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Ernest Rutherford

Ernest Rutherford, 1st Baron Rutherford of Nelson

A course in two strands: the life of the New Zealand farm boy who became the master experimenter of his age, and the physics he found by counting flashes of light: radioactive transmutation, the atomic nucleus and the first nuclear reaction.

Born 30 August 1871, Brightwater, New Zealand Died 19 October 1937, Cambridge Nobel Prize Chemistry, 1908 Level advanced undergraduate → graduate
Alpha particles passing a point nucleus, integrated numerically. Impact parameter and deflection obey b = (d/2) cot(θ/2).
Lesson 00

How to use this course

Ernest Rutherford worked with simple apparatus and a sure sense of which question to ask next. In three decades he showed that radioactive atoms change into other atoms, that the atom has a tiny massive nucleus, and that nuclei can be broken apart in the laboratory. He also ran two of the most productive physics laboratories in history, at Manchester and at the Cavendish in Cambridge.

This course follows the physics in the order he did it. Each result is derived with the tools a modern student has, and set beside the evidence Rutherford actually had.

Prerequisites for Part II
  • Classical mechanics: central forces, conservation of energy and angular momentum.
  • Basic electromagnetism: Coulomb’s law, motion of charges in a magnetic field.
  • Linear first-order differential equations.
  • Special relativity at the level of \(E=mc^2\), for reaction energies (Lessons 08–10).
  • For Lessons 09 and 10, elementary quantum mechanics: the uncertainty principle and barrier tunnelling.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Lesson 07, the scattering formula, is the mathematical centre of the course.

Reference

Timeline

1871Born 30 August at Spring Grove (now Brightwater), near Nelson, New Zealand, fourth of twelve children of James Rutherford and Martha Thompson.
1887–89Nelson College; head boy in 1889. Wins a scholarship to Canterbury College, Christchurch.
1892–94BA (1892); MA with double first-class honours in mathematics and physical science (1893); BSc (1894). Research on the magnetization of iron by high-frequency discharges.
18951851 Exhibition scholarship takes him to the Cavendish Laboratory, Cambridge, under J. J. Thomson.
1896Detects radio waves at a distance of about half a mile. Begins work with Thomson on conduction in gases ionized by X-rays.
1898Macdonald Professor of Physics, McGill University, Montreal.
1899Paper on uranium radiation names the alpha and beta rays.
1900Discovers the thorium emanation and its exponential decay. Marries Mary Georgina Newton in Christchurch.
1902–03With Frederick Soddy: radioactivity is the spontaneous transformation of atoms. Elected Fellow of the Royal Society (1903).
1904Radio-activity published. Rumford Medal. Argues before Kelvin that radioactivity changes the age of the Earth.
1907Langworthy Professor of Physics, Manchester, succeeding Arthur Schuster.
1908With Hans Geiger, counts alpha particles electrically and shows they carry charge \(2e\). Nobel Prize in Chemistry.
1909With Thomas Royds, proves alpha particles are helium. Geiger and Ernest Marsden observe alpha particles scattered backwards.
1911The scattering paper: the nuclear atom.
1912Niels Bohr works in his Manchester laboratory, March to July.
1914Knighted. Wartime research on submarine detection follows.
1917–19Alpha particles knock hydrogen nuclei out of nitrogen: the first artificial nuclear reaction, published in 1919.
1919Cavendish Professor of Experimental Physics, Cambridge, succeeding Thomson.
1920Bakerian Lecture: predicts a neutral particle of mass 1. Proposes the name proton.
1922Copley Medal.
1925–30President of the Royal Society. Order of Merit (1925).
1931Created Baron Rutherford of Nelson.
1932At the Cavendish: James Chadwick discovers the neutron; John Cockcroft and Ernest Walton split lithium with accelerated protons.
1933First president of the Academic Assistance Council for refugee scholars.
1934With Mark Oliphant and Paul Harteck, fuses deuterium nuclei, producing tritium and helium-3.
1937Dies 19 October in Cambridge, aged 66. Buried in Westminster Abbey.
Part I · Formation
Lesson 01

Nelson and Canterbury, 1871–1895

Ernest Rutherford was born on 30 August 1871 at Spring Grove, a farming settlement near Nelson at the north of New Zealand’s South Island. The place is now called Brightwater. His father, James, had come from Perth in Scotland as a child and worked as a wheelwright, flax-miller and farmer. His mother, Martha Thompson, had emigrated from Hornchurch in Essex and was a schoolteacher. Ernest was the fourth of twelve children.

He went to local schools at Foxhill and Havelock and in 1887 won a scholarship to Nelson College, where he became head boy. A further scholarship took him in 1890 to Canterbury College in Christchurch, then part of the University of New Zealand. The college was small, but its professors of mathematics and of chemistry and physics gave him a strong base. He took a BA in 1892, an MA in 1893 with a double first in mathematics and physical science, and a BSc in 1894.

First research

At Canterbury he studied how iron is magnetized by rapidly oscillating currents, the high-frequency discharges of a Leyden jar. He found that a magnetized iron needle lost some of its magnetization when hit by such oscillations, and used this to build a detector for the electromagnetic waves Heinrich Hertz had produced a few years earlier. In 1895 he won an 1851 Exhibition scholarship for research abroad. He chose the Cavendish Laboratory in Cambridge.

While in Christchurch he lodged with the family of Mary Georgina Newton. They became engaged before he left and married in Christchurch in 1900, once he had a salary large enough to support a family.

Lesson 02

Cambridge research student, 1895–1898

Rutherford arrived at the Cavendish in 1895, among the first research students admitted under new rules that let graduates of other universities work for a Cambridge degree by research. The director was J. J. Thomson. Rutherford improved his magnetic wave detector until it registered signals at about half a mile, then a record. He briefly hoped that wireless signalling might earn him money, but Guglielmo Marconi’s longer-range demonstrations overtook him, and Thomson offered something better.

Wilhelm Röntgen discovered X-rays at the end of 1895. Thomson set Rutherford to study how X-rays make gases conduct electricity. Together they showed that the radiation produces positive and negative ions in the gas, that the current saturates once every ion is collected, and that ions recombine if left alone. This was the physics of the ionization chamber, the main detector Rutherford used for the next decade. It also fed into Thomson’s 1897 identification of the electron.

In 1896 Henri Becquerel found that uranium salts emit penetrating rays without any external stimulus. Rutherford turned to them in 1898. Using the ionization method, he covered a uranium sample with successive layers of thin aluminium foil and measured the current. The first few foils cut the ionization sharply; after that it fell much more slowly. He concluded that there were at least two kinds of radiation. He called the easily absorbed kind alpha and the more penetrating kind beta. The paper appeared in the Philosophical Magazine in January 1899, by which time he had left Cambridge. In 1897 he had received a BA by research and the Coutts-Trotter Studentship of Trinity College.

Lesson 03

McGill, 1898–1907

On Thomson’s recommendation Rutherford was appointed Macdonald Professor of Physics at McGill University in Montreal in 1898, at the age of 27. The Macdonald Physics Building was new and well equipped, paid for by the tobacco manufacturer William Macdonald. Rutherford stayed nine years and did the work that won him the Nobel Prize.

In 1899 R. B. Owens, McGill’s professor of electrical engineering, noticed that the radiation from thorium compounds varied erratically when air moved in the room. Rutherford traced the effect to a radioactive gas given off by thorium, which he called the emanation, and showed in 1900 that its activity fell by half in about a minute, whatever the starting amount. He also found that surfaces exposed to the emanation became radioactive themselves, the so-called excited activity.

Frederick Soddy, a young Oxford chemist, joined McGill as a demonstrator in chemistry in 1900. From 1901 to 1903 he and Rutherford worked through the chemistry of thorium and its products. Their conclusion, that radioactive atoms spontaneously turn into atoms of other elements, was the theory of radioactive transformation (Lesson 05). Rutherford was elected a Fellow of the Royal Society in 1903 and published the first textbook of the subject, Radio-activity, in 1904. His Silliman Lectures at Yale in 1905 became Radioactive Transformations (1906).

McGill drew visitors. Otto Hahn worked with him in 1905–06. Bertram Boltwood of Yale corresponded with him about the decay series of uranium and the use of radioactivity to date rocks (Lesson 06). Rutherford’s daughter, Eileen, was born in 1901. Offers from other universities arrived regularly. In 1907 he accepted the Langworthy chair at Manchester, attracted by its laboratory and by the chance to be nearer the centre of European physics.

Part II · The physics
Lesson 04

Alpha and beta rays

For the general reader. Uranium gives off invisible rays. Rutherford sorted them by how easily they are stopped: alpha rays are halted by a sheet of paper, beta rays pass through much more. Over ten years he showed that the alpha ray is a stream of helium atoms stripped of two electrons, thrown out of the atom at high speed. The alpha particle then became his main tool for probing the atom.

Goals
  • Model absorption by thin layers and see how two components appear in the data.
  • Understand why alpha particles are hard to bend in a magnetic field.
  • Follow the three measurements that identified the alpha particle as \(\mathrm{He}^{2+}\).

Two components in the absorption curve

Rutherford measured ionization current \(I\) after the radiation passed through a total thickness \(x\) of aluminium. If each layer \(dx\) removes the same fraction \(\mu\,dx\) of what reaches it, then \(dI=-\mu I\,dx\) and \(I=I_0e^{-\mu x}\). A mixture of two kinds of radiation gives

\[ I(x)=I_\alpha e^{-\mu_\alpha x}+I_\beta e^{-\mu_\beta x},\qquad \mu_\alpha\gg\mu_\beta . \]

On a logarithmic plot this is a steep line that bends into a shallow one. The first few foils remove the alpha component; after that only the beta part remains. This is exactly the shape Rutherford reported in 1899. The exponential law is only an approximation for alpha particles, which in fact have a fairly definite range in matter, as William Henry Bragg showed in 1904. For Rutherford’s thin foils and broad sources the approximation was good enough to separate the two components.

Bending in a magnetic field

A charge \(q\) with momentum \(p\) moving perpendicular to a field \(B\) follows a circle of radius

\[ r=\frac{p}{qB}. \]

Becquerel and others soon showed that beta rays bend like fast electrons. The alpha rays seemed undeflected. In 1902–03 Rutherford, using strong fields and a stack of narrow slits, deflected them in the direction of positive charge and measured both magnetic and electric deflections. That gave the ratio of charge to mass, about half that of the hydrogen ion. This fits either \(\mathrm H_2^+\) or \(\mathrm{He}^{2+}\). He suspected helium, because helium is found trapped in uranium and thorium minerals.

Result (Rutherford and Geiger, 1908)

By counting individual alpha particles from a known radium source and separately measuring the total charge they carried, the charge per alpha particle is \(q_\alpha\approx 2e\). Combined with \(q/m\), this gives \(m_\alpha\approx 4\,m_{\mathrm H}\), the mass of a helium atom.

The counter used in this work, built with Hans Geiger in 1908, amplified the small ionization of a single alpha particle by accelerating the freed electrons in a strong field until they ionized further gas. It is the ancestor of the Geiger counter. They checked the count against a second method: scintillations, the tiny flashes each alpha particle makes on a screen of zinc sulphide, counted by eye through a microscope.

The direct proof

In 1908 Rutherford and Thomas Royds sealed radon gas in a glass tube with walls thin enough for alpha particles to pass through but too thick for gas atoms to diffuse. The tube sat inside an evacuated outer vessel. After several days they compressed the gas collected in the outer vessel into a discharge tube and saw the spectrum of helium. The control experiment, with helium itself placed in the inner tube, showed nothing. The paper appeared in 1909. Rutherford’s Nobel Lecture of December 1908 was titled The chemical nature of the alpha particles from radioactive substances.

Exercise 4.1Suppose a single foil of thickness \(x_{1/2}\) halves the alpha ionization. How many such foils reduce it to 1%? If the beta component is 1% of the total at \(x=0\) and has \(\mu_\beta=\mu_\alpha/100\), which component dominates after 10 foils?

Show solution
\(2^{-n}=0.01\) gives \(n=\log_2 100\approx 6.6\), so seven foils. After 10 foils the alpha part is \(0.99\times2^{-10}\approx 9.7\times10^{-4}\). The beta part is \(0.01\times2^{-10/100}\approx 0.0093\), about ten times larger. The curve has bent into its shallow beta tail.

Exercise 4.2Compare the radius of curvature in a 1 T field for (a) an alpha particle of kinetic energy 7.7 MeV (\(m_\alpha c^2=3727\) MeV) and (b) an electron of kinetic energy 1 MeV. Use \(r[\mathrm m]=p[\mathrm{GeV}/c]/(0.2998\,zB[\mathrm T])\).

Show solution
(a) Non-relativistically \(pc=\sqrt{2mc^2T}=\sqrt{2\cdot3727\cdot7.7}\approx 240\) MeV. With \(z=2\), \(r=0.240/(0.2998\cdot2)\approx0.40\) m. (b) Relativistically \(pc=\sqrt{T^2+2Tm_ec^2}=\sqrt{1+1.022}\approx1.42\) MeV, so \(r=0.00142/0.2998\approx4.7\) mm. The alpha particle is about 85 times stiffer, which is why its deflection was missed at first.

Exercise 4.3In a counting experiment, \(N=6.0\times10^{9}\) alpha particles deliver a total charge of \(1.92\) nC to a collector. Find the charge per particle in units of \(e=1.602\times10^{-19}\) C.

Show solution
\(q=1.92\times10^{-9}/6.0\times10^{9}=3.2\times10^{-19}\) C \(\approx2.0\,e\). With \(q/m\) half that of the hydrogen ion, the mass is about four hydrogen masses.
Lesson 05

Radioactive decay and transmutation

For the general reader. Chemists had believed for a century that elements never change into one another. In 1902 Rutherford and Soddy showed that radioactive atoms do exactly that, one atom at a time and at random, so that a fixed fraction of any sample changes in each second. The rule is simple, and it explains why some radioactive substances fade in minutes while others last for billions of years.

Goals
  • Derive the exponential decay law from a constant probability per unit time.
  • Solve the equations for a chain of decays.
  • Understand the thorium X experiment and secular equilibrium.

The thorium X experiment

Soddy and Rutherford precipitated thorium from solution with ammonia and found that most of the activity stayed in the filtrate, in a substance they called thorium X. Freshly separated thorium X was very active, and its activity decayed by half in about four days. Over the same days the purified thorium regained its activity. The two curves were mirror images, and their sum stayed constant. Their explanation, published in 1902 as The cause and nature of radioactivity, was that thorium continually produces thorium X, which decays, and that the radiation accompanies the change of one kind of atom into another.

Law of radioactive change (Rutherford and Soddy, 1902–03)

Each atom of a given radioactive species has the same probability \(\lambda\,dt\) of decaying in any interval \(dt\), independent of its age, its chemical state and the temperature. For a sample of \(N\) atoms,

\[ \frac{dN}{dt}=-\lambda N,\qquad N(t)=N_0e^{-\lambda t},\qquad A(t)=\lambda N(t). \]

The time for half the atoms to decay is \(T_{1/2}=\ln2/\lambda\), and the mean life is \(\tau=1/\lambda\).

The law follows from the assumption of a constant probability. If each atom survives an interval \(dt\) with probability \(1-\lambda\,dt\), the survival probability over time \(t\) satisfies \(P(t+dt)=P(t)(1-\lambda\,dt)\), so \(P'=-\lambda P\) and \(P=e^{-\lambda t}\). In 1910 Rutherford, Geiger and Harry Bateman showed that the number of alpha particles counted in equal intervals follows the Poisson distribution, the statistics of independent random events.

Decay chains

Let a parent (1) decay into a radioactive daughter (2). Then

\[ \frac{dN_1}{dt}=-\lambda_1N_1,\qquad \frac{dN_2}{dt}=\lambda_1N_1-\lambda_2N_2. \]

With \(N_2(0)=0\), multiply the second equation by \(e^{\lambda_2t}\) and integrate:

\[ N_2(t)=\frac{\lambda_1N_1(0)}{\lambda_2-\lambda_1}\left(e^{-\lambda_1t}-e^{-\lambda_2t}\right). \]

Bateman solved the general chain of \(n\) members in 1910; the solutions bear his name. Two limits matter. If the parent lives much longer than the daughter, \(\lambda_1\ll\lambda_2\), then after several daughter half-lives \(\lambda_2N_2\approx\lambda_1N_1\): the activities are equal. This is secular equilibrium, and it is why thorium recovered its activity in step with the decay of the separated thorium X. In 1913 Soddy and Kazimierz Fajans stated the displacement laws: alpha decay moves an element two places down the periodic table, beta decay one place up. Soddy coined the word isotope the same year.

Exercise 5.1Thorium X (radium-224) has half-life 3.6 days. After separation, its activity is \(A_0e^{-\lambda t}\) and the thorium recovers as \(A_0(1-e^{-\lambda t})\). How long until the thorium has regained 90% of its activity?

Show solution
\(e^{-\lambda t}=0.1\), so \(t=\ln10/\lambda=T_{1/2}\ln10/\ln2\approx3.32\times3.6\approx12\) days. The sum of the two activities is \(A_0\) at all times.

Exercise 5.2Radium-226 (half-life 1600 years) is in secular equilibrium with uranium-238 (half-life \(4.47\times10^9\) years) in an old ore. How many grams of radium does a tonne of uranium-238 contain?

Show solution
Equal activities give \(N_{\mathrm{Ra}}/N_{\mathrm U}=T_{\mathrm{Ra}}/T_{\mathrm U}=1600/4.47\times10^9\approx3.6\times10^{-7}\). In mass, multiply by \(226/238\): about \(3.4\times10^{-7}\), or 0.34 g per tonne. This is why Marie and Pierre Curie had to process tonnes of residue to isolate a fraction of a gram.

Exercise 5.3Show that \(N_2(t)\) in the two-member chain reaches its maximum at \(t_{\max}=\dfrac{\ln(\lambda_2/\lambda_1)}{\lambda_2-\lambda_1}\), and that at this moment the parent and daughter activities are equal.

Show solution
Setting \(dN_2/dt=0\) gives \(\lambda_1e^{-\lambda_1t}=\lambda_2e^{-\lambda_2t}\), so \(e^{(\lambda_2-\lambda_1)t}=\lambda_2/\lambda_1\). At this time the rate equation reads \(0=\lambda_1N_1-\lambda_2N_2\), so the activities are equal.
Lesson 06

Half-life and radiometric dating

For the general reader. Because each radioactive species decays at a fixed rate, a rock containing uranium is a clock. Count how much decay product has built up, and you know how long the clock has been running. Rutherford was the first to read such a clock, and his answer, hundreds of millions of years, overturned the leading physicist’s estimate of the age of the Earth.

Goals
  • Derive the age equation for a parent–daughter pair.
  • Apply it to the helium and lead methods.
  • See why radioactivity removed the conflict between Kelvin and the geologists.

Kelvin and the age of the Earth

William Thomson, Lord Kelvin, had estimated the age of the Earth from its rate of cooling. By 1897 he put it at about 20 to 40 million years, far too short for the geologists and for Charles Darwin. In 1903 Pierre Curie and Albert Laborde showed that radium releases heat continuously. In 1904 Rutherford lectured in London with Kelvin in the audience and argued that radioactive heating supplies a source of energy that the cooling calculation had left out, so that the Earth could be much older. He then proposed a direct measurement.

Age equation

A closed system starts at \(t=0\) with \(P_0\) parent atoms and no daughter atoms. If each decay eventually produces \(k\) daughter atoms, then at time \(t\), with \(P\) and \(D\) the present amounts,

\[ D=k\,P\left(e^{\lambda t}-1\right),\qquad t=\frac{1}{\lambda}\ln\!\left(1+\frac{D}{kP}\right)=\frac{T_{1/2}}{\ln2}\ln\!\left(1+\frac{D}{kP}\right). \]

Derivation. \(P=P_0e^{-\lambda t}\) and \(D=k(P_0-P)=kP(e^{\lambda t}-1)\). Intermediate members of a chain do not matter once they are in secular equilibrium, because they hold a negligible and constant number of atoms.

Helium and lead

Each uranium-238 atom emits eight alpha particles on its way to lead-206, and each alpha particle becomes a helium atom. Rutherford measured helium in uranium minerals and, around 1905, obtained ages of hundreds of millions of years, some above 500 million. He knew helium can leak out of a mineral, so these were lower limits. Boltwood, starting from Rutherford’s suggestion, used lead instead (\(k=1\)), which stays locked in the crystal. In 1907 he published ages from 410 to 2200 million years. Robert John Strutt in London reached similar conclusions with helium. Modern uranium–lead dating, refined by Arthur Holmes and later workers, gives the Earth an age of about 4.5 billion years.

Exercise 6.1A zircon crystal contains lead-206 and uranium-238 in the atomic ratio \(D/P=0.50\). Take \(T_{1/2}=4.47\times10^{9}\) years and assume no initial lead. Find its age.

Show solution
\(t=(4.47\times10^9/0.693)\ln1.5=6.45\times10^9\times0.405\approx2.6\times10^9\) years.

Exercise 6.2A uranium mineral holds 0.8 helium atoms per uranium-238 atom. Using \(k=8\), find the helium age. If half of the helium produced has escaped, what is the true age?

Show solution
\(D/kP=0.1\): \(t=6.45\times10^9\ln1.1\approx6.1\times10^8\) years. If half escaped, the true ratio is \(D/kP=0.2\) and \(t=6.45\times10^9\ln1.2\approx1.18\times10^9\) years. Leakage always makes helium ages too young.

Exercise 6.3Show that for \(t\ll T_{1/2}\) the age equation reduces to \(t\approx D/(k\lambda P)\), and interpret this as “accumulated product divided by production rate”.

Show solution
\(\ln(1+u)\approx u\) for small \(u\), so \(t\approx D/(k\lambda P)\). The production rate of daughter atoms is \(k\lambda P\), nearly constant when little parent has decayed, and the age is the accumulated stock divided by the rate. Rutherford’s first estimates used this linear form.
Lesson 07

Scattering and the nuclear atom

For the general reader. In 1909 Geiger and Marsden fired alpha particles at thin metal foil. Nearly all went straight through, but about one in several thousand bounced back. Rutherford later said this was as surprising as a heavy shell bouncing off tissue paper. In 1911 he explained it: the atom’s positive charge and almost all its mass sit in a tiny central nucleus. He worked out exactly how many particles should be deflected through each angle, and the experiments agreed.

Goals
  • Derive the relation between impact parameter and deflection angle for a Coulomb force.
  • Derive the Rutherford differential cross-section.
  • Use it to estimate nuclear sizes and to see why the Thomson atom fails.

The experiment

Geiger had found in 1908 that gold foil spreads a narrow beam of alpha particles by a degree or so. Rutherford suggested that Marsden, then an undergraduate, look for particles scattered through large angles. In 1909 Geiger and Marsden reported that about one alpha particle in 8000 striking a platinum plate was turned back towards the side it came from, and that heavier metals reflected more. In J. J. Thomson’s model the atom’s positive charge is spread through a sphere of atomic size, and the electric field is never strong enough to reverse an alpha particle in one encounter. Many small deflections can add up, but the probability of a large total angle by that route is astronomically small.

Rutherford’s 1911 paper, The scattering of α and β particles by matter and the structure of the atom, assumed instead a point charge \(Ze\) at the centre of the atom and treated a large deflection as a single encounter.

The orbit

Let a particle of charge \(ze\), mass \(m\) and speed \(v_0\) approach a fixed point charge \(Ze\) with impact parameter \(b\). Write \(\kappa=zZe^2/(4\pi\varepsilon_0)\) and \(E=\tfrac12mv_0^2\). Define the distance of closest approach in a head-on collision,

\[ d=\frac{\kappa}{E}=\frac{zZe^2}{4\pi\varepsilon_0E}. \]

The orbit is a hyperbola symmetric about the line through the nucleus and the point of closest approach. Energy conservation means the speed far away is \(v_0\) on both legs, so the momentum changes only along that symmetry axis, by \(\Delta p=2mv_0\sin(\theta/2)\), where \(\theta\) is the deflection angle. Compute the same impulse from the force. Let \(\varphi\) be the polar angle measured from the symmetry axis. The force component along the axis is \((\kappa/r^2)\cos\varphi\). Angular momentum conservation gives \(mr^2\dot\varphi=mv_0b\), so \(dt=r^2\,d\varphi/(v_0b)\). The polar angle runs from \(-(\pi-\theta)/2\) to \((\pi-\theta)/2\). Then

\[ \Delta p=\int\frac{\kappa}{r^2}\cos\varphi\,dt=\frac{\kappa}{v_0b}\int_{-(\pi-\theta)/2}^{(\pi-\theta)/2}\cos\varphi\,d\varphi=\frac{2\kappa}{v_0b}\cos\frac{\theta}{2}. \]

Equating the two expressions for \(\Delta p\):

\[ \tan\frac{\theta}{2}=\frac{\kappa}{mv_0^2b}=\frac{d}{2b},\qquad\text{i.e.}\qquad b=\frac{d}{2}\cot\frac{\theta}{2}. \]

The cross-section

Particles with impact parameter between \(b\) and \(b+db\) pass through a ring of area \(2\pi b\,db\) and come out between \(\theta\) and \(\theta+d\theta\), into solid angle \(d\Omega=2\pi\sin\theta\,d\theta\). The differential cross-section is the ratio:

\[ \frac{d\sigma}{d\Omega}=\frac{b}{\sin\theta}\left|\frac{db}{d\theta}\right|. \]

With \(b=\tfrac d2\cot\tfrac\theta2\) we have \(|db/d\theta|=\tfrac d4\csc^2\tfrac\theta2\) and \(\sin\theta=2\sin\tfrac\theta2\cos\tfrac\theta2\), so

\[ \frac{d\sigma}{d\Omega}=\frac{\tfrac d2\cot\tfrac\theta2\cdot\tfrac d4\csc^2\tfrac\theta2}{2\sin\tfrac\theta2\cos\tfrac\theta2}=\frac{d^2}{16}\,\frac{1}{\sin^4(\theta/2)}. \]
Rutherford scattering formula (1911)

For a non-relativistic particle of charge \(ze\) and kinetic energy \(E\) scattered by a fixed point charge \(Ze\), under the Coulomb force alone,

\[ \frac{d\sigma}{d\Omega}=\left(\frac{zZe^2}{4\pi\varepsilon_0\,4E}\right)^{2}\frac{1}{\sin^4(\theta/2)}. \]

A foil with \(n\) atoms per unit volume and thickness \(t\), thin enough that single scattering dominates, sends a fraction \(nt\,\frac{d\sigma}{d\Omega}\,d\Omega\) of incident particles into \(d\Omega\).

Integrating from \(\theta\) to \(\pi\), the cross-section for deflection beyond \(\theta\) is simply \(\pi b(\theta)^2=\tfrac{\pi d^2}{4}\cot^2\tfrac\theta2\). For a nucleus of finite mass \(M\) the same formula holds in the centre-of-mass frame with \(E\) replaced by the centre-of-mass energy; for alpha particles on gold the correction is about 2%.

The test

The formula makes four separate predictions: the count at fixed angle scales as \(1/\sin^4(\theta/2)\), as the foil thickness \(t\), as \(Z^2\), and as \(1/E^2\), i.e. \(1/v^4\). Geiger and Marsden tested all four in 1913, with silver and gold foils, angles up to 150°, and alpha particles slowed by mica sheets. The angular law held while the counting rate changed by a factor of many thousands. Rutherford’s estimate of the central charge, about half the atomic weight in units of \(e\), led Antonius van den Broek to suggest that it equals the position in the periodic table. Henry Moseley’s X-ray spectra in 1913–14 confirmed this: the nuclear charge is the atomic number.

Two pieces of luck helped. The classical formula coincides exactly with the quantum mechanical result for a pure Coulomb potential, so it was not upset by the quantum theory that came later. And the alpha particles available were not energetic enough to reach heavy nuclei, so no deviations from the point-charge law appeared. Deviations seen later with light nuclei gave the first measurements of nuclear size.

Exercise 7.1Use \(e^2/(4\pi\varepsilon_0)=1.44\) MeV fm. Find the distance of closest approach for 7.7 MeV alpha particles hitting gold (\(Z=79\)) and aluminium (\(Z=13\)) head-on. Compare with nuclear radii \(R\approx1.2A^{1/3}\) fm.

Show solution
Gold: \(d=2\cdot79\cdot1.44/7.7\approx29.5\) fm, while \(R_{\mathrm{Au}}\approx1.2\cdot197^{1/3}\approx7.0\) fm. The alpha never touches the nucleus and the point-charge formula holds. Aluminium: \(d=2\cdot13\cdot1.44/7.7\approx4.9\) fm, while \(R_{\mathrm{Al}}+R_\alpha\approx3.6+1.9=5.5\) fm. The alpha reaches the nuclear surface, and the formula fails at large angles. This anomalous scattering was used in the 1920s to estimate nuclear size.

Exercise 7.2Gold has \(n=5.9\times10^{28}\) atoms m\(^{-3}\). For a foil 0.4 µm thick and 7.7 MeV alpha particles, estimate the fraction scattered through more than 90°.

Show solution
\(b(90^\circ)=d/2\approx14.8\) fm, so \(\sigma=\pi b^2\approx6.9\times10^{-28}\) m\(^2\). Then \(nt\sigma=5.9\times10^{28}\cdot4\times10^{-7}\cdot6.9\times10^{-28}\approx1.6\times10^{-5}\), about one particle in 60 000. Rare, but easily seen by patient counting.

Exercise 7.3By what factor does the counting rate at 15° exceed that at 150°, for the same detector area at the same distance?

Show solution
\(\sin^4(75^\circ)/\sin^4(7.5^\circ)=(0.9659/0.1305)^4\approx(7.40)^4\approx3.0\times10^3\). This range is what Geiger and Marsden had to span with one technique.

Exercise 7.4Show that the total cross-section \(\int\frac{d\sigma}{d\Omega}\,d\Omega\) is infinite. What physical effect makes it finite in a real atom?

Show solution
Near \(\theta=0\), \(\sin^4(\theta/2)\approx\theta^4/16\) while \(d\Omega\approx2\pi\theta\,d\theta\), so the integrand behaves as \(\theta^{-3}\) and diverges. Equivalently, \(\pi b^2\to\infty\) as \(\theta\to0\): every impact parameter gives some deflection. In a real atom the electrons screen the nuclear charge beyond about the atomic radius, which cuts off large \(b\).
Lesson 08

The first artificial nuclear reaction

For the general reader. During the First World War, between bouts of military research, Rutherford sent alpha particles through nitrogen gas and saw flashes from particles that travelled much further than the alphas themselves. They were hydrogen nuclei, knocked out of nitrogen. For the first time a nucleus had been broken by a projectile fired in the laboratory. The leftover nucleus, as later photographs showed, was oxygen.

Goals
  • Follow the logic of the 1919 nitrogen experiment.
  • Compute the Q-value of a reaction from atomic masses.
  • Find the threshold energy and compare it with the Coulomb barrier.

The experiment

Marsden had noticed in 1914 that alpha particles passing through hydrogen produce fast hydrogen nuclei with a long range. Rutherford took up the work in 1917. A radioactive source sat in a box filled with gas, and a zinc sulphide screen was placed behind absorbing foils thick enough to stop every alpha particle. With oxygen or carbon dioxide in the box, few scintillations got through. With dry air, more appeared, and the excess came from nitrogen. By magnetic deflection and range he identified the particles as hydrogen nuclei. He published the results in 1919 as four papers, Collision of α particles with light atoms, in the Philosophical Magazine; the fourth was titled An anomalous effect in nitrogen.

Rutherford concluded that the hydrogen nucleus was a constituent of the nitrogen nucleus, and that the alpha particle had knocked it out. He was careful not to say what was left behind. With Chadwick he showed over the following years that alpha particles eject protons from boron, fluorine, sodium, aluminium and phosphorus as well. In 1925 Patrick Blackett settled the question. Among about 23 000 cloud-chamber photographs, showing some 415 000 alpha tracks, he found eight forked tracks. Each showed a proton and a heavy recoil, and no outgoing alpha particle. The alpha particle had been absorbed:

\[ {}^{4}_{2}\mathrm{He}+{}^{14}_{7}\mathrm{N}\longrightarrow{}^{17}_{8}\mathrm{O}+{}^{1}_{1}\mathrm H. \]
Reaction energetics

For a reaction \(a+X\to Y+b\), the Q-value is the kinetic energy released:

\[ Q=\left(m_a+m_X-m_Y-m_b\right)c^2. \]

Atomic masses may be used, because electron numbers balance. If \(Q<0\) and \(X\) is at rest, the reaction is possible only for projectile kinetic energy above the threshold

\[ T_{\mathrm{th}}=-Q\,\frac{m_a+m_X+m_Y+m_b}{2m_X}\approx-Q\left(1+\frac{m_a}{m_X}\right). \]

Derivation of the threshold. The invariant \(s=(E_{\mathrm{tot}})^2-(p_{\mathrm{tot}}c)^2\) equals \((m_a^2+m_X^2)c^4+2(T_a+m_ac^2)m_Xc^2\) in the laboratory. At threshold the products are at rest in the centre-of-mass frame, so \(s=(m_Y+m_b)^2c^4\). Solving for \(T_a\) gives the exact formula; the approximate form uses \(m_Y+m_b\approx m_a+m_X\). Physically, part of the projectile’s energy must go into motion of the centre of mass and is unavailable for the reaction.

The threshold is not the whole story. The alpha particle must also come close enough to the nitrogen nucleus to touch it, against the Coulomb repulsion. Classically this needs energy above the barrier \(V_C=zZe^2/(4\pi\varepsilon_0R)\), with \(R\) the sum of the radii.

Exercise 8.1Using the atomic masses \(m({}^4\mathrm{He})=4.002603\), \(m({}^{14}\mathrm N)=14.003074\), \(m({}^{17}\mathrm O)=16.999132\), \(m({}^1\mathrm H)=1.007825\) u and \(1\,\mathrm u\,c^2=931.5\) MeV, find \(Q\) for the nitrogen reaction.

Show solution
Initial mass 18.005677 u; final mass 18.006957 u. \(Q=-0.001280\times931.5\approx-1.19\) MeV. The reaction absorbs energy.

Exercise 8.2Find the threshold alpha energy and the Coulomb barrier for this reaction, taking \(R=1.2(4^{1/3}+14^{1/3})\) fm. Were Rutherford’s 7.7 MeV alpha particles from radium C′ sufficient?

Show solution
\(T_{\mathrm{th}}\approx1.19(1+4/14)\approx1.53\) MeV. \(R\approx1.2(1.59+2.41)\approx4.8\) fm, so \(V_C\approx2\cdot7\cdot1.44/4.8\approx4.2\) MeV in the centre-of-mass frame, about 5.4 MeV in the laboratory. The barrier, not the threshold, is the real obstacle, and 7.7 MeV alpha particles clear it. This is why the effect was seen with light elements only: for heavy nuclei the barrier exceeds the alpha energy.

Exercise 8.3From Blackett’s numbers, estimate the probability that an alpha particle traversing his chamber caused the reaction. Why did Rutherford need scintillation counting, rather than photographs, to discover the effect?

Show solution
\(8/415\,000\approx2\times10^{-5}\). Photographing enough tracks to catch a handful of events took Blackett years. A scintillation screen behind absorbers ignores the vast majority of ordinary alpha particles and records only the long-range protons, so it selects the rare events automatically.
Lesson 09

The neutron: prediction and discovery

For the general reader. After the nitrogen experiment Rutherford thought nuclei were built from hydrogen nuclei, which he named protons, plus electrons to cancel some of the charge. In 1920 he predicted a new particle: a proton and an electron bound so tightly that they behave as a single neutral object. Such a particle would pass through matter almost unseen. Twelve years later his deputy, James Chadwick, found it.

Goals
  • See why a neutral nuclear constituent was needed.
  • Reproduce Chadwick’s determination of the neutron mass from recoil velocities.
  • Understand the quantum arguments against electrons inside nuclei.

The Bakerian Lecture, 1920

In his Bakerian Lecture to the Royal Society, Nuclear constitution of atoms, Rutherford pointed out that nuclear masses are roughly twice the nuclear charges, so a nucleus of atomic number \(Z\) and mass number \(A\) seemed to need \(A\) protons and \(A-Z\) electrons. He suggested that a proton and electron could combine into a neutral unit of mass 1. Such a particle would move freely through matter, would be hard to detect and hard to confine, and might enter nuclei easily. He also suggested a heavy hydrogen of mass 2. The name proton for the hydrogen nucleus dates from the same year.

Chadwick and Rutherford searched for the neutral particle through the 1920s without success. In 1930 Walther Bothe and Herbert Becker found that beryllium bombarded by polonium alpha particles emits a very penetrating radiation, which they took to be gamma rays. In January 1932 Irène Joliot-Curie and Frédéric Joliot showed that it ejects protons of about 5 MeV from paraffin. For gamma rays to do this by Compton scattering they would need an implausibly high energy. Chadwick, with Rutherford’s prediction in mind, repeated the experiment within weeks and measured recoils from several gases. His letter Possible existence of a neutron appeared in Nature in February 1932.

Elastic head-on collision

A particle of mass \(m\) and speed \(u\) striking a stationary nucleus of mass \(M\) head-on gives it the maximum recoil speed

\[ V_{\max}=\frac{2m}{m+M}\,u. \]

Comparing maximum recoil speeds of two targets eliminates the unknown \(u\): \(\dfrac{V_1}{V_2}=\dfrac{m+M_2}{m+M_1}\).

The formula follows from conservation of momentum and kinetic energy in one dimension. Chadwick measured the maximum recoil speeds of hydrogen and nitrogen nuclei struck by the beryllium radiation and so found the mass of the incident particle. A massless gamma ray could not produce the observed nitrogen recoils with the energy that the proton recoils required.

Why not electrons in the nucleus?

Chadwick’s neutron settled a crisis that the proton–electron model had already reached. Werner Heisenberg and others soon described nuclei as made of protons and neutrons alone, with the neutron a particle in its own right rather than a compound. Beta decay then had to create its electron at the moment of decay, as in Enrico Fermi’s 1933–34 theory.

Exercise 9.1Chadwick found maximum recoil speeds of about \(3.3\times10^9\) cm/s for hydrogen and \(4.7\times10^8\) cm/s for nitrogen. Find the mass of the neutral particle in units of the proton mass (take \(M_{\mathrm H}=1\), \(M_{\mathrm N}=14\)).

Show solution
\(V_{\mathrm H}/V_{\mathrm N}=(m+14)/(m+1)=3.3/0.47\approx7.0\). Then \(m+14=7m+7\), so \(m\approx1.16\). Given the measurement errors, Chadwick concluded that the mass is close to that of the proton. A later analysis of a boron reaction gave a much more precise value.

Exercise 9.2An electron confined to a nucleus of radius 5 fm has momentum uncertainty \(\Delta p\gtrsim\hbar/\Delta x\). Using \(\hbar c=197\) MeV fm, estimate its kinetic energy. Compare with typical beta-decay energies of about 1 MeV.

Show solution
\(\Delta p\,c\approx197/5\approx39\) MeV. This is far above \(m_ec^2\), so the electron is ultra-relativistic with kinetic energy of order 40 MeV, much more than any known nuclear binding could hold or beta decay could release.

Exercise 9.3In the proton–electron model, how many spin-½ particles make up a nitrogen-14 nucleus? What total spin and statistics does that predict? Band spectra of \(\mathrm N_2\) showed in 1929 that nitrogen-14 nuclei have integer spin and obey Bose statistics. How does the neutron model fix this?

Show solution
14 protons and 7 electrons make 21 fermions: an odd number, so half-integer spin and Fermi statistics, contradicting experiment. With 7 protons and 7 neutrons there are 14 fermions, giving integer spin and Bose statistics, as observed.
Lesson 10

The Cavendish under Rutherford

For the general reader. As director of the Cavendish Laboratory from 1919, Rutherford pushed his students towards higher energies and better detectors. In 1932 two of them, John Cockcroft and Ernest Walton, built a high-voltage machine that accelerated protons and split lithium nuclei. It was the first nuclear reaction produced entirely by artificial means. Quantum tunnelling explained why it worked at energies far lower than classical physics required.

Goals
  • Compute the energy released when lithium-7 absorbs a proton.
  • Estimate the Gamow tunnelling factor and see why modest voltages suffice.
  • Calculate the energetics of deuterium fusion, discovered at the Cavendish in 1934.

Tunnelling and the accelerator

In 1928 George Gamow, and independently Ronald Gurney and Edward Condon, explained alpha decay as quantum tunnelling through the Coulomb barrier. Gamow visited Cambridge in 1929. Cockcroft realized that the same effect works in reverse: a proton can enter a light nucleus with an energy well below the barrier, with a probability that is small but not negligible. Rutherford backed the construction of a voltage multiplier.

Gamow factor

For two nuclei of charges \(z_1e\), \(z_2e\) with relative speed \(v\) and kinetic energy far below the Coulomb barrier, the probability of tunnelling to short distance is approximately

\[ P\approx e^{-2\pi\eta},\qquad \eta=\frac{z_1z_2e^2}{4\pi\varepsilon_0\hbar v}=z_1z_2\,\alpha\,\frac{c}{v}, \]

with \(\alpha\approx1/137\) the fine-structure constant.

Sketch. The WKB transmission through a barrier is \(\exp\big(-\tfrac2\hbar\int_R^{r_c}\sqrt{2\mu(V(r)-E)}\,dr\big)\), with \(r_c\) the classical turning point where \(V(r_c)=E\). For \(V=\kappa/r\) and \(R\ll r_c=\kappa/E\), the integral is \(\sqrt{2\mu E}\int_0^{r_c}\sqrt{r_c/r-1}\,dr=\tfrac{\pi}{2}r_c\sqrt{2\mu E}\). With \(E=\tfrac12\mu v^2\) the exponent is \(\tfrac{2}{\hbar}\cdot\tfrac{\pi}{2}\cdot\tfrac{\kappa}{E}\cdot\mu v=\tfrac{2\pi\kappa}{\hbar v}=2\pi\eta\).

In April 1932 Cockcroft and Walton reported that protons of a few hundred keV striking lithium produced alpha particles, seen as scintillations:

\[ {}^{1}_{1}\mathrm H+{}^{7}_{3}\mathrm{Li}\longrightarrow2\,{}^{4}_{2}\mathrm{He}. \]

They shared the Nobel Prize in Physics in 1951. Measuring the energy of the alpha particles also gave one of the first direct checks of \(E=mc^2\) in a nuclear reaction.

Deuterium fusion, 1934

After Harold Urey isolated deuterium in 1931–32, Mark Oliphant, Paul Harteck and Rutherford bombarded deuterium compounds with accelerated deuterons. In 1934 they observed two reactions, producing two new nuclei, tritium and helium-3:

\[ \mathrm D+\mathrm D\longrightarrow{}^{3}\mathrm H+p,\qquad \mathrm D+\mathrm D\longrightarrow{}^{3}\mathrm{He}+n. \]

These are the first fusion reactions made in a laboratory, and both are central to fusion research today.

Exercise 10.1With \(m({}^7\mathrm{Li})=7.016003\) u, find the Q-value of the lithium reaction. For slow protons, what energy does each alpha particle carry, and in what directions do they fly?

Show solution
\(Q=(7.016003+1.007825-2\times4.002603)\times931.5\approx0.018622\times931.5\approx17.3\) MeV. With negligible initial momentum the two alphas fly back to back, each with about 8.7 MeV. Cockcroft and Walton saw pairs of scintillations on opposite sides of the target.

Exercise 10.2Compute the Gamow factor \(e^{-2\pi\eta}\) for protons on lithium (\(z_1z_2=3\)) at 125 keV and at 500 keV. Ignore the reduced-mass correction. The classical barrier is about 1.2 MeV. Comment.

Show solution
At 125 keV, \(v/c=\sqrt{2\cdot0.125/938.3}\approx0.0163\), \(\eta=3/(137\cdot0.0163)\approx1.34\), \(2\pi\eta\approx8.4\), \(P\approx2\times10^{-4}\). At 500 keV, \(v/c\approx0.0326\), \(\eta\approx0.67\), \(P\approx e^{-4.2}\approx0.015\). Both energies are well below the barrier, yet with a beam of \(10^{13}\) or more protons per second even a small fraction gives countable reactions. The rate rises steeply with energy.

Exercise 10.3Using \(m(\mathrm D)=2.014102\), \(m({}^3\mathrm H)=3.016049\), \(m({}^3\mathrm{He})=3.016029\), \(m(p)=1.007825\) (hydrogen atom) and \(m(n)=1.008665\) u, find the Q-values of the two D–D reactions.

Show solution
Initial mass \(4.028204\) u. Tritium branch: final \(4.023874\) u, \(Q\approx0.004330\times931.5\approx4.03\) MeV. Helium-3 branch: final \(4.024694\) u, \(Q\approx0.003510\times931.5\approx3.27\) MeV.
Part III · Life and legacy
Lesson 11

Manchester, war and Cambridge, 1907–1937

Manchester

At Manchester Rutherford inherited a well-equipped laboratory from Arthur Schuster and a capable assistant in Hans Geiger. The group he built included Ernest Marsden, Charles Galton Darwin, Henry Moseley, Kazimierz Fajans, George de Hevesy and, for four months in 1912, Niels Bohr. Bohr arrived with a Danish doctorate on electron theory and left with the problem that led to his 1913 quantum model of the atom, which kept Rutherford’s nucleus and explained why its electrons do not spiral into it. Rutherford communicated Bohr’s first paper to the Philosophical Magazine. Moseley’s X-ray work fixed the meaning of atomic number. He was killed at Gallipoli in 1915, a loss Rutherford felt deeply.

Rutherford received the Nobel Prize in Chemistry in 1908, for his investigations into the disintegration of the elements and the chemistry of radioactive substances. He was amused to have been transformed from a physicist into a chemist. He was knighted in 1914. During the war he served on the Admiralty’s Board of Invention and Research and worked on detecting submarines by sound, while continuing the nitrogen experiments in the gaps.

The Cavendish

In 1919 he succeeded J. J. Thomson as Cavendish Professor at Cambridge and held the chair until his death. The laboratory grew to some thirty research students a year. Rutherford’s own direct research slowed, but the Cavendish under his direction produced the neutron, the first accelerator-driven nuclear reaction, Blackett’s cloud-chamber studies, and, with Giuseppe Occhialini, early evidence of electron–positron pairs. Peter Kapitza came from Russia in 1921 and built a programme in strong magnetic fields and low temperatures. The Royal Society Mond Laboratory was built for him and opened in 1933, with a carving of a crocodile, Kapitza’s nickname for Rutherford, on its wall. In 1934 Kapitza was prevented from leaving the Soviet Union after a visit home, and his equipment was later sent to Moscow.

Rutherford was a loud, cheerful and direct leader who preferred simple apparatus and distrusted elaborate theory. He insisted that results be understood physically. He served as President of the Royal Society from 1925 to 1930. His daughter Eileen married the theoretical physicist Ralph Fowler; she died in 1930, shortly after the birth of her fourth child.

From 1933 he was the first president of the Academic Assistance Council, founded to find posts for scholars dismissed in Germany. On 3 October 1933 he chaired a fundraising meeting at the Royal Albert Hall at which Albert Einstein was the main speaker. In September of that year a newspaper report of a speech in which he called hopes of extracting useful energy from atoms “moonshine” prompted Leó Szilárd to think of the neutron chain reaction.

DiscussionRutherford discouraged speculation about practical nuclear energy, yet his laboratory created the techniques that made it possible within a decade of his death. What does this suggest about predicting the consequences of fundamental research?

Lesson 12

Honours, death and legacy

His honours included the Rumford Medal (1904), the Copley Medal (1922), the Franklin Medal (1924), the Order of Merit (1925), the Albert Medal (1928) and the Faraday Medal (1930). In 1931 he was created Baron Rutherford of Nelson.

In October 1937 he became ill with a strangulated hernia. After emergency surgery he died on 19 October 1937 in Cambridge, aged 66. He was cremated, and his ashes were buried in the nave of Westminster Abbey near the graves of Isaac Newton and Charles Darwin.

Legacy

  • Radioactivity: the transformation theory, the exponential law and the decay chains are the basis of nuclear chemistry, nuclear medicine and radiation protection.
  • Geochronology: radiometric dating, begun with his helium measurements, gave the Earth its age and geology its absolute time scale.
  • The nuclear atom: the model underlies all of atomic physics, from Bohr’s theory onward. Scattering as a method of seeing structure became the standard technique of particle physics, from the electron scattering that measured nuclear sizes to the deep inelastic scattering that revealed quarks.
  • Nuclear physics: artificial transmutation, the prediction of the neutron and the first accelerator experiments led directly to fission, reactors and fusion research.
  • Students: scientists who trained or worked with him won many Nobel Prizes, among them Soddy, Bohr, Hahn, de Hevesy, Chadwick, Blackett, Cockcroft, Walton, Powell and Kapitza.
  • Commemoration: element 104, rutherfordium, was named in 1997; his portrait appears on the New Zealand hundred-dollar note.

Final projectChoose one experiment from Part II. Describe the apparatus and the observed quantity in Rutherford’s terms, then derive the result with modern tools, and explain what alternative interpretation the experiment ruled out.

Reference

Sources and further reading

Biography

  1. Ernest Rutherford, Wikipedia (with extensive references).
  2. Ernest Rutherford, biographical, NobelPrize.org.
  3. American Institute of Physics, Rutherford’s Nuclear World, online exhibit.
  4. A. S. Eve, Rutherford: Being the Life and Letters of the Rt. Hon. Lord Rutherford, O.M., Cambridge University Press, 1939.
  5. D. Wilson, Rutherford: Simple Genius, Hodder and Stoughton, 1983.
  6. J. Campbell, Rutherford: Scientist Supreme, AAS Publications, 1999.
  7. Otago Daily Times, Humanity strand of Rutherford legacy (the Academic Assistance Council).

Physics

  1. E. Rutherford, Uranium radiation and the electrical conduction produced by it, Phil. Mag. 47 (1899).
  2. E. Rutherford, F. Soddy, The cause and nature of radioactivity, Phil. Mag. 4 (1902).
  3. E. Rutherford, Radio-activity, Cambridge University Press, 1904; Radioactive Transformations, 1906.
  4. B. B. Boltwood, On the ultimate disintegration products of the radio-active elements, Am. J. Sci. 23 (1907).
  5. E. Rutherford, The chemical nature of the alpha particles from radioactive substances, Nobel Lecture, 1908.
  6. E. Rutherford, T. Royds, The nature of the α particle from radioactive substances, Phil. Mag. 17 (1909).
  7. H. Geiger, E. Marsden, On a diffuse reflection of the α-particles, Proc. R. Soc. A 82 (1909).
  8. E. Rutherford, The scattering of α and β particles by matter and the structure of the atom, Phil. Mag. 21 (1911), 669–688.
  9. H. Geiger, E. Marsden, The laws of deflexion of α particles through large angles, Phil. Mag. 25 (1913).
  10. E. Rutherford, Collision of α particles with light atoms, I–IV, Phil. Mag. 37 (1919).
  11. E. Rutherford, Nuclear constitution of atoms (Bakerian Lecture), Proc. R. Soc. A 97 (1920).
  12. P. M. S. Blackett, The ejection of protons from nitrogen nuclei, photographed by the Wilson method, Proc. R. Soc. A 107 (1925).
  13. J. Chadwick, Possible existence of a neutron, Nature 129 (1932), 312.
  14. J. D. Cockcroft, E. T. S. Walton, Disintegration of lithium by swift protons, Nature 129 (1932).
  15. M. L. E. Oliphant, P. Harteck, Lord Rutherford, Transmutation effects observed with heavy hydrogen, Proc. R. Soc. A 144 (1934).
  16. E. Rutherford, J. Chadwick, C. D. Ellis, Radiations from Radioactive Substances, Cambridge University Press, 1930.
  17. H. Goldstein, C. Poole, J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002, §3.10 (Rutherford scattering).
  18. K. S. Krane, Introductory Nuclear Physics, Wiley, 1988 (decay chains, reaction kinematics, Gamow factor).