Alpha-Particle Scattering and Rutherford's Nuclear Model of Atom

What this subtopic is examined on

• Describing the Geiger-Marsden experiment and drawing the $N(\theta)$ graph.
• Defining impact parameter and distance of closest approach, and using $r_0\propto1/K$.
• Deriving the total energy of the electron in Rutherford's atom and explaining the minus sign.
• Stating the two limitations of the Rutherford model — set almost every year.
• Numericals: $r_0$ for a given $\alpha$-energy on gold; the volume fraction occupied by the nucleus; K.E. and P.E. from a given total energy.

Before Rutherford: Thomson's model

• J. J. Thomson (1898), the plum-pudding or watermelon model: positive charge spread uniformly through a sphere of atomic size, with electrons embedded in it like seeds.
• In this picture the electrons sit in stable electrostatic equilibrium and the mass is distributed almost continuously through the atom.
• The electric field inside such a diffuse charge cloud is weak, so a $5.5$ MeV $\alpha$-particle passing through could suffer only tiny deflections — a few tenths of a degree, and never a reversal.
• Both models give the same atomic size, about $10^{-10}$ m; and in both, the positively charged part carries most of the mass. The difference is whether that positive charge is spread out or concentrated.

The experiment (Geiger and Marsden, 1911, at Rutherford's suggestion)

Apparatus, described in words

• A radioactive source of ${}^{214}_{83}\text{Bi}$ emits $\alpha$-particles of energy $5.5$ MeV in all directions. An $\alpha$-particle is a helium nucleus ${}^{4}_{2}\text{He}$: charge $+2e$, mass $6.64\times10^{-27}$ kg.
• The source sits in a cavity in a block of lead. Only the particles that travel along a narrow channel bored through the lead get out, so the beam emerging is narrow and parallel — the lead bricks are the collimator.
• The collimated beam falls normally on a thin gold foil of thickness $2.1\times10^{-7}$ m.
• Behind and around the foil, a detector can be rotated about the foil through the full range of scattering angles. The detector is a zinc sulphide screen viewed through a microscope: each $\alpha$-particle that strikes the screen makes a brief flash of light, a scintillation, which the observer counts.
• The whole assembly is enclosed in a vacuum chamber, otherwise air molecules would scatter and absorb the beam before it reached the foil.
• The measurement: the number of scintillations per unit time, as a function of the scattering angle $\theta$.

Why gold, and why so thin

• Gold is malleable enough to be beaten into a foil only a few hundred atoms thick. Thinness matters because the analysis assumes each $\alpha$-particle is scattered by at most one nucleus.
• Gold is heavy ($Z=79$, and the nucleus about $50$ times the mass of an $\alpha$-particle), so the target nucleus can be taken as stationary throughout the collision.

Observations

• The overwhelming majority of $\alpha$-particles pass straight through with no measurable deflection.
• Only about $0.14\%$ are scattered by more than $1^\circ$.
• About $1$ in $8000$ is deflected by more than $90^\circ$, and a very few come back along nearly the incident direction ($\theta\approx180^\circ$).
• The graph of $N$ against $\theta$ falls extremely steeply: enormous at small $\theta$, and almost flat along the axis by the time $\theta$ approaches $180^\circ$. The theoretical curve, $N(\theta)\propto1/\sin^4(\theta/2)$, fits the data points.
• The graph to draw: $\theta$ (from $0^\circ$ to $180^\circ$) on the $x$-axis, $N(\theta)$ on the $y$-axis; a curve starting very high near $\theta=0^\circ$, dropping steeply, and then hugging the axis. The assumption behind it is that the atom contains a small, dense, positively charged nucleus treated as a point charge, with the rest of the atom empty.

Conclusions

• Most $\alpha$-particles are undeflected, so most of the atom is empty space.
• To turn a $5.5$ MeV $\alpha$-particle right round needs a very large repulsive force, which requires the atom's positive charge and most of its mass to be packed into a very small central volume.
• Because only $1$ in $8000$ turns back, that volume is tiny: the nucleus is $10^{-15}$ to $10^{-14}$ m across against an atomic size of $10^{-10}$ m — a linear ratio of $10^4$ to $10^5$, and a volume fraction of order $10^{-15}$.
• The atomic electrons, being about $7000$ times lighter than an $\alpha$-particle, have no appreciable effect on its path.
• This is why Rutherford is credited with the discovery of the nucleus.

Rutherford's nuclear model of the atom

• The entire positive charge $+Ze$ and nearly all the mass are concentrated in the nucleus at the centre.
• The electrons revolve in circular orbits around the nucleus, at a distance $10^4$ to $10^5$ times the nuclear radius, held by the Coulomb attraction — the "planetary" model.
• The atom is electrically neutral: $Z$ electrons for nuclear charge $+Ze$.
• The force between an $\alpha$-particle and the nucleus, on which the whole calculation rests, is $F=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{(2e)(Ze)}{r^2}$, directed along the line joining them.
• Scale illustration NCERT gives: if the solar system were scaled to the atom's proportions ($r_{\text{orbit}}/r_{\text{centre}}=10^5$), the earth's orbit would be $7\times10^{13}$ m instead of $1.5\times10^{11}$ m — more than $100$ times further out. The atom is emptier than the solar system.

12.2.1 Alpha-particle trajectory and the impact parameter

• Impact parameter $b$: the perpendicular distance of the initial velocity vector of the $\alpha$-particle from the centre of the nucleus, measured while the particle is still far from the atom.
• The path is a hyperbola: the particle comes in along one asymptote, swings around the nucleus and leaves along the other; the angle between the two asymptotes is the scattering angle $\theta$.
• A beam contains particles with all values of $b$, so the beam scatters into all directions with different probabilities.
• Rutherford's relation between the two:

$$b=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{Ze^2\cot(\theta/2)}{E}$$

• $E$ is the kinetic energy of the $\alpha$-particle and $Z$ the atomic number of the target. Read off the limits: $\theta\to0^\circ$ needs $\cot(\theta/2)\to\infty$, i.e. $b\to\infty$ (very large impact parameter, the particle passes almost undeviated); $\theta=180^\circ$ gives $\cot 90^\circ=0$, i.e. $b=0$ (head-on collision, the particle rebounds straight back).
• Diagram to draw: several parallel arrows approaching from the left at different heights above a central line through the nucleus, each curving away from the nucleus. The lowest offset curves most sharply; the one aimed straight at the nucleus reverses. Mark $b$ as the perpendicular offset of one incoming line and $\theta$ as the angle between its incoming and outgoing directions.
• How this bounds the nuclear size: as $\theta$ increases, $b$ decreases. By finding the largest scattering angles actually produced, Rutherford identified the smallest impact parameter at which an $\alpha$-particle is still scattered rather than absorbed, and that smallest $b$ is an upper limit on the nuclear radius. Had the nucleus been larger, far more particles would have been strongly scattered than were observed. Rutherford scattering is therefore a way of setting an upper limit on the size of the nucleus.

Distance of closest approach

• Definition to write: the distance from the nucleus at which the whole of the kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy, so that the particle momentarily comes to rest before reversing.
• Derivation. Mechanical energy is conserved. Far away, the energy is purely kinetic, $E_i=K$. At the turning point, distance $r_0$, the energy is purely potential:

$$E_f=U=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{(2e)(Ze)}{r_0}$$

$$K=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{r_0}\qquad\Longrightarrow\qquad r_0=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{K}$$

• Consequences examined: $r_0\propto\dfrac{1}{K}$, so doubling the kinetic energy halves $r_0$; and $r_0\propto$ the charge of the projectile, so a proton (charge $+e$) of the same energy has half the $r_0$ of an $\alpha$-particle.
• Worked values on gold ($Z=79$), all with $\dfrac{1}{4\pi\varepsilon_0}=9\times10^9$ N m$^2$ C$^{-2}$:

• Why $r_0$ exceeds the true nuclear radius (about $7$ fm for gold): both the nucleus and the $\alpha$-particle are treated as point charges of zero size, and the $\alpha$-particle reverses direction without ever touching the nucleus — its kinetic energy runs out against the Coulomb repulsion while it is still well clear. The calculation therefore gives an upper limit on the nuclear size, and the fact that $r_0$ is so much larger than the nucleus is itself confirmation that the point-charge assumption was justified.

12.2.2 Electron orbits — deriving the total energy (a standard 3-mark derivation)

Take a hydrogen atom: an electron of charge $-e$ and mass $m$ moving with speed $v$ in a circular orbit of radius $r$ about a nucleus of charge $+e$.

• Dynamical condition — the electrostatic attraction supplies the centripetal force:

$$\dfrac{mv^2}{r}=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r^2}\qquad\Longrightarrow\qquad mv^2=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r}$$

• Kinetic energy:

$$K=\dfrac{1}{2}mv^2=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{2r}$$

• Electrostatic potential energy of the charge pair $(+e)$ and $(-e)$ a distance $r$ apart:

$$U=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{(+e)(-e)}{r}=-\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r}$$

• Total energy:

$$E=K+U=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{2r}-\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r}=-\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{2r}=-\dfrac{e^2}{8\pi\varepsilon_0 r}$$

• Significance of the negative sign: the electron is bound to the nucleus. The electron and nucleus form a bound system, and energy must be supplied from outside to pull the electron away to infinity, i.e. to ionise the atom. If $E$ were positive the electron would be free and would not follow a closed orbit.
• Two relations that follow at once and are set constantly: $K=-E$ and $U=2E=-2K$.
• NCERT's numerical check: given $E=-13.6$ eV $=-2.2\times10^{-18}$ J, the formula gives $r=5.3\times10^{-11}$ m and, from $mv^2=e^2/4\pi\varepsilon_0 r$, $v=2.2\times10^{6}$ m s$^{-1}$.

The two limitations of Rutherford's model

• Stability of the atom. An electron moving in a circle is accelerating (centripetal acceleration). Classical electromagnetic theory says an accelerating charge radiates electromagnetic waves continuously. The electron would therefore lose energy continuously, its orbit radius would shrink continuously, and it would spiral into the nucleus in about $10^{-8}$ s. Matter would collapse. But atoms are observed to be stable.
• Line spectra. As the electron spirals inward, its angular velocity — and hence the frequency of the radiation it emits — changes continuously. The model therefore predicts a continuous spectrum. What is actually observed is a discrete line spectrum, characteristic of each element.
• Point to add if the question says "which could not explain the observed features of atomic spectra": classical theory cannot pick out particular wavelengths at all, so the sharp, reproducible lines of hydrogen are entirely outside its reach.

Standard results and answers to collect

• $F=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{r^2}$; $r_0=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{K}$; $b=\dfrac{Ze^2\cot(\theta/2)}{4\pi\varepsilon_0E}$; $N(\theta)\propto1/\sin^4(\theta/2)$.
• $K=-E$, $U=2E$: a state at $-3.4$ eV has $K=+3.4$ eV and $U=-6.8$ eV; a state at $-1.51$ eV has $K=+1.51$ eV and $U=-3.02$ eV.
• Volume fraction occupied by the nucleus, atom radius $5.3\times10^{-11}$ m and nuclear diameter $1.0\times10^{-15}$ m (so $r_n=0.5\times10^{-15}$ m):

$$\dfrac{V_{\text{nucleus}}}{V_{\text{atom}}}=\left(\dfrac{r_n}{r_e}\right)^3=\left(\dfrac{0.5\times10^{-15}}{5.3\times10^{-11}}\right)^3=(9.43\times10^{-6})^3\approx8.4\times10^{-16}$$

• Solid hydrogen in place of gold foil: the target nucleus is a proton of mass $1.67\times10^{-27}$ kg, which is less than the mass of the incident $\alpha$-particle, $6.64\times10^{-27}$ kg. A light target cannot reverse a heavier projectile (momentum conservation), and its charge is only $+e$ rather than $+79e$, so the scattering angles would be small and no $\alpha$-particles would bounce back. The experiment would give no evidence for a nucleus.