The Geiger-Marsden Experiment
At the suggestion of Ernst Rutherford, in 1911 Hans Geiger and Ernst Marsden directed a beam of 5.5 MeV alpha-particles, emitted from a $^{214}_{83}\text{Bi}$ radioactive source, at a thin foil of gold of thickness $2.1 \times 10^{-7}\ \text{m}$. The alpha-particles were collimated into a narrow beam by passage through lead bricks, and the scattered particles were observed with a rotatable detector consisting of a zinc sulphide screen and a microscope.
On striking the screen, each scattered alpha-particle produced a brief flash of light, a scintillation, which could be counted through the microscope. By rotating the detector, Geiger and Marsden measured the number of scattered particles as a function of the scattering angle. The whole apparatus sat inside a vacuum chamber so that the alpha-particles would not collide with air molecules along the way.
Collimated alpha-particles strike a thin gold foil; a rotatable ZnS detector counts the scintillations at each scattering angle. After NCERT Fig. 12.1–12.2.
What the Scattering Data Showed
The expectation, on Thomson's plum-pudding model where positive charge is spread uniformly through the atom, was that the alpha-particles would suffer only small deflections. The data partly agreed: most alpha-particles passed straight through, meaning they suffered no collisions at all. But a small minority behaved very differently. About 0.14% of the incident alpha-particles scattered by more than $1^\circ$, and about 1 in 8000 deflected by more than $90^\circ$.
Rutherford argued that to throw an alpha-particle backwards, it must experience a very large repulsive force. Such a force could only arise if the greater part of the mass and all of the positive charge of the atom were concentrated tightly at its centre. An incoming alpha-particle could then approach this concentrated charge very closely, and that close encounter would produce a large deflection. The theoretical curve based on a small, dense, positively charged nucleus matched the measured angular distribution closely.
| Observation | Approximate value | What it implies |
|---|---|---|
| Alpha-particles passing nearly undeflected | Vast majority | Atom is largely empty space |
| Scattered by more than $1^\circ$ | About 0.14% | Close encounters are rare |
| Deflected by more than $90^\circ$ | About 1 in 8000 | Mass and charge are concentrated |
| Nuclear size from data | $10^{-15}$ to $10^{-14}\ \text{m}$ | Nucleus is $10^4$–$10^5$ times smaller than the atom |
The repulsive force on an alpha-particle of charge $2e$ from a gold nucleus of charge $Ze$ ($Z = 79$ for gold) is Coulombic, directed along the line joining the two:
$$F = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r^2}$$
Because the gold nucleus is about 50 times heavier than an alpha-particle, it can be treated as stationary, and because the foil is thin each alpha-particle suffers at most one scattering. Under these assumptions the trajectory of a single alpha-particle can be computed from Newton's second law together with this Coulomb force. The atomic electrons, being so light, do not appreciably affect the alpha-particles.
Impact Parameter and Scattering Angle
The path an alpha-particle traces depends on its impact parameter $b$, defined as the perpendicular distance of the particle's initial velocity vector from the centre of the nucleus. A beam carries a distribution of impact parameters, so the beam scatters in many directions with different probabilities, while all particles in the beam have nearly the same kinetic energy.
A small impact parameter gives near head-on approach and a large scattering angle; a large impact parameter gives an almost undeviated path. After NCERT Fig. 12.4.
An alpha-particle that comes close to the nucleus, with a small impact parameter, suffers large scattering. In the limiting case of a head-on collision the impact parameter is minimum and the particle rebounds straight back, $\theta \approx \pi$. For a large impact parameter the particle goes nearly undeviated, $\theta \approx 0$. The fact that only a small fraction rebound tells us that the number of head-on collisions is small, which in turn implies that the mass and positive charge occupy a very small volume. Rutherford scattering is therefore a powerful way to set an upper limit on the size of the nucleus.
Small impact parameter does NOT mean small scattering angle
The relationship is inverse, and the wording is designed to trip you. A small impact parameter means the alpha-particle aims almost straight at the nucleus, feels an enormous Coulomb repulsion, and scatters through a large angle. A large impact parameter means it stays far away and barely deflects.
Small $b \Rightarrow$ large $\theta$ (head-on, $\theta \to 180^\circ$). Large $b \Rightarrow$ small $\theta$ ($\theta \to 0$).
Distance of Closest Approach
For a head-on collision the alpha-particle decelerates, momentarily stops, and reverses. The centre-to-centre distance at that turning point is the distance of closest approach, $r_0$. It is found by conserving total mechanical energy: before the encounter the energy is purely the kinetic energy $K$ of the incoming alpha-particle, and at the stopping point it is purely the Coulomb potential energy of the alpha-particle and nucleus.
$$K = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r_0} \quad\Longrightarrow\quad r_0 = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{K}$$
Rutherford's planetary atom set the stage for the quantum fix that followed. See Bohr's Model of the Hydrogen Atom for how stationary orbits rescued the model.
In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV alpha-particle before it momentarily comes to rest and reverses direction?
By energy conservation, $K = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d}$, so $d = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}$.
With $K = 7.7\ \text{MeV} = 1.2 \times 10^{-12}\ \text{J}$, $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2$ and $e = 1.6 \times 10^{-19}\ \text{C}$:
$$d = \frac{(2)(9.0 \times 10^{9})(1.6 \times 10^{-19})^2\,Z}{1.2 \times 10^{-12}} = 3.84 \times 10^{-16}\,Z\ \text{m}$$
For gold, $Z = 79$, giving $d(\text{Au}) = 3.0 \times 10^{-14}\ \text{m} = 30\ \text{fm}$. The gold nuclear radius is therefore less than $3.0 \times 10^{-14}\ \text{m}$. This is larger than the true radius (about 6 fm) because the alpha-particle reverses without ever touching the nucleus.
What $r_0$ depends on
Reading $r_0 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}$ with $K = \tfrac{1}{2}mv^2$ shows that the closest approach is inversely proportional to the kinetic energy and directly proportional to $Z$. Faster (more energetic) alphas, or lighter alphas at fixed speed, get closer. NEET 2016 asked exactly how $r_0$ depends on the mass $m$ at fixed velocity.
$r_0 \propto \dfrac{Z}{K} = \dfrac{2Z}{mv^2}$ → at fixed $v$, $r_0 \propto \dfrac{1}{m}$. The closest approach is an upper bound on nuclear radius, not the radius itself.
Rutherford's Nuclear Model
The scattering results led Rutherford to the nuclear, or planetary, model of the atom: the entire positive charge and most of the mass are concentrated in a tiny central nucleus, with the electrons revolving around it like planets around the sun. His experiments suggested a nuclear size of about $10^{-15}\ \text{m}$ to $10^{-14}\ \text{m}$, against an atomic size of about $10^{-10}\ \text{m}$ known from kinetic theory. The nucleus is thus 10,000 to 100,000 times smaller than the atom, and most of the atom is empty space, which is why the great majority of alpha-particles pass straight through the foil. NIOS Section 24.1.1 records the same two features: charge and mass confined to a region of about $10^{-15}\ \text{m}$, with electrons orbiting at a distance so that the atom is neutral.
For a dynamically stable orbit, the electrostatic attraction between an orbiting electron and the nucleus must supply the centripetal force. For hydrogen,
$$\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{mv^2}{r}$$
The total energy of the electron then works out negative, $E = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{2r}$, signifying that the electron is bound to the nucleus. A positive total energy would mean the electron does not follow a closed orbit.
If the solar system had the same proportions as the atom (nuclear radius $\approx 10^{-15}\ \text{m}$, electron orbit $\approx 10^{-10}\ \text{m}$), would the earth be closer to or farther from the sun than it actually is?
The ratio of orbit to nucleus radius is $\dfrac{10^{-10}}{10^{-15}} = 10^{5}$. Scaling the sun's radius ($7 \times 10^{8}\ \text{m}$) by the same factor gives an orbit of $10^{5} \times 7 \times 10^{8} = 7 \times 10^{13}\ \text{m}$ — more than 100 times the actual earth-orbit radius of $1.5 \times 10^{11}\ \text{m}$. The earth would be much farther away, showing the atom contains far more empty space than the solar system.
Why the Model Fails
Rutherford's nuclear model was a decisive step, but it could not survive classical electromagnetic theory. NCERT and NIOS both list two fatal difficulties. The first concerns stability. An electron moving in a circular orbit is constantly accelerating, and an accelerating charge must radiate electromagnetic energy. Losing energy continuously, the electron should spiral inward and fall into the nucleus, so the atom could not be stable — in flat contradiction to the observed permanence of matter.
The second difficulty concerns spectra. As the electron spirals in, its orbital frequency would change continuously, and since classical theory ties the radiated frequency to the frequency of revolution, the atom would emit a continuous spread of frequencies. Real atoms, however, emit sharp, discrete line spectra. The nuclear model could explain neither the stability nor the characteristic line spectrum, which is why Bohr's quantum postulates were needed.
| Feature | Thomson's model | Rutherford's model |
|---|---|---|
| Positive charge | Spread uniformly through the atom | Concentrated in a tiny nucleus |
| Mass distribution | Nearly continuous | Highly non-uniform (mostly in nucleus) |
| Large-angle scattering | Cannot explain | Explained by close encounters with the nucleus |
| Instability cause | Electrostatic | Radiation from accelerating electrons |
| Line spectra | Not addressed | Cannot explain |
Five things to carry into the exam
- Geiger and Marsden fired 5.5 MeV alpha-particles at a gold foil ($2.1 \times 10^{-7}\ \text{m}$ thick); most passed through, about 1 in 8000 deflected past $90^\circ$.
- Impact parameter $b$ is the perpendicular distance of the initial velocity from the nucleus: small $b$ gives a large scattering angle, large $b$ gives a small one.
- Distance of closest approach: $r_0 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}$, so $r_0 \propto Z/K$; it is an upper limit on nuclear radius.
- Nuclear model: positive charge and most of the mass in a nucleus $10^4$–$10^5$ times smaller than the atom; electrons orbit it.
- Model fails because accelerating electrons must radiate (unstable atom) and would give a continuous, not line, spectrum.