The Atom: Thomson, Rutherford & α-Scattering

Rutherford turned scattering into a way of seeing, and the method never stopped. Fire a known beam at a target, measure how it deflects, and read off the structure of what it hit — that is the operating principle of every particle accelerator from the 1911 gold foil to the Large Hadron Collider, which is a Geiger–Marsden experiment run at unimaginably higher energy. In the 1960s, firing high-energy electrons at protons at Stanford revealed, by exactly Rutherford's logic of unexpected large-angle scattering, that the proton itself has hard internal lumps: the quarks. Closer to home, Rutherford backscattering spectrometry dates thin films and analyses material surfaces by bouncing ions off them and measuring the recoil, and ion-beam analysis of ancient artefacts reads their composition the same way. The nuclear density that so astonished Rutherford — $10^{17}\ \mathrm{kg\,m^{-3}}$ — turns out to be the density of a neutron star, so every time a massive star collapses, nature builds a body the size of a city out of nothing but packed nuclear matter. And the empty-atom picture underlies all of chemistry and solid-state physics: everything solid you have ever touched is, by volume, essentially vacuum threaded by electric fields.

Start with what everyone believed in 1904. Thomson had shown that atoms contain electrons — light, negative, and easily knocked loose. Since a whole atom is electrically neutral, an equal amount of positive charge had to be present too, and the natural picture was that this positive charge filled the atom as a diffuse sphere with the electrons dotted through it: the plum-pudding model. The crucial feature is that in this picture the charge is spread out. Anywhere inside such a sphere, the electric field is weak, because at any point only the charge closer to the centre than you pulls on you and it is thinly distributed. A fast, heavy alpha particle ploughing through should feel nothing but a gentle jostle — thousands of feeble nudges from the diffuse charge and the tiny electrons, each nudge deflecting it by a hundredth of a degree, adding up at most to a degree or two. On the pudding model a large deflection is not merely unlikely, it is essentially forbidden: there is no concentrated field anywhere to produce it. The alpha particle is the perfect probe. It is a helium nucleus, charge $+2e$, about $7300$ times heavier than an electron, and radioactive sources spit them out at a few million electronvolts of energy — enormous on the atomic scale. Being so heavy, it is barely deflected by the light electrons (a truck is not turned aside by hitting a mosquito). Whatever bends its path must be something at least as massive as the alpha itself and strongly charged. Rutherford, Geiger and Marsden used exactly this to X-ray the atom.

Now the experiment itself. A radioactive source (radium or its gas radon) supplies a narrow beam of alpha particles in an evacuated chamber. The beam strikes a gold leaf a few thousand atoms thick — gold because it can be beaten astonishingly thin and its heavy atoms scatter strongly. A screen coated with zinc sulphide, which flashes a tiny scintillation each time an alpha strikes it, is mounted on a swinging arm so it can be moved to any angle. In a dark room, Geiger and Marsden counted the flashes by eye, angle by angle, for hours on end.

The counts told a stark story. At small angles the flashes came thick and fast: the overwhelming majority of alphas passed through the foil deflected by less than a degree, exactly as both models predicted. But as Geiger and Marsden swung the screen round to larger and larger angles, they kept finding flashes where the pudding model said there should be none — at 90 degrees, at 120, even a handful bouncing back toward the source at more than 150 degrees. Rare, yes: only about one alpha in eight thousand turned through more than a right angle. But one is infinitely more than zero, and zero was what a diffuse charge demanded. Rutherford grasped what it meant. To reverse a heavy, fast, $+2e$ projectile you need a region of enormous electric field, which means a large charge packed into a tiny volume. Spread that same charge over the whole atom and the field everywhere is far too weak — an alpha would need to meet the entire positive charge of the atom acting as if from a single point, and from very close range. The only way to arrange that is to concentrate all the positive charge, and (since the alpha is barely slowed) nearly all the mass, into a minute central body. Rutherford called it the nucleus. The rareness of the bounce-backs is itself the proof of smallness: most alphas miss the nucleus by a mile of atomic emptiness and are barely bent; only the rare one aimed almost dead-centre comes close enough to be hurled back.

How big is the nucleus? The scattering data set an upper limit through a beautifully simple idea: the distance of closest approach. Consider the most extreme case, an alpha fired straight at the nucleus — a head-on collision, zero impact parameter. As it approaches, the electric repulsion between its charge $+2e$ and the nuclear charge $+Ze$ does negative work on it, converting its kinetic energy into electrical potential energy and slowing it. At one instant the alpha is momentarily at rest, at its closest point $r_0$; then the same repulsion drives it straight back the way it came. Because the electrons are far too light to matter and the nucleus barely recoils, energy is conserved between "far away, all kinetic" and "closest, all potential":

where: $E$ is the alpha's initial kinetic energy far from the atom, $Z$ is the atomic number (the number of protons; gold has $Z = 79$), $e = 1.60\times10^{-19}\ \mathrm{C}$ is the elementary charge, and $k = 1/4\pi\varepsilon_0 = 8.99\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}$ is the Coulomb constant. The physical reading is a simple energy trade: the closer the alpha gets, the more of its motion it has paid out as potential energy, and it can go no further than the point where the account is empty. Rearranged, this gives the closest approach directly:

A handy exam-desk form uses the packaged constant $ke^2 = 1.44\ \mathrm{MeV\,fm}$ (electronvolt-scale energy times femtometre-scale distance), so $r_0[\mathrm{fm}] = 2Z(1.44)/E[\mathrm{MeV}]$. For a typical $5\ \mathrm{MeV}$ alpha on gold, $r_0 = 2(79)(1.44)/5 = 45.5\ \mathrm{fm} = 4.55\times10^{-14}\ \mathrm{m}$. This is an upper bound on the nuclear radius: the alpha turned around at $45\ \mathrm{fm}$ without ever reaching the nucleus, and Coulomb's inverse-square law held perfectly all the way in, so the nucleus must be smaller than $45\ \mathrm{fm}$. Firing faster alphas at heavier or lighter nuclei tightened the bound toward the modern value, a nuclear radius of order $10^{-15}\ \mathrm{m} = 1\ \mathrm{fm}$. The full angular pattern that Geiger and Marsden mapped out follows from tracking many alphas with different aims. An alpha aimed to pass a distance $b$ (the impact parameter) to one side of the nucleus is turned through an angle $\theta$ given by Rutherford's relation

with $r_0$ the head-on closest approach of equation (2). Read it off: a large impact parameter (a distant miss, $b$ large) makes $\cot(\theta/2)$ large and so $\theta$ small — barely bent; a vanishing impact parameter ($b \to 0$) forces $\theta \to 180^\circ$ — the head-on bounce-back. Counting how many alphas land in each angular slice then gives the celebrated Rutherford scattering law: the number scattered into angle $\theta$ goes as

The steep $1/\sin^4(\theta/2)$ falloff is why large-angle events are so rare, and its exact agreement with the flash counts — checked in detail by Geiger and Marsden in 1913 — was the clinching confirmation of the point nucleus. The derivation of equations (3) and (4) tracks the alpha along its hyperbolic path and is beyond this article; we quote the results and use them. Equation (1), the distance of closest approach, is pure energy conservation and we derive it in full below.

Two lengths now define the atom: the nucleus at about $r_{\text{nuc}} \approx 1\times10^{-15}\ \mathrm{m}$ and the atom itself at about $r_{\text{atom}} \approx 1\times10^{-10}\ \mathrm{m}$ (a tenth of a nanometre, set by the electron cloud). Their ratio,

means the atom's radius is a hundred thousand times the nucleus's. Since volume scales as the cube of the radius, the nucleus occupies only about $(10^{5})^{-3} = 10^{-15}$ of the atom's volume — the atom is, to fifteen decimal places, empty. If the nucleus were a pea on the centre spot of a football stadium, the atom's electrons would be a faint mist out at the topmost seats. All the pushing that makes solid matter feel solid, and all of chemistry, happens in that outer mist; the nucleus holds the mass and the identity of the element (its charge $Ze$) but takes almost no room. That the mass is concentrated with the charge follows from the same experiment. The alpha, heavy as it is, is turned right around — impossible if it struck something light, so the scattering centre must be at least comparably massive, and in fact carries essentially the whole atomic mass. Packing the mass of an atom into $10^{-15}\ \mathrm{m}$ gives a nuclear density of about $2.3\times10^{17}\ \mathrm{kg\,m^{-3}}$ — a matchbox of pure nuclear matter would weigh billions of tonnes. This is the density of a neutron star, which is nothing but nuclear matter held together by gravity.

Board level — the atom is mostly empty space (CBSE and ICSE pattern)

Take the radius of an atom as $1.0\times10^{-10}\ \mathrm{m}$ and the radius of its nucleus as $1.0\times10^{-15}\ \mathrm{m}$. Find (a) the ratio of the atomic radius to the nuclear radius, (b) the fraction of the atom's volume occupied by the nucleus, and (c) describe the scale with a everyday analogy.

Step 1. Radius ratio: $\dfrac{r_{\text{atom}}}{r_{\text{nuc}}} = \dfrac{1.0\times10^{-10}}{1.0\times10^{-15}} = \boxed{1.0\times10^{5}}$. The atom's radius is a hundred thousand times the nucleus's.

Step 2. Volume fraction: volume goes as radius cubed, so $\dfrac{V_{\text{nuc}}}{V_{\text{atom}}} = \left(\dfrac{r_{\text{nuc}}}{r_{\text{atom}}}\right)^3 = (10^{-5})^3 = \boxed{10^{-15}}$. The nucleus fills one part in a thousand million million of the atom — the rest is empty.

Step 3. Analogy: if the nucleus were a pea (about $1\ \mathrm{cm}$) at the centre spot, the atom's outer edge would be $10^5$ times farther out, about $1\ \mathrm{km}$ away — the electrons would be a thin haze a kilometre off, roughly the far end of a large sports ground. Solid matter is almost entirely nothing.

Board level — distance of closest approach on gold (CBSE pattern)

A $5.0\ \mathrm{MeV}$ alpha particle is fired head-on at a gold nucleus ($Z = 79$). Using $ke^2 = 1.44\ \mathrm{MeV\,fm}$, find the distance of closest approach.

Step 1. At the closest point all kinetic energy has become electrical potential energy: $E = \dfrac{2Zke^2}{r_0}$, so $r_0 = \dfrac{2Zke^2}{E}$.

Step 2. Substitute: $r_0 = \dfrac{2\,(79)\,(1.44\ \mathrm{MeV\,fm})}{5.0\ \mathrm{MeV}} = \dfrac{227.5\ \mathrm{MeV\,fm}}{5.0\ \mathrm{MeV}} = \boxed{45.5\ \mathrm{fm}} = 4.55\times10^{-14}\ \mathrm{m}$.

Step 3. Sanity check: this is far larger than the true gold-nucleus radius ($\sim 7\ \mathrm{fm}$), so the alpha never reaches the nucleus and the inverse-square Coulomb law holds throughout — consistent with Rutherford's assumption.

Competitive level — the energy needed to touch the nucleus (JEE and NEET pattern)

The radius of a gold nucleus ($Z = 79$, mass number $A = 197$) is $R = 1.2\,A^{1/3}\ \mathrm{fm}$. (a) Find $R$. (b) What head-on kinetic energy would an alpha need for its distance of closest approach to just equal $R$? (c) Comment on why $5$–$8\ \mathrm{MeV}$ alphas from natural sources obey Rutherford's formula so well.

Step 1. Nuclear radius: $A^{1/3} = 197^{1/3} = 5.82$, so $R = 1.2\times5.82 = \boxed{7.0\ \mathrm{fm}}$.

Step 2. Set $r_0 = R$ in $r_0 = 2Zke^2/E$ and solve for $E$: $E = \dfrac{2Zke^2}{R} = \dfrac{2\,(79)\,(1.44)}{7.0} = \dfrac{227.5}{7.0} = \boxed{32.5\ \mathrm{MeV}}$.

Step 3. An alpha would need about $32\ \mathrm{MeV}$ to reach the gold nuclear surface — far more than the few-MeV alphas radioactive sources provide. So natural alphas always stop tens of femtometres short, in the region where only the pure Coulomb $1/r^2$ force acts; the strong nuclear force never gets a say. That is precisely why Rutherford's point-charge formula matched the gold data so exactly. Timing note: about 90 seconds. The cube root $197^{1/3} \approx 5.8$ is worth memorising for gold; once you see part (b) is just part (a) run backwards through the same formula, no new physics is needed and the arithmetic is a single division.

Competitive level — impact parameter for a right-angle scatter (JEE Advanced and ISC pattern)

A $5.0\ \mathrm{MeV}$ alpha ($Z_{\text{target}} = 79$, gold) is scattered through exactly $90^\circ$. Using $b = \tfrac12 r_0\cot(\theta/2)$ with the head-on closest approach $r_0 = 45.5\ \mathrm{fm}$, find the impact parameter, and state what impact parameter gives a $180^\circ$ bounce-back.

Step 1. For $\theta = 90^\circ$, the half-angle is $45^\circ$ and $\cot 45^\circ = 1$. So $b = \tfrac12\,(45.5)\,(1) = \boxed{22.8\ \mathrm{fm}}$ — half the head-on closest approach.

Step 2. For a $180^\circ$ bounce-back, $\theta/2 = 90^\circ$ and $\cot 90^\circ = 0$, so $b = 0$: only a perfectly head-on alpha returns straight back. This is the geometric statement of why exact backscatters are the rarest events of all.

Step 3. Reading the trend: as $b$ shrinks from large values toward zero, the scattering angle climbs from near $0^\circ$ toward $180^\circ$. Every alpha feels the same Coulomb law; only its aim, encoded in $b$, decides the outcome. Timing note: under a minute. The whole item hinges on two cotangent values, $\cot 45^\circ = 1$ and $\cot 90^\circ = 0$; if you know $b = \tfrac12 r_0\cot(\theta/2)$, both parts fall out by inspection with no calculator.

"The experiment measured the size of the nucleus." It measured the distance of closest approach, an upper bound on the nuclear radius. The alpha stops short of the nucleus; write "the nucleus is smaller than $r_0$", never "$r_0$ is the nuclear radius". CBSE, ICSE and JEE all set traps that reward this distinction.

Forgetting the factor of 2 in the alpha's charge. The alpha is a helium nucleus with charge $+2e$, not $+e$. The potential energy is $k(2e)(Ze)/r_0$, giving $r_0 = 2Zke^2/E$. Drop the 2 and every closest-approach answer is halved. This is the single most common arithmetic error on this topic.

Confusing atomic number and mass number. The nuclear charge in Coulomb's law is $Ze$ with $Z$ the atomic number (protons: gold $Z = 79$), while the nuclear-radius formula uses the mass number $A$ (protons plus neutrons: gold $A = 197$) as $R = 1.2\,A^{1/3}\ \mathrm{fm}$. Swapping them wrecks both calculations.

Claiming a Thomson atom gives "small" large-angle scattering. It gives essentially none. The point of the experiment is qualitative: a spread-out charge forbids large angles outright, so even one bounce-back kills the pudding model. Do not soften this to "fewer large deflections".

Board scope. The nuclear atom and the gold-foil experiment are on CBSE, ICSE, JEE Main, JEE Advanced, NEET, IB HL (Theme E) and A-Level (CIE 9702). The distance-of-closest-approach calculation is firmly examined on JEE, NEET, ISC, IB HL and A-Level; the impact-parameter formula and the full Rutherford scattering law are essentially JEE-Advanced and ISC material, quoted-only enrichment (not examined) on NEET, IB HL and A-Level. CBSE expects the closest-approach idea and the model's logic. This topic is IB HL, not SL (SL treats the nuclear atom only qualitatively), and AP Physics 2 starts its modern-physics unit at atomic energy levels and spectra — it does not examine Rutherford scattering, so it is not tagged here. IGCSE mentions the nuclear model only in words.

Texts. NCERT Physics Class 12, Chapter 12 ("Atoms"), the sections on Thomson's model, the alpha-particle scattering experiment and the distance of closest approach · H. C. Verma, Concepts of Physics Vol. 2, "Bohr's Model and Physics of the Atom" · Halliday, Resnick and Walker, "Nuclear Physics" for the scattering treatment · Eisberg and Resnick, Quantum Physics, for the full Rutherford cross-section.

Provenance. HS exam constants. Recompute: packaged constant $ke^2 = 1.44\ \mathrm{MeV\,fm}$ from $k = 8.99\times10^{9}$, $e = 1.60\times10^{-19}$; gold closest approach $r_0 = 2(79)(1.44)/5 = 45.5\ \mathrm{fm} = 4.55\times10^{-14}\ \mathrm{m}$; SI assembly $3.64\times10^{-26}/8.0\times10^{-13}$; gold nuclear radius $1.2\times197^{1/3} = 7.0\ \mathrm{fm}$; touch-energy $2(79)(1.44)/7.0 = 32.5\ \mathrm{MeV}$; right-angle impact parameter $b = \tfrac12(45.5)\cot 45^\circ = 22.8\ \mathrm{fm}$; atom-to-nucleus radius ratio $10^{-10}/10^{-15} = 10^5$ and volume fraction $10^{-15}$; nuclear density $\approx 2.3\times10^{17}\ \mathrm{kg\,m^{-3}}$; fraction of alphas beyond $90^\circ$ about $1$ in $8000$.

Prerequisites. Electron Emission & the Work Function — where the electron is established as a real, extractable constituent of the atom, the piece Thomson's model was built to house. Leads to. Atomic Spectra: Emission & Absorption Lines, the next article, which shows that atoms emit and absorb only sharp fixed colours — the clue Rutherford's stable-but-structured atom could not explain and that forces the quantum picture. For enrichment beyond the exam, the university seam The Bohr Model of Hydrogen takes the nuclear atom and quantises its orbits (a go-deeper pointer, never a dependency here).