Atomic Models and Energy Levels

From plum pudding to the nuclear atom

By 1900 scientists knew that atoms contain negative electrons, yet atoms are neutral overall. So where is the positive charge, and how is everything arranged?

Thomson's model (the "plum pudding", 1904). The atom is a sphere of positive charge spread out evenly. The tiny negative electrons are embedded in it like plums in a pudding. The total charge is zero.

Rutherford's model (1911). Geiger and Marsden, working under Rutherford, fired alpha particles ($\alpha$, positively charged) at very thin gold foil and looked for flashes on a fluorescent screen. This is the alpha-particle scattering experiment (the gold foil experiment).

Observations and what they show:

In Rutherford's model, electrons orbit a tiny, dense, positive nucleus, much like planets around the Sun. The positive charge in Thomson's model was spread thinly, so it could only ever deflect an $\alpha$ particle by a tiny angle. Large deflections were impossible, and that is why the experiment ruled it out.

Bohr's model of hydrogen

In 1913 Niels Bohr kept Rutherford's picture of electrons orbiting a nucleus, but added two bold postulates for the hydrogen atom:

The orbits are labelled by the whole number $n = 1, 2, 3, \dots$ (the principal quantum number). For hydrogen the energy of level $n$ is

The energies are negative because the electron is bound to the nucleus. We take zero energy to mean a free electron, completely separated from the atom ($n = \infty$). The lower (more negative) the energy, the more tightly bound the electron is.

The ground state is the lowest energy level ($n = 1$), the state in which the atom is normally found. Any higher level is an excited state.

Notice how the levels crowd together as $n$ increases. The gap between $n = 1$ and $n = 2$ is huge, while the gaps near the top are tiny.

Energy levels and photons

An atom changes level only by emitting or absorbing a photon.

• Emission: the electron drops from a higher level $E_2$ to a lower level $E_1$. A photon is emitted with energy $hf = E_2 - E_1$.

• Absorption: the electron jumps up. It must absorb a photon whose energy is exactly equal to the gap between the two levels.

Since $c = f\lambda$, the wavelength of the photon is $\lambda = \dfrac{c}{f} = \dfrac{hc}{E_2 - E_1}$. The equation $E_2 - E_1 = hf$ is in the Formulae and Tables booklet.

To work in joules, use $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$. Use $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$.

Spectra, ionisation and the diffraction grating

Now we can answer the opening question. Only certain energy gaps exist in an atom, so only certain photon energies, and therefore only certain colours, can be emitted. That is why a gas gives a line spectrum and not a continuous rainbow. Each element has its own set of energy levels, so its set of lines is like a fingerprint. Astronomers identify the elements in stars by their spectra.

Counting lines

If an atom is excited to level $n$, it may fall back in several steps. Each different pair of levels gives a different line. From $n = 3$ the possible transitions are $3 \to 2$, $3 \to 1$ and $2 \to 1$, so there are three lines.

Excitation and ionisation

• Excitation lifts an electron to a higher level. A photon can do this only if its energy matches a gap exactly. A bombarding electron needs only to have at least that much kinetic energy, because it can give up part of its energy and keep the rest.

• Ionisation removes the electron completely ($n \to \infty$). The ionisation energy of hydrogen is $13.6\ \text{eV}$ from the ground state. A photon with more than this energy can also ionise the atom, and the extra energy becomes the kinetic energy of the freed electron.

• The longest wavelength that can ionise corresponds to the minimum energy: $\lambda = hc/E$.

Measuring wavelengths with a diffraction grating

A diffraction grating with $N$ lines per metre has spacing $d = 1/N$. The maxima occur where $d\sin\theta = m\lambda$, with $m = 1, 2, \dots$ the order. Each wavelength is sent to a different angle, so the lines are separated.

Where the numbers come from: the Bohr orbit

Bohr treated the hydrogen electron (mass $m$, charge $-e$) as moving in a circle of radius $r$ around the proton. The electrostatic attraction supplies the centripetal force:

where $k = 8.99 \times 10^{9}\ \text{N m}^2\text{C}^{-2}$ (this is $k = \dfrac{1}{4\pi\varepsilon_0}$). His second idea was that the angular momentum is quantised: $mvr = \dfrac{nh}{2\pi}$. Only orbits that satisfy this are allowed, which gives radii $r_n = n^2 \times 5.3 \times 10^{-11}\ \text{m}$.

Because $r_n \propto n^2$, the energy $-\dfrac{ke^2}{2r_n}$ is proportional to $-\dfrac{1}{n^2}$, which is exactly the formula $E_n = -13.6/n^2$ eV. The model reproduces the hydrogen spectrum very well.

Back to the start: neon signs and street lights glow in sharp colours because the atoms in the gas have only certain energy gaps. Each electron jump releases a photon of one definite energy, and therefore one definite colour.

Summary

• Thomson: the atom is a ball of positive charge with electrons embedded in it. Rutherford (gold foil scattering): most of the atom is empty space and the positive charge and mass are concentrated in a tiny nucleus.

• Bohr's postulates: electrons occupy allowed orbits (energy levels) without radiating, and a jump between levels emits or absorbs one photon, $E_2 - E_1 = hf$.

• Hydrogen levels: $E_n = -13.6/n^2$ eV. The ground state is $n = 1$; ionisation energy $= 13.6$ eV. Use $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$.

• Wavelength of a photon: $\lambda = \dfrac{hc}{E_2 - E_1}$. A photon is absorbed only if its energy matches a gap exactly, whereas an electron needs at least the gap energy.

• Discrete levels give line spectra: bright lines (emission) or dark lines (absorption) at the same wavelengths. A grating separates them using $d\sin\theta = m\lambda$.

• The electrostatic force gives the centripetal force, so $E_k = \dfrac{ke^2}{2r}$ and $E = -\dfrac{ke^2}{2r}$.