Why Most Alpha Particles Fly Straight Through Gold Foil

An atom is far too small to see, yet scattering alpha particles from a thin gold foil shows its structure, and the sharp coloured lines in light from glowing hydrogen show how its electron is arranged.

This part covers alpha-particle scattering and the nuclear model, Bohr's postulates, the radius and energy of orbits, and the hydrogen spectrum.

When alpha particles are fired at a thin gold foil, most pass straight through, a few are deflected and a very small number bounce back — showing that an atom's positive charge and nearly all its mass sit in a tiny central nucleus.

Observations and conclusions:

- Most pass undeflected — the atom is mostly empty space
- A few deflect through large angles — strong repulsion from a concentrated positive charge
- Very few rebound — the nucleus is tiny but holds most of the mass

Distance of closest approach. An alpha particle heading straight at the nucleus stops where all its kinetic energy has become electric potential energy:

$$K=\frac{1}{4\pi {\epsilon }_0}\frac{(2e)(Ze)}{d}\Rightarrow d=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}$$

Worked example. A $7.7$ MeV alpha particle approaches a gold nucleus ($Z=79$):

$$d=\frac{9\times 10^9\times 2\times 79\times (1.6\times 10^{-19}{)}^2}{7.7\times 1.6\times 10^{-13}}\approx 3.0\times 10^{-14}\text{ m}$$

Impact parameter $b$ is the perpendicular distance of the alpha particle's initial path from the nucleus: $b=0$ means a head-on collision and a rebound, while a large $b$ gives only a small deflection.

An everyday example. Rolling marbles across a large empty courtyard with one small heavy stone in the middle — almost all roll past, and only the rare marble that hits the stone head-on bounces straight back.

The substance. The closest approach gives an upper limit on the nuclear size, since the alpha particle never actually touches the nucleus.

Rutherford's model cannot explain why atoms are stable or why they emit light only at particular wavelengths, and Bohr's postulates fix both problems by allowing electrons only in certain orbits and emitting light only when an electron jumps between them.

Limitations of Rutherford's model:

- An orbiting electron is accelerating, so electromagnetic theory says it should radiate energy and spiral into the nucleus
- Its frequency would change continuously, giving a continuous spectrum, but hydrogen shows sharp lines

Bohr's postulates:

- Stationary orbits — an electron can revolve in certain orbits without radiating energy
- Quantised angular momentum — allowed orbits have $mvr=\frac{nh}{2\pi }$, with $n=1,2,3,\dots$
- Frequency condition — a jump from a higher orbit to a lower one emits a photon with $h\nu =E_i-E_f$

Worked example. In the first orbit of hydrogen, the angular momentum is

$$L=\frac{h}{2\pi }=\frac{6.63\times 10^{-34}}{2\pi }\approx 1.06\times 10^{-34}\text{ J s}$$

A jump releasing $10.2$ eV emits light of wavelength $\frac{1240}{10.2}\approx 122$ nm, in the ultraviolet.

An everyday example. A staircase rather than a ramp — you can stand on one step or the next but not in between, just as an electron can have only certain energies.

The substance. Bohr's model works only for hydrogen and hydrogen-like ions with a single electron.

**Setting the electrostatic attraction equal to the centripetal force and combining it with Bohr's quantisation condition gives $r_n=n^2{r}_1$ with $r_1\approx 0.053$ nm, $v_n=\frac{v_1}{n}$ and $E_n=-\frac{13.6}{n^2}$ eV.

Derivation:**

- Centripetal force: $\frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}$
- Quantisation: $mvr=\frac{nh}{2\pi }$
- Solving these together:

$$r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi m{e}^2},v_n=\frac{e^2}{2{\epsilon }_{0}nh}$$

Energy. Kinetic energy is $\frac{e^2}{8\pi {\epsilon }_{0}r}$ and potential energy is $-\frac{e^2}{4\pi {\epsilon }_{0}r}$, so

$$E_n=-\frac{e^2}{8\pi {\epsilon }_0{r}_n}=-\frac{13.6}{n^2}\text{ eV}$$

The negative sign means the electron is bound; $13.6$ eV is the ionisation energy of hydrogen.

Worked example. For $n=1$:

$$r_1=\frac{(6.63\times 10^{-34}{)}^2\times 8.85\times 10^{-12}}{\pi \times 9.1\times 10^{-31}\times (1.6\times 10^{-19}{)}^2}\approx 5.3\times 10^{-11}\text{ m}$$

and $v_1\approx 2.2\times 10^6$ m s${}^{-1}$. For $n=2$, $r_2=4r_1\approx 0.21$ nm and $E_2=-3.4$ eV, so exciting the electron from $n=1$ to $n=2$ needs $10.2$ eV.

An everyday example. A satellite in a higher orbit moves more slowly — the electron's speed likewise falls as $\frac{1}{n}$.

The substance. Kinetic energy equals minus the total energy, so in the ground state $K=13.6$ eV while $U=-27.2$ eV.

**Each spectral line comes from an electron dropping between two energy levels and emitting a photon of energy $E_i-E_f$, and de Broglie's idea explains quantisation: an orbit is allowed only when a whole number of electron wavelengths fits around it, $2\pi r=n\lambda$.

Spectral series (qualitative):

- Lyman** — jumps down to $n=1$; ultraviolet
- Balmer — jumps down to $n=2$; visible
- Paschen, Brackett and Pfund — jumps down to $n=3,4,5$; infrared

Worked example. The first Balmer line, from $n=3$ to $n=2$:

$$\Delta E=-1.51-(-3.4)=1.89\text{ eV},\lambda =\frac{1240}{1.89}\approx 656\text{ nm}$$

a red line in the visible region.

de Broglie's explanation. A standing electron wave must close on itself, so

$$2\pi r_n=n\lambda =\frac{nh}{mv}\Rightarrow mv{r}_n=\frac{nh}{2\pi }$$

which is exactly Bohr's second postulate.

An everyday example. A sodium street lamp glows yellow and a neon sign glows red — each gas emits its own set of lines from its own energy levels.

The substance. A gas absorbs light at the same wavelengths it emits, because both involve the same pairs of levels.