are indeed also used to photograph the near infrared. The threshold is the same idea as in the photoelectric effect: what matters is the energy of a single photon, not the intensity of the light.

Given

atomic masses $m_\text{Pu} = 238.049560\,\text{u}$, $m_\text{U} = 234.040952\,\text{u}$ and $m_\text{He} = 4.002603\,\text{u}$ and $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$

Find

the reaction equation and the energy $Q$ of one decay

Principle

In alpha decay a helium-4 nucleus leaves the nucleus, so the nucleon number decreases by four and the atomic number by two. The combined mass of the reaction products is smaller than that of the original nucleus, and the difference of the masses corresponds to the energy released. Atomic masses include the electrons: a plutonium atom has $94$ electrons, a uranium atom $92$ and a helium atom $2$, so the masses of the electrons cancel.

Formula

${}^{238}_{94}\text{Pu} \rightarrow {}^{234}_{92}\text{U} + {}^{4}_{2}\text{He}$ and $Q = (m_\text{Pu} - m_\text{U} - m_\text{He})\,c^2$

Conversions

The mass defect is converted to energy by substituting $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$.

Substitution

so

Answer

${}^{238}_{94}\text{Pu} \rightarrow {}^{234}_{92}\text{U} + {}^{4}_{2}\text{He}$

And one decay releases $Q \approx 5.59\,\text{MeV}$. Three significant figures, because the mass defect $0.006005\,\text{u}$ has four.

Reasonableness

The energies of alpha decays are typically a few megaelectronvolts, a million times those of chemical reactions. The mass decreases by only about $0.0025\,\%$.

Given

thermal power $P = 2000\,\text{W}$, $Q = 5.594\,\text{MeV}$ of part a, $T_{1/2} = 87.7\,\text{a}$, $m_\text{Pu} = 238.049560\,\text{u}$ and $4.8\,\text{kg}$ of plutonium dioxide

Find

the mass $m$ of plutonium-238 and its fraction of the plutonium of the generator

Principle

Every decay produces heat $Q$, so the activity, that is, the decays per second, is the power divided by the energy of one decay. The activity is proportional to the number of undecayed nuclei. The mass is obtained from the number of nuclei using the mass of one atom.

Formula

$A = \dfrac{P}{Q}$, $A = \lambda N$, where $\lambda = \dfrac{\ln 2}{T_{1/2}}$, so $N = \dfrac{A}{\lambda}$ and $m = N m_\text{Pu}$

Conversions

Substitution

The plutonium in the plutonium dioxide is

so the fraction is $3.52\,\text{kg} : 4.23\,\text{kg} = 0.83$.

Answer

The generator must contain about $m \approx 3.5\,\text{kg}$ of plutonium-238, two significant figures like the mass of the fuel and the power. This is about $83\,\%$ of the plutonium of the generator.

Reasonableness

The fraction is below $100\,\%$, as it should be: in production other isotopes of plutonium come along with plutonium-238, and part of the plutonium-238 has decayed before the launch. The result also agrees with the known figure: plutonium-238 produces heat at about $2000\,\text{W} : 3520\,\text{g} \approx 0.57\,\tfrac{\text{W}}{\text{g}}$.

Find

why plutonium-238 is chosen as the fuel and what is important for radiation safety

Principle

Three things are required of a fuel: the power must not decrease much during the flight, the activity must be high in a small mass, and the radiation must be turned into heat inside the fuel without heavy shielding. The activity per nucleus is $\lambda = \tfrac{\ln 2}{T_{1/2}}$, so it is inversely proportional to the half-life.

Model answer

Polonium-210. The half-life $138\,\text{d}$ is short. After a year the fraction of the power left is $\left(\tfrac{1}{2}\right)^{365.25 / 138} \approx 0.16$, about one sixth, so it is not suitable for a flight of decades.

Uranium-238. The half-life $4.47 \cdot 10^{9}\,\text{a} : 87.7\,\text{a} \approx 5 \cdot 10^{7}$ times longer than that of plutonium-238, so the activity per nucleus is that many times smaller. Hundreds of millions of kilograms of uranium would be needed to produce the same power.

Cobalt-60. The half-life is rather short, and the strong gamma radiation penetrates matter: it would require tonnes of lead shielding and would damage the electronics and instruments of the rover. The alpha particles of plutonium-238, on the other hand, stop already in the fuel, so their energy is turned into heat where it is needed. Plutonium-238 is thus long-lived enough and still active enough.

Radiation safety. Alpha radiation does not penetrate the shell of the generator, so the external radiation is small. Plutonium is, however, very dangerous if it gets into the body, for example as inhaled dust: alpha particles give all their energy to the nearby cells. This is why the fuel is a durable ceramic, plutonium dioxide, in tight capsules that are designed to withstand an explosion and a fall in a launch accident.

Given

initial power $P_0 = 2000\,\text{W}$, electrical power $110\,\text{W}$, time $t = 14\,\text{a}$ and $T_{1/2} = 87.7\,\text{a}$

Find

the thermal power $P$ and the electrical power with the same efficiency, and the reason for the faster decrease

Principle

The thermal power is proportional to the activity, and the activity is proportional to the number of undecayed nuclei. So the thermal power decreases in the same way as the number of nuclei, according to the decay law. If the efficiency stays the same, the electrical power decreases in the same proportion.

Formula

$P = P_0 \left(\tfrac{1}{2}\right)^{t / T_{1/2}}$

Conversions

Not needed, because $t$ and $T_{1/2}$ are in the same unit.

Substitution

and the electrical power $110\,\text{W} \cdot 0.8953 = 98.5\,\text{W}$

Answer

The thermal power is about $P \approx 1790\,\text{W}$, and with the same efficiency the electrical power would be about $98\,\text{W}$. The real electrical power decreases faster, because the thermocouples age and the efficiency decreases. The decay law therefore gives only an upper limit for the electrical power.

Reasonableness

$14$ years is about a sixth of a half-life, so about $10\,\%$ of the power has been lost, clearly less than half. The generator is thus enough for a flight of decades.

Given

activity $A = 10\,\text{Bq}$, time $t = 1\,\text{a}$, $Q = 8.961 \cdot 10^{-13}\,\text{J}$ of part a, mass of the lungs $m = 1.2\,\text{kg}$, $w_\text{R} = 20$ and $w_\text{T} = 0.12$

Find

the effective dose $H_\text{E}$ in a year and a comparison with the average annual dose

Principle

Alpha particles stop within a few cells, so the energy of every decay remains in the lung tissue. The activity is decays per second, and the half-life $87.7\,\text{a}$ is so long that the activity stays approximately constant for a year. The absorbed dose is energy per mass. The radiation weighting factor of alpha radiation gives the equivalent dose of the lungs, and because only the lungs are irradiated, the effective dose is obtained with the tissue weighting factor of the lungs.

Formula

$E = AtQ$, $D = \dfrac{E}{m}$, $H_\text{T} = w_\text{R} D$ and $H_\text{E} = w_\text{T} H_\text{T}$

Conversions

Substitution

Answer

The effective dose would be about $H_\text{E} \approx 0.57\,\text{mSv}$ a year, two significant figures like the mass $1.2\,\text{kg}$. This is about a tenth of the average annual dose in Finland, $5.9\,\text{mSv}$, even though $10\,\text{Bq}$ is a very small activity. This is why plutonium is prevented from entering the body with a durable ceramic fuel and tight capsules.

Reasonableness

From the outside the same alpha emitter of $10\,\text{Bq}$ would be almost harmless, because alpha particles stop in the dead surface layer of the skin. In the lungs the factor $20$ increases the dose, and the factor $0.12$ reduces it to a risk for the whole body. The result is of the same order of magnitude as the dose limit of the population from the use of radiation, $1\,\text{mSv}$ a year.

Find

a graph whose points should lie on a straight line through the origin according to the assumption, and the conclusion from it

Graph

The count rate of the background is $R_\text{t} = \dfrac{112}{5.0\,\text{min}} = 22.4\,\tfrac{1}{\text{min}}$. It is subtracted from every reading, which gives the net count rate $R$ of the source. If the source is a point source and the radiation spreads evenly in all directions, at the distance $r$ the photons are spread over the surface of a sphere whose area is $4\pi r^2$. The fraction that hits the window of the tube is then proportional to $\tfrac{1}{r^2}$, so according to the assumption $R = \dfrac{K}{r^2}$. We choose $\tfrac{1}{r^2}$ for the horizontal axis and $R$ for the vertical axis.

Model answer

The points do not lie on a straight line through the origin. The line through the origin drawn through the most distant point passes above the nearest points, and the points form a curve bending downwards. The same can be seen from the product $Rr^2$ in the table: according to the assumption it would be constant, but apart from random variation it increases with distance from about $12\,000$ to $29\,000$. The count rate thus increases more slowly than $\tfrac{1}{r^2}$ as the source is approached. The assumption does not hold as such, at least near the source, so some part of it is wrong: for example, the distance is not the right distance from the source to the place where the radiation is detected.

Find

a graph whose points should lie on a straight line according to the model $R = \dfrac{K}{(r + d)^2}$, and the distance $d$ from it

Graph

From the model we get $\dfrac{1}{\sqrt{R}} = \dfrac{r + d}{\sqrt{K}} = \dfrac{1}{\sqrt{K}} \cdot r + \dfrac{d}{\sqrt{K}}$. So the quantity $\tfrac{1}{\sqrt{R}}$ is a first-degree function of $r$, and the line crosses the $r$ axis at $r = -d$. We choose $r$ for the horizontal axis and $\tfrac{1}{\sqrt{R}} = R^{-1/2}$ for the vertical axis; its unit is $\sqrt{\text{min}} = \text{min}^{1/2}$. For example, at $20\,\text{cm}$, $R = 71.6\,\tfrac{1}{\text{min}}$, so $\tfrac{1}{\sqrt{R}} = 11.82 \cdot 10^{-2}\,\sqrt{\text{min}}.$

Model answer

The points lie well on a straight line, so the model describes the measurement over the whole range. The line is extended as a dashed line to the $r$ axis, which it crosses at $r = -1.3\,\text{cm}$. The intercept can also be calculated from two distant points of the line, $(-1.3\,\text{cm}$; $0)$ and $(20.0\,\text{cm}$; $12.0 \cdot 10^{-2}\,\sqrt{\text{min}})$. Thus $d \approx 1.3\,\text{cm}$. The effective detection point of the tube is therefore about a centimetre behind the window, which fits the structure of a GM tube. Lines fitted by eye typically give $d = 1.0$–$1.6\,\text{cm}$. The slope of the line is $k = \dfrac{12.0 \cdot 10^{-2}\,\sqrt{\text{min}} - 0}{20.0\,\text{cm} - (-1.3\,\text{cm})} = 5.63 \cdot 10^{-3}\,\tfrac{\sqrt{\text{min}}}{\text{cm}}$, from which $K = \dfrac{1}{k^2} = 3.15 \cdot 10^{4}\,\tfrac{\text{cm}^2}{\text{min}}$.

Find

the uncertainty of the net count rates at $2.0\,\text{cm}$ and $20\,\text{cm}$, the more uncertain point and improvements

Principle

The random variation in the number of pulses is about $\sqrt{N}$, so the relative variation $\tfrac{\sqrt{N}}{N} = \tfrac{1}{\sqrt{N}}$ is the greater, the fewer pulses are counted. Subtracting the background brings its own small uncertainty. Near the source, reading the distance also matters, because the count rate depends on the square of the distance.

Model answer

$20\,\text{cm}$. $\Delta N \approx \sqrt{94} \approx 10$, and the variation of the $112$ pulses of the background, $\sqrt{112} \approx 11$ in five minutes, is $2.1\,\tfrac{1}{\text{min}}$. The net count rate is thus $R = 72 \pm 10\,\tfrac{1}{\text{min}}$, with an uncertainty of about $14\,\%$.

$2.0\,\text{cm}$. $\Delta N \approx \sqrt{3048} \approx 55$, so $R = 3026 \pm 55\,\tfrac{1}{\text{min}}$, only about $2\,\%$. At this distance the error in the distance is more important: if the distance is wrong by $1\,\text{mm}$, the distance $3.3\,\text{cm}$ to the detection point changes by about $3\,\%$ and the count rate by about $2 \cdot 3\,\% = 6\,\%$.

Conclusion. Because of random variation the most uncertain is the most distant point at $20\,\text{cm}$, while near the source it is the setting of the distance. Far from the source it is worth counting pulses for longer: a counting time four times as long halves the relative variation. The background should be measured for even longer. For the near points the source and the tube should be fixed on a stand where the distance is set accurately.

Given

$K = 3.15 \cdot 10^{4}\,\tfrac{\text{cm}^2}{\text{min}}$ and $d = 1.3\,\text{cm}$ of part b, background $R_\text{t} = 22.4\,\tfrac{1}{\text{min}}$, dose rate $3.0\,\mu\text{Sv/h}$ at $10\,\text{cm}$

Find

the distance $r$ at which the net count rate equals that of the background, and an evaluation of the reasoning

Principle

According to the model of part b the net count rate decreases like $\tfrac{1}{(r + d)^2}$. We set it equal to the count rate of the background. The dose rate of a point source decreases in the same way with the square of the distance, so it can be estimated for the same distance.

Formula

$\dfrac{K}{(r + d)^2} = R_\text{t}$, so $r = \sqrt{\dfrac{K}{R_\text{t}}} - d$, and the dose rate

Conversions

Not needed.

Substitution

Answer

The net count rate equals that of the background at a distance of about $r \approx 36\,\text{cm}$, and the dose rate there is about $0.2\,\mu\text{Sv/h}$. The reasoning is wrong. The radiation of the source does not stop; its dose rate decreases to the order of magnitude of background radiation, and with this counting time the tube cannot distinguish it from the variation of the background. Far from the source the radiation is still very weak, so distance is a good means of protection, as is limiting the time.

Reasonableness

The dose rate of the source, $0.2\,\mu\text{Sv/h}$, is of the same order of magnitude as the outdoor background $0.05$–$0.30\,\mu\text{Sv/h}$, so the pulse measurement and the dose rate tell the same thing. For comparison: working for $45\,\text{min}$ at $10\,\text{cm}$ gives about $3.0\,\mu\text{Sv/h} \cdot 0.75\,\text{h} \approx 2\,\mu\text{Sv}$, while a person in Finland receives on average $5.9\,\text{mSv} : 365 \approx 16\,\mu\text{Sv}$ a day.

Find

why a free neutron decays but a proton does not, what happens to the quarks, and which interaction causes it

Principle

A free particle can decay by itself only if the combined mass of the products is smaller than its own mass: the difference of the masses is released as the kinetic energy of the products. A neutron is $udd$ and a proton $uud$.

Model answer

Masses. In the decay of a neutron a proton, an electron and an electron antineutrino are produced: $n \rightarrow p + e^- + \bar{\nu}_e$. The charge is conserved, because $0 = (+e) + (-e) + 0$. The difference of the masses is $939.565\,\tfrac{\text{MeV}}{c^2} - 938.272\,\tfrac{\text{MeV}}{c^2} - 0.511\,\tfrac{\text{MeV}}{c^2} = 0.782\,\tfrac{\text{MeV}}{c^2}$, when the very small mass of the antineutrino is left out. The neutron is thus heavier than the products together, and about $0.782\,\text{MeV}$ of energy is released in the decay. The proton is lighter than the neutron, so a free proton cannot change into a neutron: that would require energy.

Quarks and interaction. One $d$ quark of the neutron ($udd$) changes into a $u$ quark, which produces a proton ($uud$). Only the weak interaction changes the flavour of a quark. Its mediator particle $W^-$ is produced when the $d$ quark changes, and it immediately turns into an electron and an antineutrino.

Given

mass fraction of helium-4 $Y = 0.245$

Find

the ratio $\tfrac{N_p}{N_n}$ of the numbers of protons and neutrons when the nuclei formed

Principle

Almost all the neutrons ended up in helium-4 nuclei, and each of them contains two neutrons and two protons. For every neutron that ended up in helium, one proton also ended up in helium, and the helium contains $2N_n$ nucleons. We assume the masses of the proton and the neutron to be equal and ignore the small effect of the binding energy of the nucleus on the mass. Then the mass fraction is the same as the fraction of the numbers of nucleons.

Formula

$Y = \dfrac{2N_n}{N_n + N_p}$, so $\dfrac{N_p}{N_n} = \dfrac{2 - Y}{Y}$

Conversions

$24.5\,\% = 0.245$

Substitution

Answer

When the nuclei formed, there were about $7.2$ protons for every neutron. The assumptions were that the masses of the nucleons are equal and that the number of neutrons left in deuterium, helium-3 and lithium is small.

Reasonableness

At the moment $1\,\text{s}$ there were $6$ protons per neutron, and as neutrons decay the ratio grows, so the value $7.2$ is in the right direction. If there had been exactly $7$ protons for every neutron, helium would have been $\tfrac{2}{8} = 25\,\%$ of the mass, which is close to the observed value.

Given

at the moment $t = 1\,\text{s}$ the ratio $1 : 6$, the ratio $1 : 7.16$ of part b and $T_{1/2} = 609\,\text{s}$

Find

the time from the Big Bang to the formation of nuclei

Principle

We consider a group in which at the moment $1\,\text{s}$ there is one neutron and six protons. Every neutron that decays turns into a proton, so there are always $7$ nucleons. When $N_n$ neutrons are left, there are $7 - N_n$ protons. First we solve when their ratio is the ratio of part b, and then the time from the decay law.

Formula

$\dfrac{7 - N_n}{N_n} = 7.16$, so $N_n = \dfrac{7}{1 + 7.16}$. From the decay law $N_n = N_0 \left(\tfrac{1}{2}\right)^{t / T_{1/2}}$, where $N_0 = 1$, we get

Conversions

Not needed.

Substitution

so

from the moment $1\,\text{s}$.

Answer

Nuclei began to form about $1\,\text{s} + 135\,\text{s} \approx 140\,\text{s}$, that is, about $2\,\text{min}$ after the Big Bang. Two significant figures are enough, because the initial ratio is known only approximately.

Reasonableness

Only about $14\,\%$ of the neutrons decayed, so less than one half-life passed. The result is of the order of minutes; in part d it is compared with the about three minutes of more accurate models.

Find

whether the result of part c changes only a little if the initial ratio is $1 : 5$, what the result requires, and how it agrees with the about three minutes of more accurate models

Principle

We calculate the same time with the ratio $1 : 5$: there are now $6$ nucleons for every original neutron. The sensitivity of the result tells how reliable the method is.

Model answer

Calculation. The number of neutrons left is $N_n = \dfrac{6}{1 + 7.16} = 0.735$, so $\dfrac{t}{T_{1/2}} = \dfrac{\lg \tfrac{1}{0.735}}{\lg 2} = 0.444$ and $t = 0.444 \cdot 609\,\text{s} = 270\,\text{s}$. The nuclei would form about $271\,\text{s}$, that is, about $4.5\,\text{min}$ after the Big Bang.

Evaluation. The claim is false: the time almost doubles, even though the initial ratio changed only a little. The reason is that the time depends on what fraction of the neutrons has time to decay. With the ratio $1 : 6$ only about $14\,\%$ of the neutrons decay, with the ratio $1 : 5$ about $26\,\%$. When the fraction that decays is so small, even a small change in the initial ratio changes it, and with it the time, by a large proportion. The method gives the right order of magnitude, a few minutes, but not the exact time.

Assumptions. The result of part c requires that after the moment $1\,\text{s}$ neutrons changed into protons only in the decay of free neutrons and not in other reactions, that the nuclei formed practically at a single moment, and that almost all the neutrons ended up in helium-4.

Comparison. According to more accurate models nuclei began to form after about three minutes. The about $2\,\text{min}$ of part c and the about $4.5\,\text{min}$ with the ratio $1 : 5$ are both of the same order of magnitude: minutes, not seconds or hours. The simple model thus gives the right order of magnitude.

Given

present proportions $0.720\,\%$ and $99.274\,\%$, time $t = 2.0 \cdot 10^{9}\,\text{a}$, half-lives $T_{235} = 7.04 \cdot 10^{8}\,\text{a}$ and $T_{238} = 4.47 \cdot 10^{9}\,\text{a}$

Find

the proportion of uranium-235 in the uranium $2.0$ billion years ago

Principle

Both isotopes decay all the time, each with its own half-life. We calculate how many nuclei of each there were $2.0$ billion years ago for every $100$ present-day uranium nuclei: in the past there were more nuclei by the factor $2^{t / T_{1/2}}$ than now. Uranium-235 decays faster, so there was relatively more of it earlier.

Formula

$N = N_0 \left(\tfrac{1}{2}\right)^{t / T_{1/2}}$, so $N_0 = N \cdot 2^{t / T_{1/2}}$, and the proportion $\dfrac{N_{0,235}}{N_{0,235} + N_{0,238}}$

Conversions

Not needed, because all the times are in years.

Substitution

The proportion is $\dfrac{5.16}{5.16 + 135.4} = 0.0367$.

Answer

$2.0$

Billion years ago about $3.7\,\%$ of the uranium was uranium-235. The proportion was the same as in the enriched fuel of present-day light-water reactors, so ordinary water was enough as a moderator and a chain reaction was possible. Now natural uranium contains only $0.72\,\%$ uranium-235, and a chain reaction with light water cannot be sustained.

Reasonableness

Almost three half-lives of uranium-235 have passed, so there was about seven times as much of it as now, and of uranium-238 only about a third more than now. An increase of the proportion to about five times is therefore reasonable. Two significant figures, like the age $2.0 \cdot 10^{9}\,\text{a}$.

Find

what the small proportion of uranium-235 tells us and how groundwater started and stopped the reactor

Principle

Uranium-235 undergoes fission easily when a slow neutron hits it, but the neutrons produced in fission are fast. A chain reaction continues when the multiplication factor is at least $1$, that is, every fission causes on average at least one new fission. Water slows neutrons down as they collide with the hydrogen nuclei of the water molecules.

Model answer

Small proportion. In a chain reaction it is exactly uranium-235 that undergoes fission, so it was used up from the ore. This is why its proportion decreased from the value $0.720\,\%$ found elsewhere to $0.440\,\%$. The small proportion is thus a trace of the fissions. This is also supported by the fact that the ore contains isotopes of fission products.

Starting. When groundwater seeped into the ore, it slowed down the fast neutrons produced in fission. The slow neutrons caused new fissions of uranium-235, and when the multiplication factor rose to one, the chain reaction started.

Stopping. The energy of the fissions turned into heat, and the ore heated up so much that the water boiled away. Without a moderator the neutrons stayed fast, the multiplication factor fell below one and the chain reaction stopped. When the ore cooled, water flowed back and the reaction started again. This explains the observed cycles, about half an hour of operation and a pause of a couple of hours. The reactor thus regulated itself, in the same way as a light-water reactor, in which boiling reduces the moderating ability.

Given

power of Oklo $P_\text{O} = 100\,\text{kW}$, time $t_\text{O} = 150\,000\,\text{a}$ and the thermal power of Olkiluoto 3 $P_\text{OL3} = 4300\,\text{MW}$

Find

the time $t_\text{OL3}$ in which Olkiluoto 3 produces the same energy

Principle

Thermal energy is power times time. When both produce the same energy, the times are inversely proportional to the powers.

Formula

$P_\text{O} t_\text{O} = P_\text{OL3} t_\text{OL3}$, so

Conversions

The time need not be converted to seconds, because the result is obtained directly in years.

Substitution

Answer

Olkiluoto 3 produces the same energy in about $3.5$ years. The power and the time are only order-of-magnitude estimates, so the result has at most two significant figures.

Reasonableness

The power of the natural reactor was about $43\,000$ times smaller than that of Olkiluoto 3, but it operated about $43\,000$ times longer than $3.5$ years. The energy is

With about $200\,\text{MeV}$ for one fission, this corresponds to about $1.5 \cdot 10^{28}$ fissions, that is, about $6\,\text{t}$ of uranium-235, which is of the same order of magnitude as the estimates in the literature.

Find

how the staying in place of plutonium can be concluded, and an evaluation of the claim that on the basis of Oklo the final disposal in Onkalo is certainly safe

Principle

A natural example can support a model if the conditions are similar, but it does not prove the safety of another place. We must also check what can be observed at Oklo at all: the half-life of plutonium-239 is $24\,100\,\text{a}$.

Model answer

What is seen at Oklo. In two billion years $2.0 \cdot 10^{9}\,\text{a} : 24\,100\,\text{a} \approx 83\,000$ half-lives of plutonium-239 have passed, so no plutonium is left. Its staying in place is concluded from its decay product: plutonium-239 decays by alpha decay into uranium-235, and this uranium has been found in the same places. Many fission products have also stayed near where they were formed, but some of the more mobile substances have been carried away.

Evaluation. Oklo is a good natural example of how bedrock and clay can retain many heavy elements for a very long time. The claim, however, assumes that the conditions in Onkalo are similar to those at Oklo. The bedrock, the chemistry of the groundwater and the form of the waste are different: at Oklo there was not spent fuel in canisters but ore. An observation at one place therefore does not transfer as such to another, and the word "certain" is too strong.

What safety is based on. The safety of Onkalo is based on several successive barriers: a copper canister, the bentonite clay around it and bedrock at a depth of about $430\,\text{m}$. Their operation is studied with experiments and models for hundreds of thousands of years, and STUK evaluates the calculations. Oklo supports these models but does not replace them.