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Frédéric Joliot-Curie

Jean Frédéric Joliot

A course in two strands: the life of the Paris engineer who joined Marie Curie’s laboratory, married her daughter and became the conscience and chief planner of French atomic science, and the physics and chemistry he built with Irène Curie: cloud-chamber tracks, positrons, artificial radioactivity, and the neutrons that make a chain reaction possible.

Born 19 March 1900, Paris Died 14 August 1958, Paris Nobel Prize Chemistry, 1935, with Irène Joliot-Curie Level advanced undergraduate → graduate
The first artificial radioelements, January 1934. On a logarithmic scale each decay is a straight line whose slope gives the half-life: nitrogen-13 from boron (10.0 min), phosphorus-30 from aluminium (2.50 min), aluminium-28 from magnesium (2.24 min). Modern half-lives.
Lesson 00

How to use this course

Frédéric Joliot trained as an industrial engineer and became one of the founders of nuclear physics and nuclear chemistry. With his wife Irène Curie he came close to discovering the neutron and the positron, then in January 1934 made the first radioactive atoms by artificial means. The discovery won them the 1935 Nobel Prize in Chemistry. In 1939 his team at the Collège de France showed that each uranium fission releases several neutrons, and filed patents for a nuclear reactor. After the war he directed the national research council, led the new Atomic Energy Commission and built France’s first reactor. In 1950 the government dismissed him for his communist politics and his campaign against nuclear weapons.

This course treats Joliot as he treated himself: as a physicist who thought like a chemist. Each technical lesson follows one of his experiments and gives the reader the tools to reproduce its numbers. The course pairs with the Irène Joliot-Curie course, which covers polonium, alpha-particle physics, counting statistics and the road to fission from her side, and with the Chadwick course, whose Lesson 08 analyses the neutron discovery that grew from the Joliot-Curie experiments.

Prerequisites for Part II
  • General chemistry: electrode potentials, gas laws, precipitation and solubility.
  • Special relativity: energy and momentum of particles and photons, \(E=mc^2\), Q-values of reactions.
  • The exponential decay law and the motion of a charged particle in a magnetic field.
  • For Lessons 09 and 10, the idea of a cross-section and of a neutron mean free path. The Chadwick course, Lesson 10, introduces the diffusion of neutrons.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Lesson 07 is the centre of the course. Masses are atomic masses in unified units u, with \(1\,\mathrm u\,c^2=931.5\) MeV. Joliot and Curie signed their scientific papers “Irène Curie and Frédéric Joliot”; the course uses the same names when it discusses the papers.

Reference

Timeline

1900Born 19 March in Paris, son of Henri Joliot, a merchant, and Émilie Roederer.
1908Enters the Lycée Lakanal at Sceaux.
1919Admitted to the École municipale de physique et de chimie industrielles (EPCI), Paris.
1923Graduates first in his class as an engineer, after a training period at the Arbed steelworks in Luxembourg.
1924Military service; leaves the artillery school at Poitiers as a second lieutenant.
1925On the recommendation of Paul Langevin, becomes personal assistant (préparateur) to Marie Curie at the Radium Institute.
1926Marries Irène Curie in October.
1927Daughter Hélène born. She becomes a nuclear physicist.
1930Doctorate: Étude électrochimique des radioéléments. Applications diverses.
1932January: with Irène Curie, finds that the beryllium radiation ejects fast protons from paraffin. Son Pierre born; he becomes a biologist. Matteucci Medal (with Irène).
1933Positrons from gamma rays in lead and from alpha-bombarded aluminium. October: the Solvay Conference in Brussels.
193415 January: note to the Paris Academy announcing artificial radioactivity. Chemical identification of radio-nitrogen and radio-phosphorus follows within weeks.
1935Nobel Prize in Chemistry with Irène Joliot-Curie. Maître de conférences at the Paris Faculty of Science.
1937Chair of nuclear chemistry created for him at the Collège de France. Laboratory of atomic synthesis at Ivry-sur-Seine.
1939January: physical proof of uranium fission. March–April: with Hans von Halban and Lew Kowarski, shows that fission releases neutrons, about 3.5 per fission by their estimate. Early May: patents on energy production by chain reaction.
1940Norway’s stock of heavy water reaches Paris in March. In June Halban and Kowarski carry it to England.
1941Helps found the Resistance movement Front national and becomes its president.
1942Joins the French Communist Party.
1943Elected to the Académie des sciences.
1944August: takes part in the liberation of Paris. Director of the Centre national de la recherche scientifique (CNRS).
1945Commissariat à l’énergie atomique (CEA) created in October; Joliot is its first High Commissioner.
1947Hughes Medal of the Royal Society.
194815 December: ZOE, the first French reactor, goes critical at the Fort de Châtillon.
1950Stockholm Appeal against nuclear weapons. April: dismissed as High Commissioner. President of the World Peace Council.
1951Receives the first Stalin Peace Prize.
1955Signs the Russell–Einstein Manifesto.
1956Irène dies in March. He takes over her Sorbonne chair of nuclear physics and the direction of the Curie laboratory, and builds the Orsay centre.
1958Dies in Paris on 14 August, aged 58. State funeral; buried beside Irène at Sceaux.
Part I · Formation
Lesson 01

Paris and the École de physique et chimie, 1900–1924

Jean Frédéric Joliot was born in Paris on 19 March 1900, son of Henri Joliot, a merchant, and Émilie Roederer.

In 1908 he entered the Lycée Lakanal at Sceaux, a boarding school south of Paris, and stayed nine years. He then prepared for engineering school at the École Lavoisier, a municipal school in Paris.

The EPCI

In 1919 he passed into the École municipale de physique et de chimie industrielles of the City of Paris, the school where Pierre Curie had taught and where Pierre and Marie Curie had discovered polonium and radium in a shed in 1898. The school trained engineers in both physics and chemistry, with long hours in the laboratory. One of its professors was Paul Langevin, Pierre Curie’s former student, and Joliot came to regard him as his scientific and political mentor. Joliot spent a training period at the Arbed steelworks in Luxembourg in 1922 and graduated first in his year in 1923.

He did his military service in 1924 and left the artillery school at Poitiers as a second lieutenant. An industrial career was open to him. He asked Langevin whether he could do research instead. Langevin advised him to go to Marie Curie.

Lesson 02

The Radium Institute, 1925–1930

In 1925 Joliot became Marie Curie’s personal assistant (préparateur) at the Radium Institute in the rue Pierre-Curie, on Langevin’s recommendation and with a grant from the Rothschild foundation. Curie noticed at once that he had no degree beyond his engineering diploma. At her insistence he took the baccalauréat subjects he lacked, then a licence in science, while working full time in the laboratory. He also learned radioactive techniques from her senior assistant, Irène Curie, three years his elder.

Irène was reserved where Frédéric was talkative and charming. They married in October 1926. Their daughter Hélène was born in 1927 and their son Pierre in 1932. They adopted the surname Joliot-Curie in private life but signed their papers as Irène Curie and Frédéric Joliot.

The thesis

Joliot’s doctoral thesis, defended in 1930, was on the electrochemistry of radioelements. Polonium and other radioelements were available only in quantities far too small to weigh. He studied how they deposit on electrodes from extremely dilute solutions, a topic that drew on his engineering training and that Lesson 04 analyses. The work gave him a command of radiochemical technique that proved decisive four years later.

At the same time he built instruments. He constructed a sensitive Wilson cloud chamber that could be operated inside a magnetic field. The chamber became the main tool of the couple’s work from 1929, when they began to publish together.

Lesson 03

Two discoveries missed, 1930–1933

The Radium Institute had one great asset: the world’s strongest polonium sources, prepared by Irène Curie from old radon tubes. Polonium emits alpha particles with almost no gamma radiation, which makes it ideal for nuclear experiments. In 1930 Hans Geiger’s former student Walther Bothe and Herbert Becker in Germany had found that beryllium bombarded with polonium alpha particles emits a very penetrating radiation. The Joliot-Curies repeated the experiment with sources ten times stronger.

The neutron

In January 1932 they reported to the Academy of Sciences that the beryllium radiation ejects protons from paraffin and other hydrogen-rich substances with energies of several MeV. They interpreted the effect as a Compton scattering of very energetic gamma rays on protons. When Rutherford heard of it he said he did not believe it. James Chadwick, at the Cavendish, showed within a month that the radiation consists of neutral particles of about the proton mass. The analysis that settled the matter is in the Chadwick course, Lesson 08.

The positron

In their cloud chamber in a magnetic field they had seen tracks of electrons that curved the wrong way. They took them for electrons travelling towards the source rather than away from it. In August 1932 Carl David Anderson at Caltech identified the positive electron in cosmic-ray tracks, and in early 1933 Patrick Blackett and Giuseppe Occhialini at Cambridge confirmed it. The Joliot-Curies quickly showed that hard gamma rays produce positrons in lead, and that light elements bombarded by alpha particles emit positrons along with neutrons (Lesson 06).

The Solvay Conference

In October 1933 they presented these results to the Seventh Solvay Conference in Brussels. They proposed that aluminium under alpha bombardment emits a neutron and a positron together. Lise Meitner reported that her own experiments had found no neutrons from aluminium. Most of the audience was sceptical, though Niels Bohr showed interest. Francis Perrin pointed out that the positron energies were continuous, which is hard to explain if the neutron and positron leave together. The Joliot-Curies went home to check. Three months later the check produced artificial radioactivity.

Part II · The physics and chemistry
Lesson 04

Electrochemistry of invisible quantities

For the general reader. A strong polonium source contains less than a millionth of a gram of polonium. It cannot be weighed or seen. Joliot’s thesis asked how such tiny amounts behave in solution and on electrodes. The answer depends on the ordinary laws of electrochemistry, applied at concentrations a million times lower than in any chemistry textbook.

Goals
  • Relate the activity of a radioelement to the number of atoms and the mass present.
  • Use the Nernst equation to see how deposition potentials shift at trace concentrations.
  • Understand why trace deposits do not form a bulk metal phase.

Counting atoms by their decay

A radioactive sample of \(N\) atoms with decay constant \(\lambda\) has activity \(A=\lambda N\). The half-life is \(t_{1/2}=\ln2/\lambda\). One curie (Ci) is \(3.7\times10^{10}\) decays per second, roughly the activity of one gram of radium-226. Polonium-210 has a half-life of 138.4 days, so

\[ N=\frac{A\,t_{1/2}}{\ln 2},\qquad m=\frac{N M}{N_A}. \]

A source of one millicurie contains about \(6.4\times10^{14}\) atoms, or 0.22 micrograms. Radioactive methods can detect activities a million times smaller. This was the field in which Joliot worked: chemistry with atoms counted by their decays.

The Nernst equation at high dilution

Nernst equation

For an electrode reaction \(\mathrm{M}^{n+}+n\,e^-\rightleftharpoons\mathrm M\), the equilibrium electrode potential is

\[ E=E^\circ+\frac{RT}{nF}\ln\frac{a_{\mathrm M^{n+}}}{a_{\mathrm M}}, \]

where \(a\) denotes activity (in the thermodynamic sense), \(F=96\,485\) C/mol is Faraday’s constant and \(R=8.314\) J/(mol K). At 25 °C, \(RT\ln 10/F=59.2\) mV.

Derivation. At equilibrium the electrical work \(-nFE\) equals the Gibbs free-energy change \(\Delta G=\Delta G^\circ+RT\ln Q\) of the reduction, with \(Q=a_{\mathrm M}/a_{\mathrm M^{n+}}\). Dividing by \(-nF\) gives the formula with \(E^\circ=-\Delta G^\circ/nF\).

Two consequences matter for trace amounts. First, each factor of ten in dilution lowers the equilibrium potential by \(59.2/n\) mV, so a radioelement at \(10^{-10}\) mol/L needs a potential several hundred millivolts more negative than at 1 mol/L to start depositing. Second, a deposit of a few billion atoms covers only a tiny fraction of an electrode. It does not form a bulk metal, so \(a_{\mathrm M}\) is not 1. The deposit sits on a foreign metal, and the deposition depends on the electrode material. Earlier workers had defined a “critical potential” at which a radioelement begins to deposit. Joliot studied how the deposition of polonium varies with potential, concentration and electrode, and used the results to prepare clean, thin sources and to identify chemical states of radioelements present in unweighable amounts.

Exercise 4.1Compute the number of atoms and the mass of polonium-210 (molar mass 210 g/mol) in a 1 mCi source and in the 100 mCi sources the Joliot-Curies used around 1932.

Show solution
\(\lambda=\ln2/(138.4\times86\,400\ \mathrm s)=5.80\times10^{-8}\ \mathrm s^{-1}\). For 1 mCi, \(N=3.7\times10^{7}/5.80\times10^{-8}=6.4\times10^{14}\) atoms and \(m=6.4\times10^{14}\times210/6.02\times10^{23}=2.2\times10^{-7}\) g. A 100 mCi source holds 22 micrograms, a speck invisible to the eye.

Exercise 4.2A 1 mCi polonium sample is dissolved in 10 mL. Find its molar concentration. Assuming ideal behaviour and \(n=2\), by how much does the Nernst potential differ from that of a 1 mol/L solution?

Show solution
\(c=6.4\times10^{14}/(6.02\times10^{23}\times0.010)\approx1.1\times10^{-7}\) mol/L, about 7 decades below 1 mol/L. The potential is lower by \(7\times59.2/2\approx207\) mV. Deposition therefore starts at a noticeably more negative potential than bulk tables suggest, one reason why tracer electrochemistry needed its own experiments.

Exercise 4.3A metal surface has about \(10^{15}\) atom sites per cm². If all the polonium in a 1 mCi sample were deposited on a 1 cm² disc, what fraction of a monolayer would it form? What does this imply for the activity \(a_{\mathrm M}\) of the deposit?

Show solution
\(6.4\times10^{14}/10^{15}\approx0.6\) of a monolayer. Weaker sources form a small fraction of one atomic layer. The deposited atoms are isolated on a foreign substrate, so they do not have the chemical potential of bulk polonium, and \(a_{\mathrm M}\) depends on coverage and substrate. The Nernst formula with \(a_{\mathrm M}=1\) fails, which is why the deposition potential had to be measured.
Lesson 05

Cloud chambers and penetrating rays

For the general reader. A charged particle passing through supersaturated vapour leaves a trail of droplets, as an aircraft leaves a vapour trail. C. T. R. Wilson invented the chamber that does this. In a magnetic field the trails curve, and the curvature tells the particle’s momentum and the sign of its charge. Joliot built one of the best such chambers in Europe. It showed the Joliot-Curies the protons knocked out by neutrons and, unrecognised, the first positrons.

Goals
  • Compute the cooling produced by an adiabatic expansion and why it creates supersaturation.
  • Use the magnetic rigidity \(B\rho=p/ze\) to read momenta from tracks.
  • See why the sign of a track’s curvature is ambiguous without knowing its direction.

The expansion chamber

A Wilson chamber holds a gas saturated with water or alcohol vapour. A piston suddenly enlarges the volume. For a quick, reversible adiabatic expansion of an ideal gas with \(\gamma=c_p/c_v\),

\[ TV^{\gamma-1}=\text{constant},\qquad T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}. \]

The gas cools by 20 to 30 K, and the vapour becomes supersaturated. Droplets cannot form spontaneously, because a tiny droplet has a higher vapour pressure than a flat surface (the Kelvin effect). Ions are exceptions: the charge lowers the energy of a small droplet, so droplets condense on the ions left along a charged particle’s path. The track is lit and photographed within a fraction of a second.

Magnetic rigidity

Momentum from curvature

A particle of charge \(ze\) and momentum \(p\) moving perpendicular to a uniform field \(B\) follows a circle of radius \(\rho\) with

\[ B\rho=\frac{p}{ze},\qquad p\,[\mathrm{MeV}/c]=299.8\;z\;B\,[\mathrm T]\;\rho\,[\mathrm m]. \]

The sense of rotation gives the sign of \(ze\) only if the direction of motion is known.

The relation follows from equating the Lorentz force \(zevB\) to \(p v/\rho\), which holds relativistically because the force is perpendicular to the velocity and changes only the direction of \(\mathbf p\). The numerical form uses \(1\ \mathrm{MeV}/c=5.344\times10^{-22}\) kg m/s.

What the Joliot-Curies saw, 1932

With polonium–beryllium sources they photographed tracks of protons and of helium and carbon nuclei projected by the penetrating radiation, and many fast electrons. In a field of about 1,300 gauss (0.13 T) some electron tracks had energies they estimated above 10 MeV. Others curved in the opposite sense from that expected for electrons leaving the source. They interpreted them as electrons moving back towards the source. After Anderson’s paper they saw that many were positrons. Anderson had settled the direction by placing a lead plate across his chamber: a particle loses momentum crossing the plate, so its track curves more tightly on the far side.

Exercise 5.1Air (\(\gamma=1.40\)) at 293 K is expanded suddenly by a factor \(V_2/V_1=1.25\). Find the final temperature. Saturated water vapour pressure is 2.34 kPa at 20 °C and about 0.42 kPa at −5 °C. Estimate the supersaturation ratio, ignoring the drop in vapour pressure from the expansion itself.

Show solution
\(T_2=293\times1.25^{-0.40}=268\) K, about −5 °C. The partial pressure of vapour falls with the gas by the factor \((V_1/V_2)^{\gamma}\approx0.73\) for an adiabatic process, to about 1.7 kPa. With this correction the ratio is \(1.7/0.42\approx4\); without it, \(2.34/0.42\approx5.6\). Either way the vapour is several times supersaturated, enough for ion-induced but not spontaneous condensation in water.

Exercise 5.2Find the radius of curvature of a 10 MeV electron and of a 5 MeV proton in 0.13 T.

Show solution
Electron: \(pc=\sqrt{T^2+2Tmc^2}=\sqrt{100+10.2}=10.5\) MeV, so \(\rho=10.5/(299.8\times0.13)=0.27\) m. Proton: \(pc=\sqrt{25+2\times5\times938}=97\) MeV, \(\rho=97/(299.8\times0.13)\approx2.5\) m. Proton tracks are almost straight in such a field, while the electron curvature is easy to measure in a chamber 20–30 cm wide.

Exercise 5.3A track in a field pointing into the page curves anticlockwise as seen on the photograph. Show that it could be a negative particle moving one way or a positive particle moving the other. How does a 6 mm lead plate across the chamber remove the ambiguity?

Show solution
The force \(q\mathbf v\times\mathbf B\) changes sign under either \(q\to-q\) or \(\mathbf v\to-\mathbf v\), so the circle is the same for both cases. A particle crossing lead loses energy, so \(\rho\) is smaller after the plate. The side with the tighter curvature is downstream, which fixes the direction and hence the sign of the charge.
Lesson 06

Materialisation and annihilation

For the general reader. In 1933 the Joliot-Curies showed that hard gamma rays passing near heavy nuclei turn into pairs of electrons and positrons. They called it “materialisation”: energy becoming matter, as Dirac’s theory predicted. The reverse also happens. A positron that slows down in matter meets an electron, and both vanish into gamma rays. Joliot was one of the first to detect this annihilation radiation.

Goals
  • Show that a photon cannot create a pair in empty space.
  • Derive the threshold for pair creation near a nucleus.
  • Find the energies and directions of annihilation photons.
Pair creation threshold

A photon cannot produce an electron–positron pair in free space. In the field of a nucleus of mass \(M\), initially at rest, the threshold photon energy is

\[ E_{\mathrm{th}}=2m_ec^2\left(1+\frac{m_e}{M}\right)\approx 1.022\ \text{MeV}. \]

Proof. The invariant \(s=(\sum E)^2-(\sum\mathbf pc)^2\) is the same before and after. For a lone photon, \(s=E^2-E^2=0\), but a pair has \(s\ge(2m_ec^2)^2>0\). So a third body is needed. With a nucleus at rest, \(s=(E+Mc^2)^2-E^2=M^2c^4+2EMc^2\). The final state needs \(s\ge(2m_e+M)^2c^4\), which gives \(2EMc^2\ge4m_eMc^4+4m_e^2c^4\), the result. The nucleus takes negligible energy, so the pair shares \(E-2m_ec^2\).

The experiments of 1933

Thorium C″ (thallium-208) emits a gamma ray of 2.62 MeV, the hardest then available. The Joliot-Curies placed lead in their cloud chamber and found positrons produced in it by these gamma rays, The cross-section for pair creation near a nucleus of charge \(Ze\) grows as \(Z^2\), which is why lead is so effective. In the same year Joliot, and independently Jean Thibaud, observed the gamma radiation emitted when positrons stop in matter.

Annihilation at rest

An electron and a positron at rest annihilate into at least two photons. For two photons, each has energy \(m_ec^2=511\) keV and they travel in opposite directions.

One photon is forbidden at rest by the same invariant argument: \(s=(2m_ec^2)^2\ne0\). For two photons, zero total momentum forces equal and opposite momenta and so equal energies. The back-to-back 511 keV pair is the basis of positron emission tomography, which uses artificial positron emitters of exactly the kind discovered in 1934, such as nitrogen-13 and fluorine-18.

Exercise 6.1A 2.62 MeV photon converts near a lead nucleus. What total kinetic energy is shared by the pair? If the positron takes 0.9 MeV, what is the radius of its track in 0.13 T?

Show solution
\(2.62-1.022=1.60\) MeV. For the positron, \(pc=\sqrt{0.81+2\times0.9\times0.511}=\sqrt{1.73}=1.32\) MeV, so \(\rho=1.32/(299.8\times0.13)=3.4\) cm.

Exercise 6.2Pair creation can also occur in the field of an atomic electron. Find the threshold in this case, and explain why it is higher.

Show solution
Put \(M=m_e\) in the threshold formula: \(E_{\mathrm{th}}=2m_ec^2(1+1)=4m_ec^2=2.04\) MeV. The final state now has three electrons’ worth of rest mass, and the light target recoils strongly, taking a large share of the energy. The process is called triplet production.

Exercise 6.3A positron annihilates in flight with kinetic energy \(T\) on an electron at rest, producing two photons. Show that the photon emitted straight forward has energy \(E_+=\tfrac12\big(T+2m_ec^2+pc\big)\), where \(p\) is the positron momentum.

Show solution
Energy: \(E_++E_-=T+2m_ec^2\). Momentum along the line: \(E_+-E_-=pc\) if the second photon goes straight back. Adding gives the result. As \(T\to0\), \(E_+\to m_ec^2\), recovering the rest case.
Lesson 07

Artificial radioactivity, 1934

For the general reader. In January 1934 the Joliot-Curies bombarded a thin aluminium foil with alpha particles, then took the source away. The foil kept emitting positrons, and the emission fell by half every three minutes or so, exactly like a natural radioelement. The alpha particles had turned some aluminium atoms into a radioactive form of phosphorus that does not exist in nature. For the first time, humans had made a radioactive substance.

Goals
  • Write the reactions of 1934 and compute their Q-values and thresholds.
  • Derive the growth and decay of an activity produced at constant rate.
  • See why a half-life independent of the alpha energy points to a product nucleus.

The experiment

The check suggested by the Solvay discussion was simple. If the neutron and positron leave together, both should stop the moment the alpha source is removed. Joliot placed a thin aluminium foil about a millimetre from a polonium source of about 60 mCi for ten minutes, removed the source, and put the foil in front of a Geiger–Müller counter. The counter kept clicking, and the rate fell exponentially. Boron and magnesium behaved the same way. The note, Un nouveau type de radioactivité, was presented to the Academy on 15 January 1934. A letter to Nature followed in February. The periods they reported were about 14 min for boron, 3 min 15 s for aluminium and 2 min 30 s for magnesium. The modern half-lives are 10.0, 2.50 and 2.24 min.

The reactions (Curie and Joliot, 1934) \[ \begin{aligned} {}^{27}_{13}\mathrm{Al}+{}^{4}_{2}\mathrm{He}&\longrightarrow{}^{30}_{15}\mathrm P+n, &\qquad {}^{30}_{15}\mathrm P&\longrightarrow{}^{30}_{14}\mathrm{Si}+e^++\nu,\\ {}^{10}_{5}\mathrm{B}+{}^{4}_{2}\mathrm{He}&\longrightarrow{}^{13}_{7}\mathrm N+n, &\qquad {}^{13}_{7}\mathrm N&\longrightarrow{}^{13}_{6}\mathrm{C}+e^++\nu. \end{aligned} \]

For magnesium they first proposed silicon-27. They soon showed that the main activity is aluminium-28, formed by \({}^{25}\mathrm{Mg}(\alpha,p){}^{28}\mathrm{Al}\), an electron emitter. The neutrino in these lines is the modern reading; in 1934 Fermi’s theory of beta decay was just appearing.

Q-values and thresholds

For a reaction \(a+X\to Y+b\), the Q-value is \(Q=(m_a+m_X-m_Y-m_b)c^2\). If \(Q<0\), the projectile needs at least the threshold kinetic energy

\[ T_{\mathrm{th}}=-Q\left(1+\frac{m_a}{m_X}\right) \]

in the laboratory, because part of its energy must go into the motion of the centre of mass. In addition, a charged projectile must approach the target despite the Coulomb barrier \(B\approx Z_aZ_Xe^2/(4\pi\varepsilon_0 R)\), with \(R\approx1.2\,(A_a^{1/3}+A_X^{1/3})\) fm and \(e^2/4\pi\varepsilon_0=1.44\) MeV fm. Below the barrier, the reaction proceeds only by tunnelling and its yield falls steeply.

Why a new nucleus

The Joliot-Curies varied the alpha energy with absorbing screens. The yield changed, but the half-life did not. A decay rate that does not depend on how the activity was made is a property of a definite nucleus, as for natural radioelements. They concluded that the alpha particle is captured, a neutron leaves at once, and the product nucleus emits the positron later. The continuous positron spectrum Perrin had noted at Solvay was the spectrum of a beta decay.

Growth and decay

Activation at constant rate

If nuclei of decay constant \(\lambda\) are produced at a constant rate \(R\) for a time \(t_{\mathrm{irr}}\), then after the irradiation stops the activity is

\[ A(t)=R\left(1-e^{-\lambda t_{\mathrm{irr}}}\right)e^{-\lambda t},\qquad t\ge0. \]

The activity approaches the saturation value \(R\) for \(t_{\mathrm{irr}}\gg t_{1/2}\).

Derivation. During irradiation \(dN/dt=R-\lambda N\), so \(N=(R/\lambda)(1-e^{-\lambda t})\). After it, \(N\) decays freely. Multiply by \(\lambda\) to get the activity.

Exercise 7.1Using \(m({}^{27}\mathrm{Al})=26.981539\), \(m({}^{4}\mathrm{He})=4.002603\), \(m({}^{30}\mathrm{P})=29.978314\), \(m(n)=1.008665\) u, find \(Q\) and the threshold for \({}^{27}\mathrm{Al}(\alpha,n){}^{30}\mathrm P\). Compare with the Coulomb barrier and with polonium alpha particles of 5.3 MeV.

Show solution
\(Q=(30.984142-30.986979)\times931.5=-2.64\) MeV. Threshold \(2.64\times(1+4.003/26.98)=3.04\) MeV. Barrier: \(R=1.2(3.00+1.59)=5.5\) fm, \(B=2\times13\times1.44/5.5\approx6.8\) MeV. Polonium alphas at 5.3 MeV are above threshold but below the barrier, so the reaction proceeds by tunnelling with a small yield. Strong sources were essential.

Exercise 7.2With \(m({}^{10}\mathrm B)=10.012937\) and \(m({}^{13}\mathrm N)=13.005739\) u, find \(Q\) for \({}^{10}\mathrm B(\alpha,n){}^{13}\mathrm N\). Then, using \(m({}^{30}\mathrm{Si})=29.973770\) and \(m({}^{13}\mathrm C)=13.003355\) u, find the maximum positron kinetic energies of phosphorus-30 and nitrogen-13. Why must \(2m_ec^2\) be subtracted when atomic masses are used?

Show solution
\(Q=(14.015540-14.014404)\times931.5=+1.06\) MeV, exothermic. For \(\beta^+\) decay with atomic masses, \(Q_{\beta^+}=[m(\text{parent})-m(\text{daughter})]c^2-2m_ec^2\): the daughter atom has one electron fewer than the parent, and a positron is created. Phosphorus-30: \(0.004544\times931.5-1.022=3.21\) MeV. Nitrogen-13: \(0.002384\times931.5-1.022=1.20\) MeV. The positrons have continuous spectra up to these endpoints, shared with the neutrino.

Exercise 7.3Aluminium is irradiated for 10 min and counting starts 1 min later. Phosphorus-30 has \(t_{1/2}=2.50\) min. What fraction of the saturation activity is present when counting starts? How long an irradiation gives 99% of saturation?

Show solution
\(\lambda=0.277\ \mathrm{min}^{-1}\). After 10 min, \(1-e^{-2.77}=0.937\). One minute later, \(0.937\times e^{-0.277}=0.71\). For 99%, \(e^{-\lambda t}=0.01\), \(t=\ln100/\lambda=16.6\) min, about 6.6 half-lives.
Lesson 08

The chemical proof of transmutation

For the general reader. A decay curve shows that something radioactive has been made, but not what it is. The new atoms numbered perhaps a few million, far too few to see. Joliot proved their identity with chemistry: if the activity follows phosphorus through a chemical reaction, it is phosphorus. The separations had to be finished within a few half-lives, that is, within minutes.

Goals
  • Understand the carrier method for separating unweighable amounts.
  • Estimate how many radioactive atoms were present.
  • Plan a separation against the clock.

The carrier principle

Trace atoms of an element behave chemically like the bulk element, provided they are in the same chemical form. To separate a few thousand atoms of radio-phosphorus, one adds a weighable amount of ordinary phosphorus, the carrier, and separates the phosphorus by ordinary chemistry. If the activity goes with the phosphorus, and its half-life is unchanged, the radioactive atoms are phosphorus atoms. This principle goes back to Marie and Pierre Curie, who followed radium by its activity through crystallisations of barium salts.

The separations of 1934
  • Nitrogen-13. Irradiated boron nitride was heated with caustic soda. The nitrogen left as ammonia gas, and the activity went with the gas. With boron oxide, by contrast, the activity stayed behind.
  • Phosphorus-30. Irradiated aluminium was dissolved in hydrochloric acid. The hydrogen given off carried the activity, as phosphine \(\mathrm{PH_3}\) would. In a second method, the phosphorus was precipitated as zirconium phosphate with added carrier, and the activity followed the precipitate.

The half-lives measured on the separated fractions matched those of the irradiated targets. The results were reported to the Academy in February 1934.

Joliot’s Nobel lecture took its title from this work: Chemical evidence of the transmutation of elements. Since Rutherford’s experiment of 1919, transmutation had been inferred from the emitted particles alone. Now the product nucleus could be isolated and named. The method also opened the way to radioactive tracers, which George de Hevesy had pioneered with natural radioelements; artificial isotopes of common elements such as phosphorus, carbon and iodine soon made them a standard tool of biology and medicine.

Exercise 8.1Just after irradiation, an aluminium foil gave about 150 counts per minute. Suppose the counter recorded 10% of the positrons. How many phosphorus-30 atoms were present? What mass is that?

Show solution
True activity \(A=1500\) per minute \(=25\ \mathrm s^{-1}\). With \(\lambda=\ln2/150\ \mathrm s=4.6\times10^{-3}\ \mathrm s^{-1}\), \(N=A/\lambda\approx5.4\times10^{3}\) atoms, about \(2.7\times10^{-19}\) g. Joliot called such quantities invisible; only the decay could reveal them.

Exercise 8.2A phosphorus separation takes 3 minutes from the end of irradiation to the start of counting. What fraction of the phosphorus-30 survives? For nitrogen-13 (\(t_{1/2}=9.97\) min), how long can the chemistry take for the same loss?

Show solution
\(2^{-3/2.50}=0.44\). The same fraction survives for nitrogen-13 after \(3\times9.97/2.50\approx12\) min. Short half-lives forced Joliot to design separations that a chemist would normally take an hour over into a few minutes.

Exercise 8.3Why does adding carrier not dilute the evidence? Suppose 1 mg of phosphorus carrier is added and 80% of it is recovered in the precipitate. What fraction of the radio-phosphorus is expected in the precipitate, and what would it mean if only 5% of the activity appeared there?

Show solution
If tracer and carrier are in the same chemical form they mix completely, so about 80% of the radio-phosphorus should follow, and the count rate of the precipitate measures it directly. The carrier only provides enough material to handle. If only 5% of the activity followed, the active atoms would not be phosphorus (or not in exchange with the carrier), and the identification would fail.
Lesson 09

Fission and the neutron count, 1939

For the general reader. In January 1939 news arrived from Berlin that neutrons split uranium nuclei into two lighter nuclei. Joliot confirmed it within days by catching the fragments. The decisive question was whether each split also releases new neutrons. If it releases more than one, each fission can cause more than one further fission, and a chain reaction can grow. Joliot, Hans von Halban and Lew Kowarski showed in the spring of 1939 that it does.

Goals
  • Estimate the energy released in fission from masses and from Coulomb repulsion.
  • Understand the neutron-balance method used to show that fission releases neutrons.
  • Connect the number of neutrons per fission to a multiplication factor and an exponential growth rate.

The fragments

Otto Hahn and Fritz Strassmann found barium in neutron-irradiated uranium in December 1938, and Lise Meitner and Otto Frisch explained it as a splitting of the nucleus. Joliot coated a cylinder with uranium oxide and placed it, under neutron irradiation, a few millimetres from a collecting surface. Radioactivity appeared on the collector: fragments had recoiled across the gap with energies of many MeV. His note was published in the Comptes rendus on 30 January 1939. He also photographed fission fragment tracks in a cloud chamber.

Energy of fission

For a typical fission \(n+{}^{235}\mathrm U\to{}^{141}\mathrm{Ba}+{}^{92}\mathrm{Kr}+3n\), the Q-value from atomic masses is about 173 MeV, appearing mainly as kinetic energy of the fragments. The later beta decays of the fragments add roughly 20 MeV more. Most of the energy comes from the Coulomb repulsion of the two fragments, which at contact is

\[ E_C\approx\frac{Z_1Z_2e^2}{4\pi\varepsilon_0(R_1+R_2)}. \]

Counting the neutrons

The Paris group placed a neutron source at the centre of a large tank of water and measured the density of slow neutrons at various distances with a detector activated by them. They repeated the measurement with a solution of uranyl nitrate in the tank and compared it with a control solution. In a large tank nearly every neutron is eventually captured in the water, and the capture rate is proportional to the neutron density. So

\[ \text{neutrons absorbed per second}\;\propto\;\int_0^\infty n(r)\,4\pi r^2\,dr . \]

With uranium present the integral grew, so uranium released more neutrons than it absorbed. The letter Liberation of neutrons in the nuclear explosion of uranium appeared in Nature in March 1939. A second letter, in April, estimated about 3.5 neutrons per fission. The modern value for slow-neutron fission of uranium-235 is \(\bar\nu\approx2.4\). At Columbia, Enrico Fermi’s group and Leo Szilard with Walter Zinn reached similar conclusions at about the same time.

Multiplication and growth

Let \(k\) be the average number of neutrons in one generation produced by the neutrons of the previous generation, and \(\ell\) the mean generation time. Then the neutron population grows as

\[ n(t)=n_0\,k^{t/\ell}\approx n_0\,e^{(k-1)t/\ell}\qquad (k\ \text{close to }1). \]

A self-sustaining chain reaction needs \(k\ge1\).

In early May 1939 Joliot, Halban and Kowarski filed three secret patents in the name of the national research fund, the forerunner of the CNRS. They covered a device for producing energy by a chain reaction in uranium with a moderator, and the use of the same principle in explosives. Francis Perrin, who joined the team, published a first estimate of the critical mass of uranium oxide, tens of tonnes without reflector.

Exercise 9.1Using \(m({}^{235}\mathrm U)=235.043930\), \(m({}^{141}\mathrm{Ba})=140.914411\), \(m({}^{92}\mathrm{Kr})=91.926156\) and \(m(n)=1.008665\) u, compute \(Q\) for the fission above. Compare with the Coulomb energy at contact, with \(R=1.2A^{1/3}\) fm.

Show solution
Left side: \(235.043930+1.008665=236.052595\). Right side: \(140.914411+91.926156+3\times1.008665=235.866562\). \(Q=0.186033\times931.5=173\) MeV. Coulomb: \(R_1+R_2=1.2(5.20+4.51)=11.6\) fm, \(E_C=56\times36\times1.44/11.6\approx250\) MeV. The fragments are deformed and somewhat separated at scission, so the real kinetic energy, about 170 MeV, is lower than this crude contact estimate.

Exercise 9.2One gram of uranium-235 fissions completely. Using 200 MeV per fission, find the energy released and compare with burning one tonne of coal (about 30 GJ).

Show solution
\(N=6.02\times10^{23}/235=2.56\times10^{21}\) nuclei. \(E=2.56\times10^{21}\times200\times1.602\times10^{-13}\ \mathrm J=8.2\times10^{10}\) J, about 82 GJ, nearly three tonnes of coal.

Exercise 9.3In a fast-neutron assembly \(\ell\approx10^{-8}\) s. How long does it take a population to grow by a factor \(10^{20}\) if \(k=2\)? If instead \(k=1.001\) and \(\ell=10^{-3}\) s, as in a thermal reactor including delayed neutrons, what is the e-folding time?

Show solution
\(2^{t/\ell}=10^{20}\) gives \(t/\ell=66\), \(t\approx0.7\) microseconds. In the second case the e-folding time is \(\ell/(k-1)=1\) s, slow enough for control rods to act. The difference between these two regimes is the difference between a bomb and a reactor.
Lesson 10

Heavy water, multiplication and ZOE

For the general reader. Natural uranium is mostly uranium-238, which captures neutrons without splitting. A chain reaction in natural uranium needs a moderator that slows neutrons without swallowing them. Ordinary water captures too many. Heavy water, made with the heavy isotope of hydrogen, captures almost none. That is why Joliot’s team brought Norway’s entire stock of heavy water to Paris in 1940, and why the first French reactor, ZOE, used heavy water when it started in 1948.

Goals
  • Compute the reproduction factor \(\eta\) of natural uranium.
  • Compare moderators by their slowing power and their absorption.
  • State the four-factor formula and use it to see why heavy water allows a small natural-uranium reactor.
Four-factor formula

For an infinite lattice of natural uranium in a moderator, the multiplication factor is

\[ k_\infty=\eta\,\varepsilon\,p\,f, \]

where \(\eta\) is the number of fission neutrons per thermal neutron absorbed in the fuel, \(\varepsilon\) the fast-fission factor, \(p\) the probability of escaping resonance capture while slowing down, and \(f\) the fraction of thermal neutrons absorbed in the fuel. A finite reactor has \(k_{\mathrm{eff}}=k_\infty P_{\mathrm{NL}}\), with \(P_{\mathrm{NL}}\) the probability that a neutron does not leak out.

Reproduction factor. For thermal neutrons, with \(\sigma_f\) and \(\sigma_a\) the fission and absorption cross-sections, and atom fractions \(x_5=0.0072\), \(x_8=0.9928\),

\[ \eta=\bar\nu\,\frac{x_5\,\sigma_f^{(5)}}{x_5\,\sigma_a^{(5)}+x_8\,\sigma_a^{(8)}} . \]

With \(\bar\nu=2.43\), \(\sigma_f^{(5)}=585\) b, \(\sigma_a^{(5)}=684\) b and \(\sigma_a^{(8)}=2.68\) b, \(\eta\approx1.35\). There are only 0.35 spare neutrons per absorption to cover all losses. A moderator that absorbs even a few percent of the thermal neutrons makes the reactor impossible.

Choosing a moderator

A good moderator slows neutrons in few collisions and absorbs few. The first property is measured by the mean logarithmic energy loss \(\xi\) (Chadwick course, Lesson 10) times the macroscopic scattering cross-section \(\Sigma_s\). The figure of merit is the moderating ratio \(\xi\Sigma_s/\Sigma_a\). Thermal absorption cross-sections are about 0.332 b for hydrogen, 0.0005 b for deuterium, 0.0002 b for oxygen and 0.0035 b for carbon.

Heavy water and ZOE

Halban and Kowarski had shown in the summer of 1939 that a uranium–water mixture came close to a chain reaction but could not sustain one. In February 1940 Raoul Dautry’s armaments ministry sent the banker and agent Jacques Allier to Norway. The Norsk Hydro company handed over its whole stock of heavy water, about 185 kg, which reached Paris in March. In June, as German troops approached, Halban and Kowarski took it to England on the cargo ship Broompark. Their wartime experiments in Cambridge, with the same heavy water, showed that a natural-uranium heavy-water reactor would work.

ZOE (for zero power, uranium oxide and heavy water, eau lourde) was built at the Fort de Châtillon at Fontenay-aux-Roses under Joliot as High Commissioner, with Kowarski in charge of construction. It held about three tonnes of uranium oxide in about five tonnes of heavy water, and went critical on 15 December 1948. Its power was raised to 150 kW by 1953. It produced radioisotopes for laboratories and trained a generation of French reactor engineers. It was shut down in 1976, and its installation is preserved.

Exercise 10.1Verify \(\eta\approx1.35\) for natural uranium. What is \(\eta\) for pure uranium-235, and what does the comparison say about enrichment?

Show solution
Numerator \(2.43\times0.0072\times585=10.24\). Denominator \(0.0072\times684+0.9928\times2.68=4.92+2.66=7.58\). \(\eta=1.35\). For pure uranium-235, \(\eta=2.43\times585/684=2.08\). Enrichment raises \(\eta\) and leaves far more room for losses, which is why light-water reactors, whose moderator absorbs strongly, need enriched fuel.

Exercise 10.2Compare the thermal absorption cross-section per molecule of \(\mathrm{H_2O}\) and \(\mathrm{D_2O}\). Given \(\xi\Sigma_s\approx1.35\ \mathrm{cm^{-1}}\) for light water and \(0.18\ \mathrm{cm^{-1}}\) for heavy water, and \(\Sigma_a=0.022\) and \(3.3\times10^{-5}\ \mathrm{cm^{-1}}\) respectively, find the two moderating ratios.

Show solution
\(\mathrm{H_2O}\): \(2\times0.332+0.0002=0.664\) b. \(\mathrm{D_2O}\): \(2\times0.0005+0.0002=0.0012\) b, about 550 times less. Moderating ratios: \(1.35/0.022\approx61\) for light water, \(0.18/3.3\times10^{-5}\approx5500\) for heavy water. Light water slows neutrons better per centimetre, but heavy water wastes almost none of them, which is decisive when \(\eta\) is only 1.35. (Commercial heavy water contains a little light water, which lowers its ratio in practice.)

Exercise 10.3Suppose a heavy-water lattice has \(\eta=1.35\), \(\varepsilon=1.02\), \(p=0.90\) and \(f=0.95\). Find \(k_\infty\). How large may the leakage probability \(1-P_{\mathrm{NL}}\) be for the reactor to be critical?

Show solution
\(k_\infty=1.35\times1.02\times0.90\times0.95=1.18\). Criticality needs \(P_{\mathrm{NL}}\ge1/1.18=0.85\), so up to about 15% of neutrons may leak. This margin is why a heavy-water reactor with only three tonnes of natural uranium oxide, like ZOE, could go critical, while a graphite pile needed tens of tonnes of uranium.
Part III · Life and legacy
Lesson 11

War, Resistance and the CEA, 1939–1950

The Nobel Prize in Chemistry for 1935 went jointly to Frédéric Joliot and Irène Joliot-Curie for their synthesis of new radioactive elements. Joliot became maître de conférences at the Faculty of Science that year, and in 1937 the Collège de France created a chair of nuclear chemistry for him. He equipped his laboratory with one of the first cyclotrons in Western Europe and built the Laboratory of Atomic Synthesis at Ivry-sur-Seine. His politics moved left with the times: he joined the Socialist Party in 1934 and the League for the Rights of Man in 1936.

The phoney war and the defeat

From November 1939 his team worked for the armaments ministry. After the experiments with heavy water and uranium described in Lessons 09 and 10, the German offensive of May 1940 ended the programme in France. Joliot sent Halban and Kowarski to England with the heavy water and the documents, and hid the uranium oxide. He chose to stay. The Germans occupied his laboratory and placed it under Wolfgang Gentner, a German physicist who had worked with him in Paris before the war. Gentner protected Joliot and his laboratory during the Occupation.

Resistance

In 1941 Joliot helped found the Front national, a Resistance movement organised by the communists but open to all, and became its president. In 1942 he joined the French Communist Party. Langevin had been arrested in 1940, and his son-in-law, the physicist Jacques Solomon, was shot by the Germans in 1942. During the liberation of Paris in August 1944 he worked at the Prefecture of Police, where his team made incendiary bottles for the fighters.

The CEA and ZOE

After the Liberation, Joliot became director of the CNRS. In October 1945 Charles de Gaulle created the Commissariat à l’énergie atomique, with Joliot as High Commissioner and Raoul Dautry as administrator-general. Irène Joliot-Curie and Francis Perrin were among the commissioners. Joliot recruited Kowarski, Bertrand Goldschmidt and Jules Guéron from the Allied programme in Canada, and the team built ZOE in under two years. He insisted that French atomic research should be for peaceful purposes. As the Cold War hardened, a communist at the head of the atomic programme became unacceptable to France’s allies and to its government. After a speech at the Communist Party congress in April 1950 in which he declared that progressive scientists would never contribute to a war against the Soviet Union, the government of Georges Bidault dismissed him. Francis Perrin succeeded him.

DiscussionJoliot argued that the 1939 patents and the French physics behind them gave France a claim to a share in nuclear energy after the war. The Anglo-American allies treated French scientists in their projects with suspicion, partly because of those patents and partly because of Joliot’s politics. Weigh the scientific, national and political arguments on each side.

Lesson 12

Peace, Orsay and legacy, 1950–1958

From 1949 Joliot led the international peace movement sponsored by the communist parties, and in 1950 he became president of the World Peace Council. The movement’s Stockholm Appeal of March 1950 called for the unconditional banning of atomic weapons and gathered hundreds of millions of signatures, many in Eastern Europe. Critics in the West saw it as serving Soviet policy, since the Soviet Union then had far fewer atomic weapons than the United States. He received the first Stalin Peace Prize in 1951. In 1955 he was one of the eleven signatories of the Russell–Einstein Manifesto, which led to the Pugwash Conferences.

He kept his Collège de France chair and continued to direct research. When Irène died in March 1956 he took over her chair of nuclear physics at the Sorbonne and the direction of the Curie laboratory. He devoted his last two years to building the nuclear physics centre at Orsay, south of Paris, which she had planned, with a 160 MeV synchrocyclotron. He suffered from liver disease and died on 14 August 1958 in Paris, aged 58. His illness, like Irène’s, is often linked to radiation exposure. De Gaulle, just returned to power, ordered a state funeral. He is buried beside Irène in the cemetery at Sceaux.

His honours included the Matteucci Medal (1932, with Irène), the Nobel Prize in Chemistry (1935), membership of the Académie des sciences (1943) and the Académie de médecine, foreign membership of the Royal Society, the Hughes Medal (1947) and the rank of commander of the Legion of Honour, awarded for his Resistance work.

Legacy

  • Artificial radioactivity: the 1934 discovery made radioisotopes into a laboratory product. Tracers, nuclear medicine and positron emission tomography all descend from it.
  • Nuclear chemistry: fast radiochemical separation and the carrier method became the standard way to identify new nuclides, including the transuranium elements.
  • The chain reaction: the 1939 neutron measurements were among the first evidence that fission could be self-sustaining, and the Paris patents were among the first on nuclear reactors.
  • French science: the CEA, ZOE and the Orsay campus, now part of Université Paris-Saclay, are his institutional legacy. His doctoral students included Georges Charpak, Nobel laureate in physics in 1992, and the Japanese physicist Toshiko Yuasa.

Final projectWrite a referee’s report on the 15 January 1934 note. Using the half-lives, Q-values and thresholds of Lesson 07 and the chemistry of Lesson 08, list the alternative explanations a sceptic could have offered (contamination, prompt emission, instrumental effects) and show which observation excludes each one.

Reference

Sources and further reading

Biography

  1. Frédéric Joliot: Biographical, NobelPrize.org.
  2. Frédéric Joliot-Curie, Wikipedia; and the French article Frédéric Joliot-Curie, with detailed chronology.
  3. P. M. S. Blackett, Jean Frédéric Joliot 1900–1958, Biographical Memoirs of Fellows of the Royal Society 6 (1960).
  4. M. Pinault, Frédéric Joliot-Curie, Odile Jacob, 2000.
  5. S. R. Weart, Scientists in Power, Harvard University Press, 1979.
  6. M. Goldsmith, Frédéric Joliot-Curie: A Biography, Lawrence and Wishart, 1976.

Physics and chemistry

  1. I. Curie, F. Joliot, Émission de protons de grande vitesse par les substances hydrogénées sous l’influence des rayons γ très pénétrants, C. R. Acad. Sci. 194, 273 (1932).
  2. I. Curie, F. Joliot, Un nouveau type de radioactivité, C. R. Acad. Sci. 198, 254 (1934).
  3. F. Joliot, I. Curie, Artificial production of a new kind of radio-element, Nature 133, 201 (1934).
  4. F. Joliot, Chemical evidence of the transmutation of elements, Nobel lecture, 12 December 1935.
  5. H. von Halban, F. Joliot, L. Kowarski, Liberation of neutrons in the nuclear explosion of uranium, Nature 143, 470 (1939); Number of neutrons liberated in the nuclear fission of uranium, Nature 143, 680 (1939).
  6. F. Guerra, M. Leone, N. Robotti, Neutron physics in the early 1930s, arXiv:physics/0510044 (detailed account of the Curie–Joliot papers of 1932–1934).
  7. From fission to the first French reactor, Europhysics News 40/6 (2009).
  8. ZOE reactor, Wikipedia.
  9. K. S. Krane, Introductory Nuclear Physics, Wiley, 1988 (reactions, beta decay, fission and reactors).