courseshub.world · Lives in Mathematics and Physics

Irène Joliot-Curie

Irène Curie

A course in two strands: the life of Marie and Pierre Curie’s elder daughter, who ran X-ray units at the front at eighteen, made the world’s strongest polonium sources and served in France’s first government to include women, and the physics and chemistry of her work: alpha particles, the statistics of decay, artificial radioactivity and the 3.5-hour substance that led to fission.

Born 12 September 1897, Paris Died 17 March 1956, Paris Nobel Prize Chemistry, 1935, with Frédéric Joliot Level advanced undergraduate → graduate
Bragg curve of a polonium alpha particle (5.30 MeV) in air: energy lost per centimetre against distance travelled. The loss grows as the particle slows and peaks just before it stops, near 3.9 cm. Range straggling, a subject of Irène Curie’s thesis, rounds off the end. Sketched from Geiger’s range–energy law, which fails in the last millimetres, so the peak is capped there.
Lesson 00

How to use this course

Irène Curie grew up inside the science of radioactivity. Her parents discovered polonium and radium the year after her birth. She learned physics from her mother and her mother’s friends, worked beside her mother in the First World War, and joined her laboratory in 1918. Her thesis on the alpha rays of polonium made her an expert in the particle that became the main probe of the nucleus. With her husband Frédéric Joliot she found the evidence that led to the neutron and to positron pair production, and in 1934 the first artificial radioactivity. On her own, with Pavle Savić, she found in 1938 a radioactive substance in uranium that behaved like lanthanum. It was a fission product, and it set Otto Hahn and Fritz Strassmann on the path that led to fission.

This course pairs with the Frédéric Joliot-Curie course, which covers the cloud chamber, pair production, artificial radioactivity, the chemistry of transmutation and the chain reaction. Here the technical lessons follow her own specialities: polonium, the physics of alpha particles, counting statistics, the theory of alpha decay and alpha reactions, and the radiochemistry of the heavy elements. A final technical lesson treats radiation dose, which shaped her life and her death.

Prerequisites for Part II
  • The exponential decay law and first-order linear differential equations.
  • Elementary probability: the Poisson and normal distributions.
  • Basic quantum mechanics: the WKB approximation for barrier penetration.
  • General chemistry: solubility, precipitation and crystallisation.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Lesson 09 is the centre of the course. Masses are atomic masses in unified units u, with \(1\,\mathrm u\,c^2=931.5\) MeV, and \(e^2/4\pi\varepsilon_0=1.44\) MeV fm. Irène signed her papers “Irène Curie” throughout her career.

Reference

Timeline

1897Born 12 September in Paris, elder daughter of Pierre Curie and Marie Skłodowska-Curie.
1898Her parents announce polonium (July) and radium (December).
1904Sister Ève born.
190619 April: Pierre Curie killed in a street accident. The girls are partly raised by their paternal grandfather, Eugène Curie.
c. 1907Taught in the “Coopérative”, a small cooperative school run by Marie Curie, Paul Langevin, Jean Perrin and other scientists.
1911Accompanies her mother to Stockholm for the Nobel Prize in Chemistry.
1914Completes secondary school at the Collège Sévigné; begins at the Faculty of Science in Paris.
1914–18Radiology nurse with her mother’s mobile X-ray units; trains military radiographers.
1918Joins the Radium Institute as her mother’s assistant.
1921Travels with her mother to the United States.
1925March: doctorate, Recherches sur les rayons α du polonium.
1926October: marries Frédéric Joliot.
1927Daughter Hélène born.
1932January: with Joliot, fast protons ejected by the beryllium radiation. Son Pierre born. Chef de travaux at the Curie laboratory.
1933Positrons from gamma rays and from alpha-bombarded light elements. Seventh Solvay Conference.
193415 January: artificial radioactivity announced. 4 July: Marie Curie dies.
1935Nobel Prize in Chemistry with Frédéric Joliot.
1936June–September: Under-Secretary of State for Scientific Research in Léon Blum’s Popular Front government.
1937Professor at the Faculty of Science of Paris.
1938With Pavle Savić, a 3.5-hour activity in irradiated uranium resembling lanthanum.
1939Officer of the Legion of Honour.
1944June: crosses into Switzerland with her children.
1946Director of the Radium Institute (Curie laboratory). Commissioner of the Commissariat à l’énergie atomique, for six years.
1948ZOE, the first French reactor, goes critical on 15 December.
1951Her candidacy for the Académie des sciences is rejected.
1956Dies of leukaemia in Paris, 17 March, aged 58. State funeral.
Part I · Formation
Lesson 01

A Curie childhood, 1897–1914

Irène Curie was born in Paris on 12 September 1897. Her mother, Marie Skłodowska-Curie, was then beginning the doctoral research on uranium rays that led, with her father Pierre Curie, to the discovery of polonium and radium in 1898. Her sister Ève was born in 1904. Pierre Curie was killed on 19 April 1906 when he was struck by a horse-drawn wagon in a Paris street. His father, the physician Eugène Curie, lived with the family and helped raise the two girls. He was a strong influence on Irène.

The Coopérative

Marie Curie thought French schools gave children too many hours at a desk and too little science. With a group of friends from the Sorbonne she set up a teaching cooperative for their own children, about ten pupils in all. Each parent taught a subject; the teachers included Marie Curie, Paul Langevin and Jean Perrin. The lessons were practical and informal, with much time in the laboratory. Irène later finished her secondary education at the Collège Sévigné, a private school in Paris, and in 1914 began studying at the Faculty of Science.

In December 1911 she went with her mother to Stockholm, where Marie Curie received her second Nobel Prize, in chemistry. Irène was fourteen.

Lesson 02

X-rays at the front, 1914–1918

When war broke out in August 1914, Marie Curie saw that army surgeons had almost no X-ray equipment for locating bullets and shell fragments in the wounded. Through the Red Cross she organised radiology cars, soon nicknamed “petites Curies”, and fixed radiology posts in hospitals near the front. Irène trained as a nurse and from 1916 worked as a radiographer, first with her mother and then alone, at hospitals in Belgium and northern France, including Furnes, Ypres and Amiens. She was eighteen, and she received the Médaille militaire in 1918.

Her work combined physics and medicine. She set up X-ray tubes, generators and screens, located fragments by triangulating from two exposures, and taught surgeons to use the images. From 1917 she trained nurses in radiology at the Edith Cavell hospital in Paris while resuming her own studies. Protection against X-rays was rudimentary. The doses she received in these years were the first of many.

Locating a fragment

A standard method was to take two exposures with the tube moved sideways by a known distance \(d\). A fragment at height \(h\) above the plate, with the tube focus at height \(H\), shows a shift \(s\) of its shadow. By similar triangles

\[ s=\frac{d\,h}{H-h},\qquad h=\frac{sH}{d+s}. \]

A few millimetres of shift fix the depth of the fragment to about a centimetre, enough to guide the surgeon. Lesson 10 returns to the doses involved.

Lesson 03

The Radium Institute and the thesis, 1918–1926

In 1918 Irène became an assistant (préparatrice) in her mother’s laboratory at the Radium Institute, and in 1920 her assistant with degrees in physics and mathematics. Her first research measured the atomic weight of chlorine from different minerals, a test of whether isotopic composition varied in nature. In 1924, with Catherine Chamié, she measured the decay constant of radon.

Her doctoral thesis, defended in March 1925, was titled Recherches sur les rayons α du polonium. Oscillation de parcours, vitesse d’émission, pouvoir ionisant. It studied three things: the fluctuations in the range of individual alpha particles from polonium (range straggling), their speed of emission, and their ionising power along the track. The thesis is the subject of Lesson 05. She also studied the ionisation curve of alpha particles in air with the Czech physicist František Běhounek.

In 1925 a young engineer, Frédéric Joliot, arrived as her mother’s assistant, and Irène taught him radioactive techniques. She was reserved, direct and indifferent to social convention; he was outgoing and charming. They married in October 1926. Their daughter Hélène was born in 1927 and their son Pierre in 1932. From 1929 they published together. The story of their joint work of 1932–1934, the neutron and positron they did not identify and the artificial radioactivity they did, is told in the Frédéric Joliot-Curie course, Lessons 03 and 05–08.

Part II · The physics and chemistry
Lesson 04

Polonium: making the sources

For the general reader. The Joliot-Curies’ discoveries depended on one material: polonium, the element her parents had found in 1898. Polonium emits alpha particles and almost no gamma rays, so it bombards a target cleanly. It cannot be bought in a jar. It must be extracted from a long chain of radioactive decays, and its supply depends on how long ago its ancestors were made. Irène Curie prepared the strongest polonium sources in the world from the residues of old radon tubes.

Goals
  • Solve the Bateman equations for a three-member chain.
  • Find how polonium grows in a sample of radium D and how long to wait before extracting it.
  • Understand secular and transient equilibrium.

The chain

Radon-222 (3.82 days) decays through short-lived products to lead-210, historically called radium D, which has a half-life of 22.2 years. Radium D decays to bismuth-210 (radium E, 5.01 days), which decays to polonium-210 (138.4 days), which decays to stable lead-206. Hospitals used sealed radon tubes for cancer therapy. After the radon had decayed, the tubes held radium D, which kept producing polonium for decades.

Bateman solution

For a chain \(1\to2\to3\to\) with decay constants \(\lambda_1,\lambda_2,\lambda_3\), all distinct, and only \(N_1(0)\) non-zero,

\[ N_3(t)=\lambda_1\lambda_2N_1(0)\left[\frac{e^{-\lambda_1t}}{(\lambda_2-\lambda_1)(\lambda_3-\lambda_1)}+\frac{e^{-\lambda_2t}}{(\lambda_1-\lambda_2)(\lambda_3-\lambda_2)}+\frac{e^{-\lambda_3t}}{(\lambda_1-\lambda_3)(\lambda_2-\lambda_3)}\right]. \]

Derivation. The equations are \(\dot N_1=-\lambda_1N_1\), \(\dot N_2=\lambda_1N_1-\lambda_2N_2\), \(\dot N_3=\lambda_2N_2-\lambda_3N_3\). Solve in turn with an integrating factor: \(N_2=\frac{\lambda_1N_1(0)}{\lambda_2-\lambda_1}(e^{-\lambda_1t}-e^{-\lambda_2t})\), then substitute into the third equation. Each term \(e^{-\lambda_it}\) is driven with coefficient fixed by the others, and the constant multiplying \(e^{-\lambda_3t}\) is set by \(N_3(0)=0\). Laplace transforms give the same result in one line: \(\tilde N_3(s)=\lambda_1\lambda_2N_1(0)/[(s+\lambda_1)(s+\lambda_2)(s+\lambda_3)]\), whose partial fractions are the three terms.

Secular equilibrium

If \(\lambda_1\ll\lambda_2,\lambda_3\), then for \(t\gg1/\lambda_2,1/\lambda_3\) the activities of all members are equal: \(\lambda_3N_3\approx\lambda_2N_2\approx\lambda_1N_1\).

A source of radium D therefore produces as many polonium atoms per second as it holds radium D decays per second, once a few polonium half-lives have passed. Polonium can be separated chemically, for example by spontaneous deposition on silver or nickel from acid solution, and the radium D left behind regenerates it.

Exercise 4.1Fresh radium D, free of its daughters, is stored. Using the Bateman formula with half-lives 22.2 y, 5.01 d and 138.4 d, the polonium activity relative to the radium D activity is about 0.11 after 30 days, 0.37 after 100 days and 0.84 after one year. Explain why the growth is governed mainly by the polonium half-life, and estimate the activity ratio after two years.

Show solution
Bismuth-210 reaches equilibrium within a few weeks, after which polonium is fed at an almost constant rate, so \(A_3/A_1\approx1-e^{-\lambda_3t}\) with a small delay. After two years, \(1-2^{-730/138.4}\approx0.97\); the full formula gives about 0.99 of the (slightly decayed) radium D activity. Waiting more than about two years gains almost nothing.

Exercise 4.2Bismuth-210 is separated pure from a radium D sample. Show that the polonium activity it produces peaks at \(t_m=\ln(\lambda_2/\lambda_3)/(\lambda_2-\lambda_3)\). Evaluate \(t_m\) and the peak polonium activity as a fraction of the initial bismuth activity.

Show solution
\(A_3=\lambda_3N_3=\frac{\lambda_2\lambda_3N_2(0)}{\lambda_2-\lambda_3}(e^{-\lambda_3t}-e^{-\lambda_2t})\); setting the derivative to zero gives \(\lambda_3e^{-\lambda_3t}=\lambda_2e^{-\lambda_2t}\), hence \(t_m\). With \(\lambda_2=0.138\) and \(\lambda_3=0.00501\) per day, \(t_m=\ln27.6/0.133\approx24.9\) days. The peak \(A_3/A_2(0)=\frac{\lambda_3}{\lambda_2-\lambda_3}(e^{-\lambda_3t_m}-e^{-\lambda_2t_m})\approx0.032\). Only 3% of the bismuth activity reappears as polonium activity, because most bismuth atoms become polonium atoms that decay slowly.

Exercise 4.3In the 1930s the Curie laboratory had polonium sources of up to a few hundred millicuries. How much radium D (in curies, at equilibrium) is needed to produce 200 mCi of polonium, and how much lead-210 mass is that?

Show solution
At equilibrium the activities are equal, so 200 mCi of radium D, that is \(7.4\times10^9\) decays per second. With \(\lambda=\ln2/(22.2\times3.156\times10^7\ \mathrm s)=9.9\times10^{-10}\ \mathrm s^{-1}\), \(N=7.5\times10^{18}\) atoms, or \(2.6\) mg of lead-210. It had to be gathered from many used radon tubes.
Lesson 05

Alpha particles in matter

For the general reader. An alpha particle ploughs through air, knocking electrons off molecules as it goes. It slows down, and the slower it goes the more it ionises, until it stops abruptly after a few centimetres. All alpha particles from polonium start with the same energy, yet they do not all stop at exactly the same place. Irène Curie’s thesis measured how much their ranges vary, how fast they leave the nucleus, and how their ionising power changes along the path.

Goals
  • State the Bethe formula for the stopping power of a heavy charged particle.
  • Derive the range from the stopping power and Geiger’s empirical rule.
  • Understand range straggling as a random-walk effect.

Stopping power

Bethe formula (non-relativistic form)

A particle of charge \(ze\) and speed \(v=\beta c\), with \(\beta\ll1\), passing through a medium with \(n_e\) electrons per unit volume and mean excitation energy \(I\), loses energy at the rate

\[ -\frac{dE}{dx}=\frac{4\pi n_e z^2}{m_ev^2}\left(\frac{e^2}{4\pi\varepsilon_0}\right)^2\ln\frac{2m_ev^2}{I}. \]

For air, \(I\approx86\) eV.

Sketch. An electron at impact parameter \(b\) receives the transverse impulse \(\Delta p=2ze^2/(4\pi\varepsilon_0bv)\), hence energy \(\Delta p^2/2m_e\). Integrating over a cylinder of electrons, \(2\pi n_e b\,db\,dx\), gives \(\frac{4\pi n_ez^2}{m_ev^2}(e^2/4\pi\varepsilon_0)^2\ln(b_{\max}/b_{\min})\). The cut-offs come from the maximum energy transfer \(2m_ev^2\) and the binding of the electrons, measured by \(I\); the quantum calculation by Hans Bethe in 1930 fixes the logarithm. The loss depends on the particle only through \(z^2/v^2\), so slower particles lose energy faster. This produces the Bragg peak in the figure at the top of the page.

Range

The mean range is \(R(E_0)=\int_0^{E_0}dE/(-dE/dx)\). Neglecting the slow variation of the logarithm, \(-dE/dx\propto1/E\), so \(R\propto E^2\propto v^4\). The logarithm makes the dependence weaker. Hans Geiger found empirically that for alpha particles in air

\[ R\propto v^3,\qquad R\,[\mathrm{cm}]\approx0.318\,E^{3/2}\,[\mathrm{MeV}],\quad 4\lesssim E\lesssim7\ \text{MeV}. \]

Polonium-210 alpha particles have 5.30 MeV and a mean range of about 3.9 cm in air at 15 °C.

Straggling

Energy loss happens in many independent collisions, each random. The total number of collisions along a path fluctuates, so identical particles stop at slightly different depths. By the central limit theorem the stopping points are close to normally distributed about the mean range. For polonium alphas in air the standard deviation is about 1% of the range. Measuring this distribution precisely, with a cloud chamber and with ionisation methods, was the first part of Irène Curie’s thesis. The same theory, extended by Niels Bohr in 1915, tests the statistical model of the energy loss.

Exercise 5.1Evaluate the Bethe formula for a 5.30 MeV alpha particle (\(m_\alpha c^2=3727\) MeV) in air of density \(1.205\times10^{-3}\) g/cm\(^3\), using \(Z/A=0.499\) and \(I=86\) eV. Use the form \(-\frac1\rho\frac{dE}{dx}=0.307\,\frac{Z}{A}\,\frac{z^2}{\beta^2}\ln\frac{2m_ec^2\beta^2}{I}\) MeV cm\(^2\)/g.

Show solution
\(\beta^2=2T/mc^2=2.85\times10^{-3}\). \(2m_ec^2\beta^2=1.022\times10^6\times2.85\times10^{-3}=2.91\) keV, so the logarithm is \(\ln(2910/86)=3.52\). Then \(0.307\times0.499\times4/0.00285\times3.52=757\) MeV cm\(^2\)/g, or about 0.91 MeV/cm in air. A rough range estimate, \(5.3/0.91\approx5.8\) cm, is too long because the loss grows as the particle slows; the true range is 3.9 cm.

Exercise 5.2Using Geiger’s rule, find the ranges in air of the alpha particles of polonium-210 (5.30 MeV) and polonium-212 (thorium C′, 8.78 MeV). The measured values are 3.9 and 8.6 cm. What are their speeds?

Show solution
\(0.318\times5.30^{1.5}=3.88\) cm and \(0.318\times8.78^{1.5}=8.27\) cm, close to the measurements. Speeds: \(v=c\sqrt{2T/mc^2}\) gives \(1.60\times10^9\) cm/s and \(2.06\times10^9\) cm/s. The ratio of ranges, 2.2, is close to the cube of the speed ratio, \(1.29^3=2.1\).

Exercise 5.3Model the energy loss as \(N\) independent collisions with mean \(\bar\epsilon\) and variance \(\sigma_\epsilon^2\). Show that the relative fluctuation of the energy lost in a fixed path scales as \(N^{-1/2}\). If a polonium alpha makes about \(10^5\) ionising collisions and the spread of individual losses is comparable to their mean, estimate the relative range straggling.

Show solution
Total loss has mean \(N\bar\epsilon\) and variance \(N\sigma_\epsilon^2\), so the relative spread is \(\sigma_\epsilon/(\bar\epsilon\sqrt N)\). With \(N\approx10^5\) (5.3 MeV divided by about 35 eV per ion pair, times a few collisions per ion pair) and \(\sigma_\epsilon\approx\bar\epsilon\), the spread is about 0.3%. Large-energy collisions give a long tail, so the true straggling, about 1%, is larger than this crude estimate.
Lesson 06

The statistics of decay

For the general reader. Each radioactive nucleus decays at a random moment, without memory of how long it has lived. A counter clicking near a source therefore never clicks at a perfectly steady rate. To measure a half-life, or to decide whether a weak new activity is real, one has to understand these fluctuations. Irène Curie’s measurement of the decay constant of radon in 1924 and the Joliot-Curies’ decay curves of 1934 rested on this statistics.

Goals
  • Derive the Poisson distribution of counts from the decay law.
  • Estimate the uncertainty of a measured half-life.
  • Correct counting rates for dead time.
Poisson counting

If each of \(N\) nuclei decays independently with probability \(p=1-e^{-\lambda\Delta t}\) during an interval \(\Delta t\), with \(N\) large and \(p\) small, the number of decays \(k\) follows the Poisson law

\[ P(k)=\frac{\mu^k e^{-\mu}}{k!},\qquad \mu=Np\approx N\lambda\Delta t,\qquad \operatorname{Var}k=\mu. \]

Derivation. The count is binomial, \(\binom Nk p^k(1-p)^{N-k}\). For \(N\to\infty\) with \(Np=\mu\) fixed, \(\binom Nk p^k\to\mu^k/k!\) and \((1-p)^{N-k}\to e^{-\mu}\). The mean and variance of the binomial are \(Np\) and \(Np(1-p)\to\mu\). A count of \(k\) thus has standard error about \(\sqrt k\), and its relative error is \(1/\sqrt k\).

Measuring a half-life

With counts \(C_1\) and \(C_2\) in equal intervals centred at times \(t_1\) and \(t_2\), the estimate is \(\hat\lambda=\ln(C_1/C_2)/(t_2-t_1)\). Propagating the Poisson errors,

\[ \frac{\sigma_\lambda}{\lambda}=\frac{1}{\lambda(t_2-t_1)}\sqrt{\frac1{C_1}+\frac1{C_2}} . \]

In practice one fits a straight line to \(\ln C\) against \(t\) for many intervals, weighting each point by its count, and subtracts the background measured with no source. Two activities with different half-lives give a curve whose tail is the longer-lived component; subtracting its extrapolation reveals the shorter one. Irène Curie relied on such analyses in all her work on neutron-induced activities.

Dead time

After each count a Geiger–Müller counter is insensitive for a dead time \(\tau\), typically \(10^{-4}\) s. If the recorded rate is \(m\), the counter is dead for a fraction \(m\tau\) of the time, so the true rate is

\[ n=\frac{m}{1-m\tau}. \]

Exercise 6.1A source gives 1000 counts in a minute and, 30 minutes later, 600 counts in a minute. Background is negligible. Find the half-life and its standard error.

Show solution
\(\lambda=\ln(1000/600)/30=0.0170\) per min, \(t_{1/2}=40.7\) min. \(\sigma_\lambda/\lambda=\sqrt{1/1000+1/600}/\ln(5/3)=0.0516/0.511=0.10\). So \(t_{1/2}=41\pm4\) min. Much longer counting, or many points, is needed for a 1% result.

Exercise 6.2A counter with \(\tau=100\) µs records 300 counts per second. Find the true rate. At what recorded rate is the correction 10%?

Show solution
\(n=300/(1-0.03)=309\) per second. The correction \(1/(1-m\tau)-1\) reaches 10% when \(m\tau=0.091\), that is \(m\approx910\) per second.

Exercise 6.3A background of 20 counts per minute is known accurately. A sample counted for 10 minutes gives 260 counts. Is there a significant activity? Give the net rate and its standard error.

Show solution
Expected background in 10 min: 200 counts. Net \(60\pm\sqrt{260}=60\pm16\) counts, or \(6.0\pm1.6\) per minute, nearly four standard errors above zero. In 1938 the 3.5-hour activity in uranium (Lesson 09) had to be established against backgrounds from other activities in exactly this way.
Lesson 07

Barriers, tunnelling and alpha reactions

For the general reader. Why does polonium-210 live for months while polonium-212 lives for less than a millionth of a second, though their alpha particles differ in energy by less than a factor of two? And why did the alpha particles of polonium produce artificial radioactivity in light elements but not in heavy ones? The same quantum effect answers both: an alpha particle must tunnel through the electrical wall around the nucleus, and the probability depends exponentially on its energy.

Goals
  • Derive the Gamow factor for a Coulomb barrier.
  • Reproduce the Geiger–Nuttall relation.
  • Use the barrier to see why polonium alphas induce reactions only in light nuclei.
Gamow factor (Gamow; Gurney and Condon, 1928)

An alpha particle of kinetic energy \(Q\) (in the centre-of-mass frame) and reduced mass \(\mu\), inside a nucleus of radius \(R\) with daughter charge \(Z\), meets the Coulomb barrier \(V(r)=2Ze^2/(4\pi\varepsilon_0r)\) for \(r>R\). Its WKB transmission probability is \(T=e^{-2G}\), with

\[ 2G=\frac{2\sqrt{2\mu}}{\hbar}\int_R^{b}\sqrt{V(r)-Q}\,dr=4\eta\left[\arccos\sqrt x-\sqrt{x(1-x)}\right], \]

where \(b=2Ze^2/(4\pi\varepsilon_0Q)\) is the classical turning point, \(x=R/b\), and \(\eta=2Ze^2/(4\pi\varepsilon_0\hbar v)=2Z\alpha_{\mathrm{fs}}/\beta\) is the Sommerfeld parameter.

Derivation. Substitute \(r=b\cos^2\theta\). Then \(V-Q=Q\tan^2\theta\) and \(dr=-2b\cos\theta\sin\theta\,d\theta\), so the integral becomes \(2b\sqrt Q\int_0^{\theta_0}\sin^2\theta\,d\theta\) with \(\cos^2\theta_0=x\). Evaluate and use \(\sqrt{2\mu Q}\,b/\hbar=\eta\). For \(x\ll1\), \(2G\approx2\pi\eta-8\eta\sqrt x\), and since \(\eta\propto Z/\sqrt Q\), \(\ln T\approx-c_1Z/\sqrt Q+c_2\sqrt{ZR}\).

Geiger–Nuttall relation

The decay constant is \(\lambda\approx f\,e^{-2G}\), where \(f\approx v/2R\) is the rate at which the alpha particle strikes the barrier. Because \(2G\) is dominated by \(2\pi\eta\propto Z/\sqrt Q\), for a given element

\[ \log_{10}t_{1/2}\approx a\,\frac{Z}{\sqrt Q}+b . \]

Alpha-induced reactions

The same factor controls the reverse process. An alpha particle approaching a nucleus of charge \(Z_X\) below the barrier height \(B=2Z_Xe^2/(4\pi\varepsilon_0R)\) enters only by tunnelling. Polonium alphas, at 5.3 MeV, are above the barrier for beryllium (about 2.6 MeV) and boron (about 3.2 MeV), somewhat below it for aluminium (about 6.8 MeV) and hopelessly below it for heavy elements. This is why the Joliot-Curies found artificial radioactivity only in light elements, and why neutrons, which feel no barrier, became the tool for the heavy elements.

Exercise 7.1For polonium-210 (\(Q=5.41\) MeV, daughter \(Z=82\), \(A=206\)) take \(R=1.2(206^{1/3}+4^{1/3})=9.0\) fm. Show that \(b=43.7\) fm, \(\eta\approx22.0\) and \(2G\approx61.2\). With \(f\approx9\times10^{20}\ \mathrm s^{-1}\), estimate the half-life and compare with 138 days.

Show solution
\(b=2\times82\times1.44/5.41=43.7\) fm, \(x=0.206\). \(\mu c^2=3727\times191\,900/(3727+191\,900)\approx3656\) MeV, so \(\beta=\sqrt{2\times5.41/3656}=0.0544\) and \(\eta=164/(137.0\times0.0544)=22.0\). \(\arccos\sqrt{0.206}=1.100\), \(\sqrt{0.206\times0.794}=0.404\), so \(2G=88.0\times0.696=61.2\). \(\lambda=9\times10^{20}e^{-61.2}\approx2.3\times10^{-6}\ \mathrm s^{-1}\), \(t_{1/2}\approx3.4\) days. This is 40 times shorter than the real 138 days, a good result for a formula that spans more than twenty orders of magnitude; the missing factor reflects the probability that an alpha particle is preformed in the nucleus.

Exercise 7.2Polonium-212 has \(Q=8.95\) MeV and \(t_{1/2}=0.30\) µs. With the same method, \(2G\approx32.3\). Compare the predicted ratio of the half-lives of polonium-210 and polonium-212 with the observed ratio.

Show solution
Predicted ratio: \(e^{61.2-32.3}=e^{28.9}\approx3.6\times10^{12}\), slightly changed by the different \(f\). Observed: \(138.4\times86\,400/0.30\times10^{-6}=4.0\times10^{13}\). The tunnelling factor explains essentially all of the factor \(10^{13}\); a 66% increase in energy changes the lifetime by thirteen orders of magnitude.

Exercise 7.3Compute the Coulomb barriers for alpha particles on \({}^9\mathrm{Be}\), \({}^{27}\mathrm{Al}\) and \({}^{63}\mathrm{Cu}\), with \(R=1.2(A^{1/3}+4^{1/3})\) fm. Which can polonium alphas reach classically?

Show solution
Be: \(R=1.2(2.08+1.59)=4.4\) fm, \(B=8\times1.44/4.4=2.6\) MeV. Al: \(R=5.5\) fm, \(B=26\times1.44/5.5=6.8\) MeV. Cu: \(R=1.2(3.98+1.59)=6.7\) fm, \(B=58\times1.44/6.7=12.5\) MeV. Only beryllium (and boron) are classically accessible at 5.3 MeV. For copper the tunnelling probability is negligible, and polonium alphas produce no reactions.
Lesson 08

Neutrons on heavy elements, 1934–1938

For the general reader. Within weeks of artificial radioactivity, Enrico Fermi in Rome bombarded every element he could find with neutrons instead of alpha particles. Neutrons have no charge and enter even the heaviest nuclei. In uranium Fermi found several new activities and thought he had made elements beyond uranium. For four years laboratories in Rome, Berlin and Paris tried to sort out what the activities in irradiated uranium and thorium were. Irène Curie led the Paris effort.

Goals
  • Understand neutron capture and the \(1/v\) law.
  • Follow the beta-decay chains expected after neutron capture in uranium and thorium.
  • See why the activities were so hard to assign.

Capture and the 1/v law

A slow neutron absorbed by a nucleus \({}^A Z\) forms \({}^{A+1}Z\) in an excited state, which usually emits a gamma ray: the \((n,\gamma)\) reaction. The product often has too many neutrons and decays by \(\beta^-\) emission to element \(Z+1\).

The \(1/v\) law

Far from resonances, the capture cross-section of slow neutrons varies as

\[ \sigma_{n\gamma}(v)\propto\frac1v,\qquad\text{so the capture rate per nucleus }\ n\,v\,\sigma\ \text{ is independent of }v. \]

A neutron spends time proportional to \(1/v\) within range of the nucleus, and at low energy the quantum amplitude for capture is roughly constant, so the probability goes as \(1/v\). This is why Fermi found in October 1934 that neutrons slowed in paraffin were far more effective.

Uranium and thorium

Fermi’s expectation for uranium was \({}^{238}\mathrm U(n,\gamma){}^{239}\mathrm U\xrightarrow{\beta^-}{}^{239}93\xrightarrow{\beta^-}{}^{239}94\). Ida Noddack objected in 1934 that the nucleus might break into large fragments, but no one tested her suggestion. Hahn, Meitner and Strassmann in Berlin assigned the activities to a series of “transuranic” elements with chemistry like rhenium, osmium and iridium. From 1935 Irène Curie worked on the thorium activities with Hans von Halban and Peter Preiswerk, and from 1937 on uranium with Pavle Savić, a young Yugoslav physicist working at the Radium Institute.

The assignments were hard for three reasons. Many activities were produced at once, with similar half-lives. The amounts were invisible, so every identification relied on carriers (Frédéric Joliot-Curie course, Lesson 08). And the chemistry of the elements beyond uranium was unknown; chemists assumed they would continue the transition-metal groups of the periodic table. In fact neptunium and plutonium, discovered at Berkeley in 1940 and 1941, behave more like uranium.

Exercise 8.1A thermal neutron at 0.025 eV has speed 2200 m/s. Uranium-238 has a thermal capture cross-section of 2.7 b. Using the \(1/v\) law, what would the capture cross-section be at 1 eV, if no resonances intervened? Why is this estimate wrong for uranium-238 near 6.7 eV?

Show solution
\(v\propto\sqrt E\), so \(\sigma(1\ \mathrm{eV})=2.7\times\sqrt{0.025/1}=0.43\) b. Uranium-238 has a huge capture resonance at 6.7 eV, where the cross-section reaches thousands of barns. The \(1/v\) law holds only well below the first resonance.

Exercise 8.2Uranium-239 has \(t_{1/2}=23.5\) min and its daughter neptunium-239 has \(t_{1/2}=2.36\) d. A uranium sample is irradiated for one hour. Find the ratio of the activities of uranium-239 and neptunium-239 at the end of the irradiation, assuming a constant production rate.

Show solution
Uranium-239: \(A_1=R(1-e^{-\lambda_1t})\) with \(\lambda_1t=2.65\), so \(A_1=0.93R\). Neptunium: \(N_2\) grows from the decays of uranium-239. To first order, the number of uranium-239 decays during the hour is about \(R(t-(1-e^{-\lambda_1t})/\lambda_1)=R(60-31.6)\) min \(=28.4R\) min, and \(\lambda_2=0.000204\) per min, so \(A_2\approx0.0058R\). The ratio is about 160. The long-lived daughter is barely visible at first, one reason neptunium was missed until 1940.
Lesson 09

The 3.5-hour substance and fission

For the general reader. In 1937 Irène Curie and Pavle Savić found in irradiated uranium an activity with a half-life of about 3.5 hours. It did not behave like any element near uranium. In 1938 they showed that it followed lanthanum, element 57, through chemical separations. Lanthanum sits less than two-thirds of the way up the periodic table from uranium. They did not take the final step of concluding that the uranium nucleus had split. Their results led Otto Hahn and Fritz Strassmann to repeat their own experiments, and in December 1938 these found barium. The split nucleus, fission, had been found.

Goals
  • Understand co-precipitation and fractional crystallisation of trace elements.
  • Use the Doerner–Hoskins law to test whether an activity belongs to a carrier element.
  • Follow the chemical reasoning that led from the 3.5-hour body to barium.

Separating unweighable elements

An invisible radioactive species is identified by the element it follows. When a carrier is precipitated or crystallised, a trace element that is chemically identical to the carrier is incorporated uniformly. A trace element that is merely similar is incorporated in a different proportion. Repeated fractional crystallisation magnifies the difference.

Doerner–Hoskins law

When a carrier crystallises slowly from solution, with the surface of the growing crystal in equilibrium with the solution, the trace fraction \(f_t\) and carrier fraction \(f_c\) remaining in solution obey

\[ \ln f_t=\lambda_{\mathrm{DH}}\,\ln f_c , \]

where \(\lambda_{\mathrm{DH}}\) is the logarithmic distribution coefficient. An activity that is an isotope of the carrier has \(\lambda_{\mathrm{DH}}=1\) in every fraction.

Derivation. If each newly deposited layer takes up trace and carrier in the ratio \(\lambda_{\mathrm{DH}}\) times their ratio in solution, then \(dT/T=\lambda_{\mathrm{DH}}\,dC/C\), where \(T\) and \(C\) are the amounts left in solution. Integrating gives the law.

The 3.5-hour body

Irène Curie and Savić first proposed that the activity was an isotope of thorium. In Berlin, Strassmann looked for thorium and found none. In the spring of 1938 Curie and Savić reported that the substance had properties like those of the rare earths, and in September 1938 that it followed lanthanum, from which they could separate it only partially by fractional crystallisation. They did not conclude that the uranium nucleus had split. In modern terms the activity was a mixture of fission products, mainly isotopes of lanthanum and yttrium with half-lives near 3.5 hours, which explains the partial separation.

Barium

Hahn and Strassmann now repeated their own experiments, looking for products resembling radium, the heavier homologue of barium. They found activities that followed a barium carrier and assigned them to radium. In December 1938 they tried to separate the “radium” from barium by fractional crystallisation of barium salts and failed. Mixed with genuine radium isotopes, the genuine radium separated and the new activities did not. They concluded, against their training as nuclear chemists, that the activities were barium. Their paper appeared on 6 January 1939. Lise Meitner and Otto Frisch interpreted the result as the splitting of the nucleus into two roughly equal parts. Part of the Paris lanthanum activity came from the decay of fission barium.

Exercise 9.1Barium chloride carrier is crystallised from a solution containing a trace activity, in four steps, each time leaving half the remaining carrier in solution. After all four steps \(f_c=1/16\). Find \(f_t\) if \(\lambda_{\mathrm{DH}}=1\), and if \(\lambda_{\mathrm{DH}}=1.5\), a trace that is enriched in the crystals, as radium is in barium salts.

Show solution
\(\lambda_{\mathrm{DH}}=1\): \(f_t=1/16=0.063\), the same as the carrier, in every fraction. \(\lambda_{\mathrm{DH}}=1.5\): \(f_t=(1/16)^{1.5}=1/64=0.016\). Radium is enriched in the crystals, and the specific activity of successive crystal fractions decreases. Hahn and Strassmann saw constant specific activity for the new species, while added radium tracers behaved as expected. The new species was barium.

Exercise 9.2The fission \(n+{}^{235}\mathrm U\to{}^{140}\mathrm{Ba}+{}^{94}\mathrm{Kr}+2n\) conserves charge and nucleon number. Check this, and explain why the fragments are beta-active, starting from the neutron-to-proton ratios of uranium-236 and of stable nuclei near barium.

Show solution
Charge: \(92=56+36\). Nucleons: \(236=140+94+2\). Uranium-236 has \(N/Z=144/92=1.57\). The fragments inherit about the same ratio: barium-140 has \(84/56=1.50\), krypton-94 has \(58/36=1.61\). Stable nuclei near barium have \(N/Z\approx1.4\) and near krypton about 1.3. The fragments are neutron-rich and decay by chains of \(\beta^-\) emission, for example barium-140 (12.8 d) to lanthanum-140 (1.68 d) to stable cerium-140.

Exercise 9.3A fission chain \(A\to B\to\) stable has \(t_{1/2}(A)=12.8\) d and \(t_{1/2}(B)=1.68\) d. After a long irradiation stops, how does the activity of \(B\) change? Explain how a chemist who separated \(B\) without knowing about \(A\) could be misled.

Show solution
Transient equilibrium: after a few days \(A_B/A_A=\lambda_B/(\lambda_B-\lambda_A)=1/(1-1.68/12.8)=1.15\), and both decay with the parent’s 12.8-day half-life. A chemist who separates \(B\) sees it decay with 1.68 days, but if the separation is incomplete, or if \(A\) is carried along, the apparent half-life is wrong. Interlocking chains like this one made the 1937–1938 activities so confusing.
Lesson 10

Radiation dose

For the general reader. Irène Curie was exposed to radiation from the age of eighteen: X-rays in wartime hospitals, then radium, radon and polonium in the laboratory. In 1946 a sealed capsule of polonium burst on her bench. She developed leukaemia and died at 58, as her mother had died of an illness linked to radiation at 66. This lesson introduces the quantities used to measure radiation dose and applies them to the sources she handled.

Goals
  • Define absorbed dose and equivalent dose.
  • Compute the dose rate from an internal alpha emitter.
  • Use the inverse-square law and attenuation for a gamma source.
Dose quantities

The absorbed dose \(D\) is the energy deposited per unit mass, measured in grays: \(1\ \mathrm{Gy}=1\ \mathrm{J/kg}\). The equivalent dose is \(H=w_RD\), in sieverts, with radiation weighting factor \(w_R=1\) for X-rays, gamma rays and electrons and \(w_R=20\) for alpha particles. For an emitter of activity \(A\) uniformly distributed in a mass \(m\) that absorbs all its energy \(\bar E\) per decay,

\[ \dot D=\frac{A\,\bar E}{m}. \]

Alpha particles deposit all their energy within about 40 µm of tissue, so an alpha emitter outside the body is almost harmless while the same emitter inside it is extremely dangerous. Gamma rays penetrate. For a point gamma source the fluence falls as \(1/r^2\) and is attenuated by shielding as \(e^{-\mu x}\). The radiation weighting factors are modern conventions, set by the International Commission on Radiological Protection; in the 1920s and 1930s protection was based on rules of thumb and on avoiding reddening of the skin.

Exercise 10.1One microcurie of polonium-210 is uniformly distributed in 1 kg of tissue. Find the absorbed dose rate and the equivalent dose over 30 days, ignoring decay and excretion.

Show solution
\(A=3.7\times10^4\ \mathrm s^{-1}\), \(\bar E=5.3\ \mathrm{MeV}=8.5\times10^{-13}\) J. \(\dot D=3.1\times10^{-8}\) Gy/s. Over 30 days (\(2.6\times10^6\) s), \(D=0.081\) Gy and \(H=20\times0.081=1.6\) Sv, a severe dose, from less than a nanogram of polonium (\(2.2\times10^{-10}\) g).

Exercise 10.2A gamma source gives a dose rate of 1 mGy/h at 1 m. What is the dose rate at 20 cm, and behind 5 cm of lead at 1 m if \(\mu=0.5\ \mathrm{cm^{-1}}\)? Which is the more effective protection, distance or this shield?

Show solution
At 20 cm: \(1\times(100/20)^2=25\) mGy/h. Behind lead at 1 m: \(e^{-2.5}=0.082\), so 0.08 mGy/h. Moving from 20 cm to 1 m gains a factor 25; the 5 cm of lead gains a further factor 12. Handling sources with long tongs and working behind lead bricks combine both.
Part III · Life and legacy
Lesson 11

Nobel, government and war, 1935–1945

Marie Curie died on 4 July 1934, six months after seeing the first artificial radioelements. In 1935 Irène Joliot-Curie and Frédéric Joliot shared the Nobel Prize in Chemistry for their synthesis of new radioactive elements. Irène and Marie Curie remain the only mother and daughter to have won Nobel Prizes.

In government

In June 1936 Léon Blum formed the Popular Front government and appointed three women as under-secretaries of state, although French women did not yet have the vote. Irène Joliot-Curie became Under-Secretary of State for Scientific Research. She had agreed to serve only for a short time, and she resigned in September after about three months, handing over to Jean Perrin. In that time she helped prepare the organisation of state-funded research that led to the CNRS. She remained committed to women’s rights and to education throughout her life, and was active in the Union des femmes françaises after the war.

In 1937 she became professor at the Faculty of Science in Paris. From then on she and Frédéric worked largely in separate laboratories, she at the Radium Institute and he at the Collège de France and Ivry. The work with Savić described in Lesson 09 belongs to these years. She was made an officer of the Legion of Honour in 1939.

Occupation

Tuberculosis, first contracted in the war years, forced her into sanatoria repeatedly from the 1930s. During the Occupation Frédéric led the Resistance movement Front national. In June 1944, with Frédéric in hiding, she crossed into Switzerland through the Jura with their two children, and returned after the Liberation.

DiscussionIrène Joliot-Curie accepted a government post in 1936 on condition that it be short. What can a working scientist achieve in a brief political appointment, and what are the risks to her science and to her independence?

Lesson 12

The CEA, Orsay and legacy, 1945–1956

In 1946 she became director of the Radium Institute’s Curie laboratory, her mother’s former post. She was one of the commissioners of the new Commissariat à l’énergie atomique for six years and took part in the construction of ZOE, the first French reactor, which went critical in December 1948. Her term ended in 1951, the year after Frédéric’s dismissal from the commission. She was not a member of the Communist Party, but she was active in the peace movement and a member of the World Peace Council.

In 1951 she stood for election to the Académie des sciences, which had rejected her mother in 1911. She was turned down, and said she would stand again at every opportunity in order to make the point that women should be admitted. The Académie elected its first woman, as corresponding member, in 1962: Marguerite Perey, who had discovered francium as Marie Curie’s assistant.

In her last years she planned a large new centre for nuclear physics at Orsay, south of Paris, with a synchrocyclotron of 160 MeV. Frédéric completed it after her death. She was diagnosed with leukaemia and died in Paris on 17 March 1956, aged 58. The government gave her a state funeral. She is buried at Sceaux, beside Frédéric, who died two years later. Her daughter Hélène Langevin-Joliot became a nuclear physicist and her son Pierre Joliot a biologist who studied photosynthesis.

Legacy

  • Artificial radioactivity: with Frédéric Joliot, the discovery that radioactive atoms can be made at will, the basis of radioisotope production, tracers and nuclear medicine.
  • Polonium and alpha physics: her sources and measurements underpinned the experiments that revealed the neutron and positron pair production.
  • The road to fission: the 3.5-hour lanthanum-like activity of 1938 was the first chemical sign that neutrons produce mid-weight elements from uranium.
  • Women in science and public life: one of the first three women in a French government, and a persistent public challenge to the exclusion of women from the Académie des sciences.

Final projectWrite the paper Curie and Savić might have written in October 1938 had they concluded that the uranium nucleus splits. Use Lessons 07–09: state the chemical evidence, compute the energy released in a split into lanthanum and a partner, and say what further test (for example, recoil of fragments) would confirm the claim.

Reference

Connections

People linked to Irène Joliot-Curie across this collection. A name without a link has no page here yet. See Irène Joliot-Curie in the network of all 337 lives →

Teachers and advisors

Influenced by

Influenced

Colleagues and collaborators

  • Jean Perrin
  • Hans von Halban 1935-1937
  • Marguerite Perey
  • Pavle Savić 1937-1939

Rivals

Family

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. Irène Joliot-Curie: Biographical, NobelPrize.org.
  2. Irène Joliot-Curie, Wikipedia; and the French article Irène Joliot-Curie.
  3. Irène Joliot-Curie, research profile, Lindau Nobel Laureate Meetings mediatheque.
  4. Irène Joliot-Curie, biography, Science Story Telling project (Flensburg).
  5. È. Curie, Madame Curie, Gallimard, 1938 (for the family background).

Physics and chemistry

  1. I. Curie, Recherches sur les rayons α du polonium. Oscillation de parcours, vitesse d’émission, pouvoir ionisant, Annales de Physique (1925).
  2. I. Curie, F. Joliot, Un nouveau type de radioactivité, C. R. Acad. Sci. 198, 254 (1934); Artificial production of a new kind of radio-element, Nature 133, 201 (1934).
  3. I. Joliot-Curie, Artificial production of radioactive elements, Nobel lecture, 12 December 1935.
  4. I. Curie, P. Savitch, Sur le radioélément de période 3,5 heures formé dans l’uranium irradié par les neutrons, and Sur la nature du radioélément de période 3,5 heures formé dans l’uranium irradié par les neutrons, C. R. Acad. Sci. 206 (1938).
  5. F. Guerra, M. Leone, N. Robotti, Neutron physics in the early 1930s, arXiv:physics/0510044.
  6. Discovery of nuclear fission, Wikipedia (with references to the Curie–Savić and Hahn–Strassmann papers).
  7. G. Gamow, Zur Quantentheorie des Atomkernes, Z. Physik 51, 204 (1928).
  8. K. S. Krane, Introductory Nuclear Physics, Wiley, 1988; W. R. Leo, Techniques for Nuclear and Particle Physics Experiments, Springer, 1994.