Why Hot Hydrogen Glows Only in a Few Sharp Colours

Atoms are far too small to see, yet firing particles at them and studying the light they give out reveals their structure. Those two clues produced the nuclear atom and the energy levels that explain why each element glows in its own set of colours.

This lesson covers Rutherford's scattering experiment, Bohr's model of hydrogen, and the hydrogen spectral series.

Rutherford's experiment showed that an atom's positive charge and almost all its mass are packed into a tiny central nucleus, with electrons moving around it through mostly empty space.

The experiment. A narrow beam of alpha particles was aimed at a very thin gold foil, and a movable zinc sulphide screen counted the flashes at different angles.

Observations and conclusions:

- Most alpha particles passed straight through — the atom is mostly empty
- A few were deflected through large angles — a strong repulsion acts from a small, dense positive region
- A very small fraction bounced almost straight back — only head-on approaches get so close to that concentrated charge

Distance of closest approach. In a head-on collision, the alpha particle stops when its kinetic energy has all become electric potential energy:

$$r_0=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}$$

Worked example. A 5.0 MeV alpha particle approaches gold, with $Z=79$ and $K=5.0\times 1.6\times 10^{-13}$ J:

$$r_0=\frac{9.0\times 10^9\times 2\times 79\times (1.6\times 10^{-19}{)}^2}{8.0\times 10^{-13}}=4.6\times 10^{-14}\text{ m}$$

The nucleus must be smaller than this, while the atom is about $10^{-10}$ m across.

An everyday example. If an atom were a cricket stadium, its nucleus would be roughly a mustard seed at the centre.

The substance. Rutherford's model could not explain stability — an orbiting electron should radiate energy and spiral into the nucleus, which is the gap Bohr's postulates filled.

**Bohr postulated that electrons move in certain stable orbits without radiating, where angular momentum is $mvr=\frac{nh}{2\pi }$, and emit or absorb light only when jumping between orbits; this gives $r_n=0.053n^2$ nm, $v_n=\frac{2.19\times 10^6}{n}$ m s${}^{-1}$ and $E_n=-\frac{13.6}{n^2}$ eV.

The postulates:**

- Electrons revolve in certain orbits without radiating energy
- In these orbits, angular momentum is a whole-number multiple of $h/2\pi$
- A jump from $E_2$ to $E_1$ emits a photon with $hf=E_2-E_1$

The derivation. The electric force provides the centripetal force, $\frac{m{v}^2}{r}=\frac{e^2}{4\pi {\epsilon }_0{r}^2}$. Combining this with $mvr=nh/2\pi$ gives

$$r_n=\frac{{\epsilon }_0{n}^2{h}^2}{\pi m{e}^2}{r}_1=\frac{8.85\times 10^{-12}\times (6.63\times 10^{-34}{)}^2}{\pi \times 9.1\times 10^{-31}\times (1.6\times 10^{-19}{)}^2}=5.3\times 10^{-11}\text{ m}$$

The total energy, kinetic plus potential, is $E_n=-\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}$, which works out to $-13.6/n^2$ eV.

Worked example — the second orbit.

$$r_2=4\times 0.053=0.212\text{ nm}{v}_2=\frac{2.19\times 10^6}{2}=1.09\times 10^6{\text{ m s}}^{-1}{E}_2=-\frac{13.6}{4}=-3.4\text{ eV}$$

An everyday example. Sodium vapour street lamps glow yellow because the electrons in sodium atoms drop between two particular energy levels.

The substance. The energy is negative because the electron is bound — 13.6 eV must be supplied to free it from the ground state.

**The Rydberg formula $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, with $R=1.097\times 10^7$ m${}^{-1}$, gives every hydrogen line, and each series is set by the lower level $n_1$.

The five series:

- Lyman**, $n_1=1$ — ultraviolet
- Balmer, $n_1=2$ — visible
- Paschen, $n_1=3$ — infrared
- Brackett, $n_1=4$ — infrared
- Pfund, $n_1=5$ — infrared

Worked example 1 — the red Balmer line, from $n=3$ to $n=2$:

$$\frac{1}{\lambda }=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{9}\right)=1.524\times 10^6{\text{ m}}^{-1}\Rightarrow \lambda =656\text{ nm}$$

Worked example 2 — the first Lyman line, from $n=2$ to $n=1$: $\frac{1}{\lambda }=1.097\times 10^7\times \frac{3}{4}$, so $\lambda =122$ nm.

Series limit. Letting $n_2$ become infinite, the Balmer limit is $\lambda =4/R=365$ nm.

Number of lines. An electron falling from level n can produce $\frac{n(n-1)}{2}$ different lines — 6 lines from $n=4$.

An everyday example. Diwali fireworks glow in different colours because each metal salt's atoms emit their own set of spectral lines.

The substance. The first line of each series has the longest wavelength, and the series limit the shortest, because the energy gaps shrink as n rises.