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James Chadwick

Sir James Chadwick

A course in two strands: the life of the cotton spinner’s son from Cheshire who survived four years in a German internment camp and became Rutherford’s deputy, and the physics he settled with careful counting: the continuous beta spectrum, the charge of the nucleus, the strong force and the neutron.

Born 20 October 1891, Bollington, Cheshire Died 24 July 1974, Cambridge Nobel Prize Physics, 1935 Level advanced undergraduate → graduate
Largest share of a neutron’s energy that a head-on elastic collision can give a nucleus of mass number A: 4A/(1+A)². Hydrogen takes it all; nitrogen takes about a quarter. Comparing such recoils weighed the neutron.
Lesson 00

How to use this course

James Chadwick is remembered for one discovery, the neutron, made in about three weeks in early 1932. That discovery rested on twenty years of patient work on the problems at the edge of Ernest Rutherford’s programme. Chadwick showed that beta rays come out with a continuous range of energies, measured the charge of the nucleus directly, found the first signs of a nuclear force much stronger than electricity, and helped Rutherford break apart light nuclei. After 1932 he weighed the neutron, built a cyclotron at Liverpool, wrote the final draft of the British report that set the atomic bomb in motion, and led the British scientists in the Manhattan Project.

The course follows the physics in the order Chadwick met it. Each experiment is analysed with the conservation laws and the scattering formula, so that the reader can reproduce the numbers on which his conclusions rested.

Prerequisites for Part II
  • Classical mechanics: elastic collisions in one and two dimensions, centre-of-mass frame.
  • The Rutherford scattering formula. The Rutherford course derives it in its Lesson 07.
  • Special relativity: energy and momentum of a photon, \(E=mc^2\) and reaction Q-values.
  • For Lessons 06 and 10, elementary quantum ideas (de Broglie wavelength) and the diffusion equation.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Lesson 08, the neutron, is the centre of the course. Masses are atomic masses in unified units u unless stated otherwise, with \(1\,\mathrm u\,c^2=931.5\) MeV.

Reference

Timeline

1891Born 20 October in Bollington, Cheshire, son of John Joseph Chadwick, a cotton spinner, and Anne Mary Knowles.
1908Enters the Victoria University of Manchester, meaning to read mathematics; enrols in physics by mistake and stays.
1911First-class honours in physics. Begins research under Rutherford.
1912First paper, with Rutherford, on comparing radium standards. Sole-author work on gamma-ray absorption.
1913MSc. 1851 Exhibition scholarship takes him to Hans Geiger at the Physikalisch-Technische Reichsanstalt, Berlin.
1914With Geiger’s counter, shows that the beta-ray spectrum is continuous. Interned as an enemy alien in November.
1914–18Held at Ruhleben camp near Berlin; research with Charles Drummond Ellis.
1919Returns to Manchester, then follows Rutherford to the Cavendish Laboratory, Cambridge.
1920Measures nuclear charges of platinum, silver and copper by alpha scattering: they equal the atomic numbers.
1921With Étienne Biéler, finds that alpha–hydrogen collisions violate Coulomb’s law at short range. PhD (June); Fellow of Gonville and Caius College.
1921–24With Rutherford, proton emission from boron, fluorine, sodium, aluminium, phosphorus and other light elements.
1923Assistant Director of Research at the Cavendish.
1925Marries Aileen Stewart-Brown; Pyotr Kapitza is best man. Twin daughters born 1927.
1927Fellow of the Royal Society.
1930Radiations from Radioactive Substances, with Rutherford and Ellis.
1932Discovers the neutron: letter to Nature in February, full paper in May. Hughes Medal.
1934With Maurice Goldhaber, splits the deuteron with gamma rays and weighs the neutron precisely.
1935Nobel Prize in Physics. Lyon Jones Professor of Physics, University of Liverpool, from 1 October.
1939The Liverpool cyclotron runs from July.
1941Writes the final draft of the MAUD Report on the feasibility of a uranium bomb.
1943–46Head of the British Mission to the Manhattan Project; at Los Alamos from early 1944.
1945Knighted (New Year Honours). Witnesses the Trinity test, 16 July.
1948Master of Gonville and Caius College, Cambridge, until 1958.
1950–51Copley Medal (1950); Franklin Medal (1951).
1970Companion of Honour.
1974Dies in his sleep in Cambridge, 24 July, aged 82.
Part I · Formation
Lesson 01

Bollington and Manchester, 1891–1913

James Chadwick was born on 20 October 1891 in Bollington, a mill town in Cheshire on the edge of the Pennines. His father, John Joseph Chadwick, was a cotton spinner and his mother, Anne Mary Knowles, had been a domestic servant. He attended Bollington Cross primary school before the family moved to Manchester.

He won a scholarship to Manchester Grammar School, but the family could not meet the remaining costs. He went instead to the Central Grammar School for Boys in Manchester, a municipal school with good science teaching. At sixteen he won two scholarships to the Victoria University of Manchester and entered in 1908.

Physics by mistake

Chadwick meant to read mathematics. By a mistake at enrolment he was registered for physics instead, did not correct it, and found the subject interesting enough to stay. Ernest Rutherford had arrived as Langworthy Professor in 1907. Chadwick graduated with first-class honours in 1911 and began research under Rutherford, who set him to compare radium standards. Their joint paper appeared in 1912. He also worked alone on the absorption of gamma rays in gases. In that paper he argued that the ionization of the upper atmosphere could not come only from radioactivity in the ground, a conclusion that pointed towards cosmic rays.

He took the MSc in 1913 and won an 1851 Exhibition scholarship for research abroad. Most holders went to Cambridge. Chadwick chose Berlin, where Hans Geiger, Rutherford’s former assistant, had just set up a radioactivity laboratory at the Physikalisch-Technische Reichsanstalt.

Lesson 02

Berlin and Ruhleben, 1913–1918

In Berlin Chadwick learned to use the counters Geiger was developing and worked alongside Walther Bothe, then Geiger’s assistant. Geiger gave him a problem that Hahn, Meitner and Otto von Baeyer had studied with photographic plates: the energy spectrum of the beta rays from radium B and C. Their plates showed sharp lines. Chadwick put a counter at the focus of a magnetic spectrometer and counted electrons point by point. He found a few weak lines sitting on a strong continuous distribution. The paper, published in 1914, is the subject of Lesson 04.

Internment

When war broke out in August 1914 Chadwick was still in Berlin. In November he was arrested as an enemy alien and sent to the civilian internment camp at Ruhleben, a racecourse on the western edge of Berlin, where several thousand British civilian men were held. He stayed there for four years. Conditions were harsh, and food was scarce by 1917; Chadwick’s digestion suffered for the rest of his life.

The prisoners organised their own lectures and courses. Chadwick taught, and with help from sympathetic German physicists he obtained materials for a small laboratory in a stable. With Charles Drummond Ellis, a young army cadet who had been visiting Germany when war came, he studied the ionization of phosphorus vapour and the photochemical reaction between carbon monoxide and chlorine. Ellis decided in the camp to become a physicist. He later became one of Chadwick’s closest collaborators at Cambridge.

After the Armistice of November 1918 the prisoners were released. Chadwick returned to Manchester in poor health and with little money.

Lesson 03

The Cavendish apprenticeship, 1919–1927

Rutherford gave him part-time teaching at Manchester, and Chadwick began the measurement of nuclear charge described in Lesson 05. In 1919 Rutherford succeeded J. J. Thomson as Cavendish Professor at Cambridge and took Chadwick with him. Chadwick held the Clerk Maxwell Studentship and enrolled at Gonville and Caius College for the PhD, which he received in June 1921. The first half of his thesis dealt with atomic numbers and the second with the forces inside the nucleus. In November 1921 he was elected a Fellow of Caius.

Rutherford’s deputy

From 1923 Chadwick was Assistant Director of Research at the Cavendish. In practice he ran the laboratory day to day. He chose and placed research students, found them apparatus, read their papers and kept Rutherford’s programme of nuclear work moving. Students recruited in these years included John Cockcroft, Norman Feather and Mark Oliphant. His own research with Rutherford, on the disintegration of light elements by alpha particles, is the subject of Lesson 07.

In August 1925 he married Aileen Stewart-Brown, the daughter of a Liverpool stockbroker. Pyotr Kapitza was best man. Their twin daughters, Joanna and Judith, were born in February 1927, the year Chadwick was elected a Fellow of the Royal Society. In 1930 he published, with Rutherford and Ellis, Radiations from Radioactive Substances, the standard reference of the field for a generation.

Throughout the 1920s he and Rutherford talked about the neutral particle Rutherford had proposed in his 1920 Bakerian Lecture. Chadwick tried several times to produce it, by passing electric discharges through hydrogen and by looking for unusual radiation from alpha bombardment. None of the attempts worked, but they prepared him to recognise the particle at once in 1932.

Part II · The physics
Lesson 04

The continuous beta spectrum

For the general reader. When a nucleus emits a beta ray, an electron, the obvious guess is that every electron from the same kind of nucleus carries the same energy, as alpha particles do. Chadwick showed in 1914 that this is false: the electrons come out with every energy from zero up to a maximum. The puzzle lasted sixteen years and was solved only when Wolfgang Pauli proposed that an unseen particle, the neutrino, carries off the missing energy.

Goals
  • Show that a two-body decay gives a single electron energy.
  • Understand why photographic plates exaggerated lines and why a counter did not.
  • Explain the lines as internal conversion and the continuum as a three-body decay.

Two-body kinematics

Two-body decay at rest

If a particle of mass \(M\) at rest decays into two particles of masses \(m_1\) and \(m_2\), the energy of particle 1 is fixed:

\[ E_1=\frac{M^2+m_1^2-m_2^2}{2M}\,c^2 . \]

Every decay of the same kind gives the same \(E_1\).

Derivation. The two products have equal and opposite momenta \(p\). Write \(E_2=Mc^2-E_1\) and square: \(E_2^2=p^2c^2+m_2^2c^4=E_1^2-m_1^2c^4+m_2^2c^4\). Substituting \(E_2=Mc^2-E_1\) gives \(M^2c^4-2Mc^2E_1=-m_1^2c^4+m_2^2c^4\), which is the formula. With three products the momenta need only sum to zero, so the energy is shared in a continuous way.

Plates against counters

In a magnetic spectrometer electrons of momentum \(p\) bend on a circle of radius \(p/(eB)\) and land at a definite position. A photographic plate records all positions at once. Its blackening is not proportional to the number of electrons, and the eye picks out sharp lines against a diffuse background. Chadwick replaced the plate by a counter at a fixed position and stepped the magnetic field, so that each reading was a count of electrons in a known momentum interval. The count rate rose and fell smoothly over a wide range, with small line peaks on top.

The lines were real. Ellis showed in the early 1920s that they are internal conversion electrons: a gamma ray from the nucleus gives its energy to an inner atomic electron, which leaves with energy \(E_\gamma-B\), where \(B\) is that electron’s binding energy. These electrons do have sharp energies. The continuum is the beta decay proper.

Where does the energy go?

Ellis and William Wooster settled the question in 1927 with a calorimeter. They absorbed all the beta rays of radium E in a thick-walled vessel and measured the heat. The heat per decay matched the average energy of the continuous spectrum, not its maximum. So the missing energy was not hidden in some gamma radiation. Niels Bohr was prepared to give up exact energy conservation in single nuclear events. In December 1930 Pauli proposed instead a light neutral particle emitted with the electron. Enrico Fermi built the theory of beta decay on this idea in 1933–34, and the neutrino was detected directly in 1956.

Exercise 4.1Show that for a heavy nucleus emitting an electron in a two-body decay, \(M\approx M_2\gg m_e\), the electron kinetic energy is very nearly the full energy release \(Q=(M-M_2-m_e)c^2\). What fraction goes to the recoiling nucleus?

Show solution
Write \(M=M_2+m_e+Q/c^2\). The momenta are equal, so the recoil kinetic energy is about \(p^2/(2M_2)\), with \(p\) of order \(\sqrt{Q^2+2Qm_ec^2}/c\). For \(Q\approx1\) MeV and \(M_2c^2\approx2\times10^5\) MeV this is a few eV, a fraction of order \(10^{-5}\). Practically all of \(Q\) goes to the electron, so a two-body decay would give an electron line at \(Q\).

Exercise 4.2Radium B (lead-214) decays to bismuth-214, which emits a gamma ray of 351.9 keV. The K-shell binding energy of bismuth is 90.5 keV and the L-shell binding is about 16 keV. At what kinetic energies do the conversion lines appear?

Show solution
K line: \(351.9-90.5=261.4\) keV. L line: about \(351.9-16=336\) keV. The separation between such lines, equal to a difference of X-ray binding energies, identified the element in which the conversion happened. Ellis used this to show that the gamma rays are emitted after the beta decay, from the daughter nucleus.

Exercise 4.3Model a beta spectrum by the kinetic-energy distribution \(N(T)\propto T(T_{\max}-T)\) on \(0\le T\le T_{\max}\). Find the mean energy. If the calorimeter had found \(T_{\max}\) per decay, what would that have meant?

Show solution
\(\langle T\rangle=\int_0^{T_{\max}}T^2(T_{\max}-T)\,dT\big/\int_0^{T_{\max}}T(T_{\max}-T)\,dT=(T_{\max}^4/12)/(T_{\max}^3/6)=T_{\max}/2\). A calorimeter reading of \(T_{\max}\) per decay would have meant every nucleus releases the same energy, with the deficit carried by radiation that the vessel absorbed. Ellis and Wooster found about the mean instead, so the energy escaped the calorimeter entirely, as a neutrino does.
Lesson 05

Weighing the nuclear charge

For the general reader. By 1913 physicists suspected that the charge of the nucleus, in units of the electron charge, equals the element’s place in the periodic table. Henry Moseley’s X-ray work supported this, but only indirectly. In 1919–20 Chadwick measured the charge directly, by counting how many alpha particles a thin metal foil turns aside. For platinum, silver and copper the answer matched the atomic number to within about one and a half per cent.

Goals
  • Invert the Rutherford formula to obtain \(Z\) from a counting experiment.
  • Understand the annular-foil geometry that made an absolute measurement possible.
  • Estimate how many counts a 1% measurement of \(Z\) requires.

The idea

The Rutherford formula gives the fraction of incident alpha particles scattered into a solid angle \(d\Omega\) at angle \(\theta\) by a foil with \(n\) atoms per unit volume and thickness \(t\):

\[ \frac{dN}{N_0}=n t\left(\frac{zZe^2}{4\pi\varepsilon_0\,4E}\right)^2\frac{d\Omega}{\sin^4(\theta/2)} . \]

Geiger and Marsden had tested the dependence on angle, thickness and energy in 1913, but the absolute number was hard to measure, because the incident and scattered fluxes differ by many orders of magnitude and were counted with different geometries.

Charge from an absolute scattering measurement

If the ratio \(f=dN/N_0\) is measured for a known foil, energy and solid angle, then

\[ Z=\frac{4\pi\varepsilon_0\,4E}{z e^2}\,\sin^2\!\frac{\theta}{2}\,\sqrt{\frac{f}{n t\,d\Omega}} . \]

A relative error \(\delta f/f\) in the measured ratio gives a relative error \(\tfrac12\,\delta f/f\) in \(Z\).

The annular foil

Chadwick placed the source and the zinc sulphide screen on a common axis and the scattering foil in the shape of a ring between them, centred on the axis. Every particle that reached the screen by scattering in the ring was deflected through nearly the same angle. Removing the ring and putting a smaller opening on the axis let him count the direct beam with the same screen and the same observer. Both counts were then made under like conditions, and their ratio gave \(f\) directly. The geometry also collected particles from a whole ring of foil, which raised the scattered count to a level that could be measured with good statistics.

His results agreed with the atomic numbers of platinum (78), silver (47) and copper (29) within the experimental error, about 1 to 1.5%. Rutherford and others had assumed the equality since 1913. Chadwick’s experiment measured it directly.

Exercise 5.1Counts follow Poisson statistics, so \(N\) counts have relative error \(1/\sqrt N\). How many scattered particles must be counted to measure \(Z\) to 1%, ignoring all other errors? If the direct beam is also counted with the same number of particles, how does the answer change?

Show solution
\(\delta Z/Z=\tfrac12\,\delta f/f=1/(2\sqrt N)=0.01\) gives \(N=2500\). With two independent counts of \(N\) each, \(\delta f/f=\sqrt{2/N}\), so \(\delta Z/Z=\sqrt{2/N}/2=0.01\) gives \(N=5000\) of each. Chadwick and his helpers counted scintillations by eye in a dark room, a minute at a time, so this was weeks of work.

Exercise 5.2Foils of silver and copper contain the same number of atoms per unit area and are used in the same apparatus. What ratio of scattered counts do you expect? If the measured ratio carries a 3% uncertainty, what range of \(Z_{\mathrm{Ag}}/Z_{\mathrm{Cu}}\) does that allow?

Show solution
\((47/29)^2\approx2.63\). A 3% uncertainty in the count ratio gives 1.5% in the charge ratio, so \(Z_{\mathrm{Ag}}/Z_{\mathrm{Cu}}=1.62\pm0.02\), excluding the next-nearest integer ratios such as \(46/29\) or \(47/30\) only marginally. Absolute measurements on each element were needed to fix the integers.
Lesson 06

Anomalous scattering and the strong force

For the general reader. Rutherford’s scattering law assumes that only the electric force acts between an alpha particle and a nucleus. In 1921 Chadwick and Étienne Biéler fired fast alpha particles at hydrogen and found far more hard collisions than electricity alone allows, in some cases a hundred times more. They concluded that at very short distances a new force acts, much stronger than the electric repulsion. This was the first quantitative glimpse of the strong nuclear force.

Goals
  • Work out the kinematics of alpha–proton collisions.
  • Find the distance of closest approach in the centre-of-mass frame.
  • See why a failure of the Coulomb law at a few femtometres means a new force.

Kinematics

An alpha particle (mass \(m\approx4\)) hits a proton (mass \(M\approx1\)) at rest. In an elastic collision the proton’s velocity at recoil angle \(\phi\) from the beam is \(V=\tfrac{2m}{m+M}u\cos\phi\). The fastest protons go forward with \(V=1.6\,u\) and kinetic energy \(\tfrac{4mM}{(m+M)^2}T=0.64\,T\). Because the alpha particle is the heavier body, it can be deflected through at most \(\arcsin(M/m)\approx14.5^\circ\). Ernest Marsden had already noticed in 1914 that such protons travel about four times as far in air as the alpha particles that hit them.

Closest approach in the centre-of-mass frame

For a projectile of kinetic energy \(T\) and mass \(m\) striking a target of mass \(M\) at rest, the centre-of-mass energy is \(E_{\mathrm{cm}}=TM/(m+M)\). In a head-on collision under the Coulomb force alone, the separation shrinks to

\[ d=\frac{zZe^2}{4\pi\varepsilon_0\,E_{\mathrm{cm}}} . \]

The Rutherford formula holds in the centre-of-mass frame provided the force is Coulombic at all separations reached.

For alpha particles on protons, \(E_{\mathrm{cm}}=T/5\). With 7.7 MeV alpha particles, \(d\approx1.9\) fm. That is smaller than the alpha particle itself. Charles Galton Darwin had computed in 1914 how many protons of each energy and angle the Coulomb law predicts.

The experiment

Chadwick and Biéler sent alpha particles of various ranges through a thin film of paraffin wax or hydrogen gas and counted the recoil protons as a function of angle and alpha energy. For slow alpha particles the counts agreed with Darwin’s formula. For the fastest ones the number of protons projected was much larger, by a factor of order 100 in some conditions, and the angular distribution was wrong. The Coulomb law was failing at distances of a few times \(10^{-13}\) cm.

They described the alpha particle as a body of definite size and shape, and the interaction near contact as a field of force of very great intensity. The language was cautious, but the conclusion was the one that matters: something stronger than electricity holds nuclei together and acts only at nuclear distances. A real theory needed quantum mechanics and, from 1935, Hideki Yukawa’s meson.

Exercise 6.1Check the numbers: find \(E_{\mathrm{cm}}\) and \(d\) for 7.7 MeV alpha particles on protons, using \(e^2/(4\pi\varepsilon_0)=1.44\) MeV fm. Compare with the sum of radii, taking \(R_\alpha\approx1.7\) fm and \(R_p\approx0.85\) fm.

Show solution
\(E_{\mathrm{cm}}=7.7/5=1.54\) MeV and \(d=2\times1.44/1.54\approx1.87\) fm. The sum of the radii is about 2.5 fm, so the particles overlap in head-on collisions. Departures from the Coulomb law are therefore expected. For gold, by contrast, \(d\approx29\) fm, well outside the nucleus.

Exercise 6.2Show that an alpha particle colliding elastically with a stationary proton cannot be deflected by more than \(\arcsin(1/4)\). Hint: in the centre-of-mass frame the alpha speed is \(v'=u/5\) and the centre of mass moves at \(V_c=4u/5\).

Show solution
The lab velocity is \(\vec V_c+\vec v'\), where \(\vec v'\) has fixed length \(u/5\) and any direction. The tip of the lab velocity lies on a circle of radius \(u/5\) centred at distance \(4u/5\) along the beam. The largest angle is reached when the lab velocity is tangent to this circle: \(\sin\theta_{\max}=(u/5)/(4u/5)=1/4\), so \(\theta_{\max}\approx14.5^\circ\). Large-angle alpha scattering from hydrogen is impossible, whatever the force.
Lesson 07

Disintegrating light elements

For the general reader. In 1919 Rutherford found that alpha particles knock hydrogen nuclei out of nitrogen. From 1920 Chadwick and he set out to find which other elements could be broken up in the same way. They found about a dozen, all light, and learned to tell real disintegration protons from the much more common protons knocked out of hydrogen impurities. A rival group in Vienna claimed many more disintegrations. Chadwick’s visit there in 1927 helped show that their counting was unreliable.

Goals
  • Use the Q-equation to relate proton energy, angle and reaction energy.
  • Understand how the range and direction of protons distinguish disintegration from contamination.
  • See why only light elements could be disintegrated with natural alpha particles.

The method

A radioactive source, usually radium C′ with 7.7 MeV alpha particles, irradiated a thin target. Absorbing foils in front of a zinc sulphide screen stopped all alpha particles and all slower protons, so only protons with a long range could produce scintillations. Hydrogen is present as a contaminant almost everywhere, in water and grease, and elastic recoil protons from it can reach well beyond the alpha range. Rutherford and Chadwick dealt with this in two ways. They looked for protons with ranges longer than any recoil proton could have, and they looked for protons emitted backwards, which elastic recoils from hydrogen cannot produce.

Between 1921 and 1924 they found disintegration protons from boron, nitrogen, fluorine, sodium, aluminium and phosphorus, and later from neon, magnesium, silicon, sulphur, chlorine, argon and potassium. Aluminium gave protons with ranges far beyond those of the incident alpha particles. Chadwick and Ellis confirmed by magnetic deflection that the particles were protons.

The Q-equation (non-relativistic)

For a reaction \(a+X\to Y+b\) with \(X\) at rest, projectile energy \(T_a\), and \(b\) emitted at angle \(\theta\) with energy \(T_b\),

\[ Q=T_b\left(1+\frac{m_b}{m_Y}\right)-T_a\left(1-\frac{m_a}{m_Y}\right)-\frac{2}{m_Y}\sqrt{m_am_bT_aT_b}\,\cos\theta . \]

Derivation. Momentum conservation gives \(\vec p_Y=\vec p_a-\vec p_b\), so \(p_Y^2=p_a^2+p_b^2-2p_ap_b\cos\theta\). Write \(p^2=2mT\) for each particle and use \(Q=T_b+T_Y-T_a\) with \(T_Y=p_Y^2/(2m_Y)\). Collecting terms gives the formula. At \(\theta=90^\circ\) the cross term vanishes, which is why measurements at right angles were the simplest to interpret.

Cambridge and Vienna

From 1923 Hans Pettersson and Gerhard Kirsch at the Institute for Radium Research in Vienna reported disintegration protons from many elements, including carbon and oxygen, where Cambridge found none. The dispute lasted several years. In December 1927 Chadwick visited Vienna and watched their observers count. He concluded that the scintillation counts depended on the observers’ expectations. The Cambridge results stood. The lesson that counting must be checked by blind tests and by independent methods shaped Chadwick’s caution in 1932, and it was one reason the Cavendish moved to electrical counters with valve amplifiers.

Exercise 7.1Using \(m({}^{27}\mathrm{Al})=26.981538\), \(m({}^{4}\mathrm{He})=4.002603\), \(m({}^{30}\mathrm{Si})=29.973770\), \(m({}^{1}\mathrm{H})=1.007825\) u, find \(Q\) for \({}^{27}\mathrm{Al}(\alpha,p){}^{30}\mathrm{Si}\). Then find the proton energy at \(90^\circ\) for 7.7 MeV alpha particles, leaving the residual nucleus in its ground state.

Show solution
Initial mass 30.984141 u, final 30.981595 u, so \(Q=0.002546\times931.5\approx2.37\) MeV. At \(90^\circ\): \(T_p(1+1/30)=Q+T_\alpha(1-4/30)=2.37+6.67=9.04\), giving \(T_p\approx8.75\) MeV. This exceeds the alpha energy itself, which is why aluminium protons outranged the incident alpha particles by a wide margin.

Exercise 7.2Show that in an elastic collision with a particle at rest, a target of equal or smaller mass can never recoil backwards. Why does this make backward protons a clean signature of disintegration?

Show solution
The target receives an impulse along the line of centres at contact, which always has a positive component along the incident direction. More formally, the recoil velocity is \(\tfrac{2m}{m+M}u\cos\phi\) with \(0\le\phi\le90^\circ\) for any masses. Hydrogen contaminants can only throw protons forwards, so a proton at more than \(90^\circ\) must come from a reaction, which releases energy and can send the proton in any direction.

Exercise 7.3Estimate the Coulomb barrier for an alpha particle on aluminium and on copper, using \(R=1.2(A_1^{1/3}+A_2^{1/3})\) fm. Why did natural alpha particles disintegrate nothing heavier than potassium?

Show solution
Aluminium: \(R\approx1.2(1.59+3.00)\approx5.5\) fm and \(V\approx2\cdot13\cdot1.44/5.5\approx6.8\) MeV. Copper (\(Z=29\), \(A=63\)): \(R\approx1.2(1.59+3.98)\approx6.7\) fm and \(V\approx12.5\) MeV. Natural alpha particles have at most about 8.8 MeV, so for heavier targets they cannot reach the nucleus classically and tunnelling is weak. A neutral projectile would avoid the barrier altogether.
Lesson 08

The neutron, 1932

For the general reader. In 1930 two German physicists found that beryllium struck by alpha particles gives off a very penetrating radiation, which they took for high-energy gamma rays. In January 1932 Irène Joliot-Curie and Frédéric Joliot showed that this radiation knocks protons out of paraffin wax with great force. Chadwick saw that gamma rays could not do this. In about three weeks he showed that the radiation consists of neutral particles with nearly the mass of a proton: the neutron.

Goals
  • Compute the gamma-ray energy needed to give a proton 5.7 MeV by Compton scattering.
  • Show that the same gamma rays cannot explain the nitrogen recoils.
  • Check that the neutron hypothesis fits all recoils and the energy balance of the beryllium reaction.

The puzzle

Walther Bothe and Herbert Becker reported in 1930 that light elements, beryllium above all, emit a penetrating radiation under polonium alpha bombardment. Its absorption in lead suggested gamma rays of a few MeV. In January 1932 the Joliot-Curies reported to the Paris Academy that the radiation ejects protons from hydrogen-containing materials, with energies up to about 5 MeV. They interpreted this as a Compton effect on protons.

Chadwick used a polonium source deposited on a silver disc, placed beside a disc of pure beryllium in an evacuated vessel. The radiation passed into a shallow ionization chamber connected to a valve amplifier and an oscillograph. Each recoil nucleus in the chamber produced a pulse whose height measured its ionization. He put sheets of paraffin and other materials in front of the chamber and filled it with different gases. His letter, Possible existence of a neutron, was sent to Nature on 17 February 1932. The full paper, The existence of a neutron, was received by the Royal Society on 10 May.

Maximum energy transfer by a photon

A photon of energy \(E\) scattering from a free particle of mass \(M\) at rest gives it at most the kinetic energy

\[ T_{\max}=\frac{2E^2}{Mc^2+2E}, \]

reached when the photon is scattered straight back.

Derivation. The Compton formula for a target of mass \(M\) gives the scattered photon energy \(E'=E/(1+(E/Mc^2)(1-\cos\vartheta))\). At \(\vartheta=\pi\), \(E'=E/(1+2E/Mc^2)\), and the target takes \(T=E-E'=2E^2/(Mc^2+2E)\).

Chadwick’s argument

The fastest protons from paraffin had a range just over 40 cm of air, a speed of \(3.3\times10^9\) cm/s and an energy of about 5.7 MeV. Solving \(T_{\max}=5.7\) MeV with \(Mc^2=938\) MeV gives \(E\approx55\) MeV. Nitrogen recoils in an expansion chamber, photographed with Norman Feather, had energies up to about 1.2 MeV. Gamma rays would need about 90 MeV for that. Two different targets required two different photon energies, both far above anything known from nuclear processes. And no known reaction of beryllium with a 5.3 MeV alpha particle could supply photons of 55 or 90 MeV.

Neutron hypothesis (Chadwick, 1932)

The radiation consists of neutral particles of mass close to the proton mass, produced by the reaction

\[ {}^{9}_{4}\mathrm{Be}+{}^{4}_{2}\mathrm{He}\longrightarrow{}^{12}_{6}\mathrm C+{}^{1}_{0}n . \]

A neutron of speed \(u\) gives a nucleus of mass number \(A\) a maximum recoil energy \(\tfrac{4A}{(1+A)^2}T_n\).

With neutrons of speed \(3.3\times10^9\) cm/s and mass 1, a nitrogen nucleus can receive up to about 1.4 MeV, matching the observed recoils. Comparing the maximum hydrogen and nitrogen speeds gave a mass of about 1.15, close to 1 within the errors (the calculation is Exercise 9.1 of the Rutherford course). Chadwick also measured recoils from helium, lithium, beryllium, carbon, air and argon, and all fitted. The neutrons lost very little energy passing through lead or brass, as a neutral particle should, and their collision radius in lead was of nuclear size, a few times \(10^{-13}\) cm.

Chadwick first pictured the neutron as Rutherford had, a proton and an electron in close combination. Within months Dmitri Ivanenko and Werner Heisenberg treated it as an elementary constituent of the nucleus. Robert Bacher and Edward Condon pointed out in 1932 that a neutron of spin \(\tfrac12\) resolves the anomaly of the nitrogen-14 spin.

Exercise 8.1Solve \(2E^2/(Mc^2+2E)=5.7\) MeV for the photon energy, with \(Mc^2=938.3\) MeV. Then find the largest recoil energy that photon could give a nitrogen-14 nucleus (\(Mc^2\approx13\,040\) MeV).

Show solution
\(2E^2-11.4E-5348=0\) gives \(E=\big(11.4+\sqrt{11.4^2+8\cdot5348}\big)/4\approx54.6\) MeV. For nitrogen, \(T_{\max}=2(54.6)^2/(13\,040+109)\approx0.45\) MeV. The observed nitrogen recoils reached about 1.2 MeV, nearly three times more. The gamma-ray hypothesis fails.

Exercise 8.2Using \(m({}^9\mathrm{Be})=9.012183\), \(m({}^4\mathrm{He})=4.002603\), \(m({}^{12}\mathrm C)=12\) and \(m(n)=1.008665\) u, find \(Q\) for the beryllium reaction. With polonium alpha particles of 5.3 MeV, estimate the largest neutron energy, emitted forward. Compare with the 5.7 MeV recoil protons.

Show solution
\(Q=(13.014786-13.008665)\times931.5\approx5.70\) MeV. For forward emission, the Q-equation with \(\theta=0\) and masses 4, 1, 12 gives \(T_n\left(1+\tfrac1{12}\right)-\tfrac{2}{12}\sqrt{4\cdot5.3\,T_n}=5.70+5.3\left(1-\tfrac4{12}\right)=9.23\). Writing \(x=\sqrt{T_n}\): \(1.0833x^2-0.767x-9.23=0\), so \(x\approx3.29\) and \(T_n\approx10.8\) MeV. Neutrons up to about 11 MeV are possible when the carbon is left in its ground state. Many reactions leave carbon-12 in its 4.4 MeV excited state, giving lower energies. A 5.7 MeV recoil proton needs only a 5.7 MeV neutron, so the energy balance works easily, unlike the 55 MeV photon.

Exercise 8.3For neutrons of 5.7 MeV, find the maximum recoil energies of nitrogen-14 and argon-40. Chadwick estimated about 1.4 MeV and 0.54 MeV. Why could a gamma ray of fixed energy never give the same pattern of recoils across targets?

Show solution
\(4A/(1+A)^2\times5.7\): nitrogen \(56/225\times5.7\approx1.42\) MeV; argon \(160/1681\times5.7\approx0.54\) MeV. For a massive projectile the transferred fraction falls roughly as \(4/A\). For a photon of energy much less than \(Mc^2\), \(T_{\max}\approx2E^2/(Mc^2)\) also falls as \(1/A\), but its scale is fixed by \(E^2/Mc^2\), which is tiny unless \(E\) is enormous. A projectile with mass comparable to the targets is the only way to give heavy recoils a sizeable share of a few MeV.
Lesson 09

The neutron’s mass and the deuteron

For the general reader. Was the neutron a proton and an electron stuck together, or a new particle? The answer depends on its mass to one part in a thousand. Chadwick first estimated it from a reaction in boron. In 1934, with his student Maurice Goldhaber, he split the nucleus of heavy hydrogen with gamma rays and obtained a much better value. The neutron came out heavier than a hydrogen atom. A bound proton and electron would have to be lighter, so the neutron had to be a particle in its own right.

Goals
  • Reproduce Chadwick’s 1932 mass estimate from the boron reaction.
  • Derive the neutron mass from deuteron photodisintegration.
  • See why \(m_n>m_{\mathrm H}\) rules out a proton–electron compound and allows free neutron decay.

The boron estimate

In his May 1932 paper Chadwick used the reaction \({}^{11}\mathrm B+{}^{4}\mathrm{He}\to{}^{14}\mathrm N+n\). Energy conservation gives

\[ m_n=m({}^{11}\mathrm B)+m({}^{4}\mathrm{He})+T_\alpha-m({}^{14}\mathrm N)-T_{\mathrm N}-T_n , \]

with kinetic energies written in mass units. With the masses then available on the oxygen-16 scale, and the measured maximum neutron energy, he found \(m_n\approx1.0067\), probably between 1.005 and 1.008. The hydrogen atom was 1.0078 on the same scale, so the neutron appeared slightly lighter, with a binding energy of 1 to 2 MeV if it were a proton–electron compound. The error came from the mass of nitrogen-14, then poorly known.

The photodisintegration of the deuteron

Goldhaber, a research student from Germany at the Cavendish, suggested in 1934 that the gamma rays of thorium C″, of energy 2.62 MeV, might split the deuteron. Chadwick and Goldhaber filled an ionization chamber with heavy hydrogen, irradiated it, and saw the protons. Their letter to Nature, A nuclear photo-effect, appeared in August 1934.

Photodisintegration of the deuteron

For \(\gamma+\mathrm d\to p+n\) with photon energy \(E_\gamma\) above the binding energy \(B\), the proton and neutron share the kinetic energy \(E_\gamma-B\) (minus a small recoil correction), nearly equally because their masses are nearly equal. With atomic masses,

\[ m_n=m(\mathrm D)-m({}^1\mathrm H)+\frac{B}{c^2},\qquad B\approx E_\gamma-2T_p . \]

From the measured proton energy and the known gamma energy they obtained the deuteron binding energy and a neutron mass of 1.0084 to 1.0090, heavier than the hydrogen atom. The modern values are \(B=2.2246\) MeV and \(m_n=1.008665\) u. It followed that a free neutron has enough mass to decay into a proton, an electron and a neutrino, and so should be radioactive. Free neutron decay was observed after the Second World War. Its mean life is about 880 seconds.

Exercise 9.1Reproduce the boron estimate with Chadwick’s numbers: \(m({}^{11}\mathrm B)=11.00825\), \(m({}^4\mathrm{He})=4.00106\), \(m({}^{14}\mathrm N)=14.0042\), and kinetic energies \(T_\alpha=0.00565\), \(T_n=0.0035\), \(T_{\mathrm N}=0.00061\) mass units.

Show solution
\(m_n=11.00825+4.00106+0.00565-14.0042-0.00061-0.0035=1.00665\approx1.0067\). An error of 0.003 in the nitrogen mass, which was about its uncertainty, moves \(m_n\) by the same amount, enough to change the sign of the comparison with hydrogen.

Exercise 9.2Using \(m(\mathrm D)=2.014102\), \(m({}^1\mathrm H)=1.007825\), \(m(n)=1.008665\) u, find the deuteron binding energy. With 2.614 MeV photons, what kinetic energy does each nucleon receive?

Show solution
\(B=(1.007825+1.008665-2.014102)\times931.5\approx0.002388\times931.5\approx2.224\) MeV. Each nucleon gets about \((2.614-2.224)/2\approx0.19\) MeV. Such slow protons give short tracks and small pulses, which made the measurement delicate.

Exercise 9.3Find the energy released in \(n\to p+e^-+\bar\nu\) from \(m(n)-m({}^1\mathrm H)\). Explain why atomic masses can be used here, and why this decay rules out the neutron being a bound state of a proton and an electron.

Show solution
\(Q=(1.008665-1.007825)\times931.5\approx0.78\) MeV. The hydrogen atomic mass includes one electron, which balances the electron produced; the 13.6 eV atomic binding is negligible. A bound state is always lighter than its free constituents by its binding energy. The neutron is heavier than a proton plus an electron, so it cannot be their bound state.
Lesson 10

Neutrons, fission and critical mass

For the general reader. Because it has no charge, a neutron can walk into any nucleus, even uranium. Within a few years of Chadwick’s discovery, neutrons had been used to make new radioactive elements and, in 1938, to split uranium. If each fission releases more than one neutron, a chain reaction is possible. In 1941 Chadwick drafted the British report that concluded a bomb of uranium-235 was feasible with a few kilograms to tens of kilograms of material.

Goals
  • Compute the energy loss of neutrons in elastic collisions and the number of collisions needed to slow them.
  • Set up the one-group diffusion model of a bare critical sphere.
  • Estimate a critical radius and see the limits of the estimate.

Slowing down

In 1934 Enrico Fermi’s group in Rome found that neutrons slowed by paraffin or water are far more effective at inducing radioactivity. Slow neutrons spend longer near each nucleus, and many capture cross-sections grow roughly as \(1/v\). The slowing happens by elastic collisions, the process Chadwick had used to weigh the neutron.

Elastic moderation

A neutron of energy \(E\) scattering elastically and isotropically (in the centre-of-mass frame) from a nucleus of mass number \(A\) leaves with energy uniformly distributed in \([\alpha E,E]\), where \(\alpha=\left(\tfrac{A-1}{A+1}\right)^2\). The mean loss of \(\ln E\) per collision is

\[ \xi=1+\frac{\alpha\ln\alpha}{1-\alpha}\qquad(\xi=1\text{ for hydrogen}). \]

The number of collisions to slow from \(E_0\) to \(E_1\) is about \(\ln(E_0/E_1)/\xi\).

Chain reaction and critical size

Otto Hahn and Fritz Strassmann found barium among the products of neutron-irradiated uranium in December 1938, and Lise Meitner and Otto Frisch explained it as fission. In March 1940 Frisch and Rudolf Peierls, in Birmingham, argued that a lump of nearly pure uranium-235 would sustain a fast-neutron chain reaction with a critical mass of the order of a kilogram. The MAUD Committee, of which Chadwick was a leading member, examined the idea. His Liverpool group measured fast-neutron cross-sections of uranium.

One-group diffusion model of a bare sphere

Let \(\phi\) be the neutron flux, \(D=1/(3\Sigma_{\mathrm{tr}})\) the diffusion coefficient, \(\Sigma_a\) and \(\Sigma_f\) the macroscopic absorption and fission cross-sections, and \(\nu\) the neutrons per fission. A steady state requires

\[ D\nabla^2\phi+(\nu\Sigma_f-\Sigma_a)\phi=0 . \]

For a sphere with \(\phi=0\) on the extrapolated surface \(R_e\), the fundamental solution is \(\phi\propto\sin(\pi r/R_e)/r\), and criticality requires

\[ \left(\frac{\pi}{R_e}\right)^2=\frac{\nu\Sigma_f-\Sigma_a}{D}\equiv B^2,\qquad R=R_e-0.71\,\lambda_{\mathrm{tr}} . \]

Sketch. In spherical symmetry \(\nabla^2\phi=\tfrac1{r^2}(r^2\phi')'\). With \(\phi=u/r\) the equation becomes \(u''+B^2u=0\), so \(u=\sin Br\) for regularity at the origin. The flux vanishes first at \(Br=\pi\). The transport mean free path \(\lambda_{\mathrm{tr}}=1/\Sigma_{\mathrm{tr}}\) sets the extrapolation distance. Diffusion theory is poor when the sphere is only a few mean free paths across, as here, so the answer is an order-of-magnitude estimate. Reflectors around the core reduce the critical mass considerably.

The final MAUD Report, which Chadwick drafted in July 1941, concluded that a uranium-235 bomb was practicable and could be ready within a few years. The British passed it to the United States, where it helped turn American research into a weapons programme. The bare-sphere critical mass of nearly pure uranium-235 is now known to be about 50 kg; with a good reflector it is much less.

Exercise 10.1How many collisions are needed, on average, to slow a 2 MeV fission neutron to 0.025 eV in hydrogen, deuterium (\(A=2\)) and carbon (\(A=12\))?

Show solution
\(\ln(2\times10^6/0.025)\approx18.2\). Hydrogen: \(\xi=1\), about 18 collisions. Deuterium: \(\alpha=1/9\), \(\xi=1+\tfrac{(1/9)\ln(1/9)}{8/9}\approx0.725\), about 25. Carbon: \(\alpha=(11/13)^2\approx0.716\), \(\xi\approx0.158\), about 115. Hydrogen slows neutrons fastest but also captures them, which is why heavy water and graphite were chosen for early reactors.

Exercise 10.2Take uranium-235 metal with \(N=0.048\times10^{24}\) atoms/cm\(^3\) and representative one-group fast values \(\sigma_f=1.235\) b, \(\sigma_c=0.065\) b, \(\sigma_{\mathrm{tr}}=6.8\) b, \(\nu=2.6\) (1 b \(=10^{-24}\) cm\(^2\)). Estimate the critical radius and mass of a bare sphere (density 18.7 g/cm\(^3\)).

Show solution
\(\Sigma_f=0.0593\), \(\Sigma_a=0.0624\), \(\Sigma_{\mathrm{tr}}=0.326\) cm\(^{-1}\), so \(D=1.02\) cm and \(\nu\Sigma_f-\Sigma_a=0.0918\) cm\(^{-1}\). Then \(B^2=0.0898\) cm\(^{-2}\), \(R_e=\pi/B\approx10.5\) cm, and \(R\approx10.5-0.71\times3.07\approx8.3\) cm. The mass is \(\tfrac43\pi(8.3)^3\times18.7\approx45\) kg. The accepted value, about 50 kg, is close, partly by luck. The estimate shows why the Frisch–Peierls figure of about a kilogram, based on a much larger assumed fission cross-section, was too optimistic, and why the answer turned on measured cross-sections like those from Liverpool.
Part III · Life and legacy
Lesson 11

Liverpool and the bomb, 1935–1946

The Nobel Prize in Physics for 1935 was awarded to Chadwick alone, for the discovery of the neutron. By then he had decided to leave Cambridge. Rutherford opposed building large accelerators at the Cavendish, and Chadwick believed nuclear physics now needed them. In March 1935 he accepted the Lyon Jones Chair of Physics at the University of Liverpool and took it up on 1 October. He found an old laboratory with little equipment and set about rebuilding it.

The Liverpool cyclotron

Chadwick raised money for a cyclotron on the Berkeley model and sent staff to learn from Ernest Lawrence. The machine ran from July 1939. Its total cost was about £5,200, and Chadwick paid part of the shortfall from his Nobel Prize money. Joseph Rotblat, who arrived from Warsaw in 1939 to work with the cyclotron, became a lecturer in Chadwick’s department; he later left the Manhattan Project on grounds of conscience and shared the 1995 Nobel Peace Prize. Otto Frisch moved to Liverpool in 1940 after writing the Frisch–Peierls memorandum.

MAUD and the Manhattan Project

Chadwick joined the MAUD Committee in 1940. Its members had to decide whether a bomb was possible in time to affect the war. Chadwick later said that once he realised it was possible, he could not sleep without sleeping pills. He wrote the final draft of the MAUD Report in July 1941. After the Quebec Agreement of 1943 placed British work within the American project, Chadwick became technical adviser to the Combined Policy Committee and head of the British Mission to the Manhattan Project. He was the only British scientist given access to the whole project.

He worked at Los Alamos from early 1944, under the cover name James Chaffee, and in Washington. He gained the confidence of General Leslie Groves, the project’s military head, and of J. Robert Oppenheimer, and used it to keep British scientists, among them Peierls, Frisch, William Penney and Klaus Fuchs, at the centre of the work. He was knighted in the New Year Honours of 1945 and watched the Trinity test in the New Mexico desert on 16 July 1945. After the war he served on the British Advisory Committee on Atomic Energy and as scientific adviser to the British delegation at the United Nations Atomic Energy Commission.

DiscussionChadwick discovered a particle whose most famous application was a weapon he then helped build. Compare his position with Rutherford’s scepticism about nuclear energy (Rutherford course, Lesson 11) and with Rotblat’s decision to leave Los Alamos. Which judgements were scientific, and which were moral?

Lesson 12

Caius, honours and legacy

In 1948 Chadwick left Liverpool and research to become Master of Gonville and Caius College, Cambridge. The mastership was not easy. Disputes with a group of Fellows over college appointments and governance marked his last years there, and he resigned in 1958. He and his wife retired to North Wales, then returned to Cambridge in 1969 to be near their daughters. From 1957 to 1962 he was a part-time member of the United Kingdom Atomic Energy Authority. He died in his sleep in Cambridge on 24 July 1974, aged 82.

His honours included the Hughes Medal (1932), the Nobel Prize in Physics (1935), a knighthood (1945), the Copley Medal and the Faraday Medal (both 1950), the Franklin Medal (1951), the Prussian order Pour le Mérite for Sciences and Arts (1966) and appointment as a Companion of Honour (1970).

Legacy

  • The neutron: with it, the nucleus became a system of protons and neutrons. Isotopes, nuclear spins and the existence of the deuteron all made sense, and nuclear physics acquired its basic constituents.
  • A new projectile: neutrons led to artificial radioactivity by neutron capture, slow-neutron physics, fission, reactors and weapons. Neutron scattering became a major tool of condensed-matter physics and structural biology.
  • The beta spectrum: the 1914 measurement began the chain of reasoning that led to the neutrino and the weak interaction.
  • The strong force: anomalous alpha–proton scattering was the first quantitative sign of a short-range force stronger than electromagnetism.
  • Method: Chadwick’s neutron paper is a model of how to eliminate hypotheses by kinematics: each claim tested against several targets, with energy and momentum bookkeeping throughout.
  • Institutions: the Liverpool cyclotron and the British role in the Manhattan Project shaped post-war British nuclear physics and the British atomic energy programme.

Final projectWrite Chadwick’s 1932 argument as a referee would check it. For each target he used (hydrogen, helium, lithium, beryllium, carbon, nitrogen, argon), compute the maximum recoil energy under the neutron hypothesis and under the gamma-ray hypothesis, and show which observations each hypothesis can and cannot explain.

Reference

Connections

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

Teachers and advisors

Students

  • Maurice Goldhaber 1933-1936
  • Étienne Biéler 1921

Influenced by

Influenced

Colleagues and collaborators

  • J. Robert Oppenheimer 1944-1945
  • Charles Drummond Ellis 1914-1930
  • John Cockcroft
  • Joseph Rotblat 1939-1944
  • Mark Oliphant
  • Norman Feather 1932
  • Otto Frisch 1940-1945
  • Rudolf Peierls 1940-1945
  • Walther Bothe 1913-1914

Rivals

Friends

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. James Chadwick, Wikipedia (with extensive references).
  2. James Chadwick, biographical, NobelPrize.org.
  3. H. Massey, N. Feather, James Chadwick, 20 October 1891 – 24 July 1974, Biographical Memoirs of Fellows of the Royal Society 22 (1976), 11–70.
  4. A. Brown, The Neutron and the Bomb: A Biography of Sir James Chadwick, Oxford University Press, 1997.
  5. G. Ecker, James Chadwick: ahead of his time, arXiv:2007.06926.
  6. Chadwick, James, Complete Dictionary of Scientific Biography, via Encyclopedia.com.
  7. M. Gowing, Britain and Atomic Energy 1939–1945, Macmillan, 1964.

Physics

  1. J. Chadwick, Intensitätsverteilung im magnetischen Spektrum der β-Strahlen von Radium B + C, Verh. Dtsch. Phys. Ges. 16 (1914), 383–391.
  2. J. Chadwick, The charge on the atomic nucleus and the law of force, Phil. Mag. 40 (1920), 734–746.
  3. J. Chadwick, E. S. Biéler, The collisions of α particles with hydrogen nuclei, Phil. Mag. 42 (1921), 923–940.
  4. E. Rutherford, J. Chadwick, The artificial disintegration of light elements, Phil. Mag. 42 (1921), 809–825.
  5. C. D. Ellis, W. A. Wooster, The average energy of disintegration of radium E, Proc. R. Soc. A 117 (1927), 109–123.
  6. E. Rutherford, J. Chadwick, C. D. Ellis, Radiations from Radioactive Substances, Cambridge University Press, 1930.
  7. J. Chadwick, Possible existence of a neutron, Nature 129 (1932), 312.
  8. J. Chadwick, The existence of a neutron, Proc. R. Soc. A 136 (1932), 692–708.
  9. J. Chadwick, M. Goldhaber, A nuclear photo-effect: disintegration of the diplon by γ rays, Nature 134 (1934), 237–238.
  10. J. Chadwick, The neutron and its properties, Nobel Lecture, 1935.
  11. R. H. Stuewer, Artificial disintegration and the Cambridge–Vienna controversy, in P. Achinstein, O. Hannaway (eds.), Observation, Experiment, and Hypothesis in Modern Physical Science, MIT Press, 1985.
  12. K. S. Krane, Introductory Nuclear Physics, Wiley, 1988 (reaction kinematics, neutron physics).
  13. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3rd ed., Prentice Hall, 2001 (moderation and one-group diffusion).