17.3.1 Bohr’s Atom

In 1896, J. J. Thompson discovered the electron. In 1911, E. Rutherford discovered the nucleus, and went on to propose the Rutherford model: that each atom contains a miniature solar system, with electrons orbiting around a massive nucleus.

It’s all beautiful except one big problem: Maxwell’s Equation predicts that a time-varying electric field must result in an electromagnetic wave. If the electrons in an atom are going in circles, their acceleration must result in the emission of EM waves from the atom. The electrons should be losing energy and spiralling towards the nucleus (just like satellites losing energy must eventually crash down to Earth). In fact, based on calculations of the rate of emission, the collapse of the atom should be over and done with in the blink of an eye. The Rutherford’s atom did not look very stable. Backtrack to 1885, when J J Balmer discovered the formula to describe the hydrogen spectrum. You see, most light sources have a continuous spectrum with no break or gap throughout its frequency (and wavelength) range. For example, when sunlight is passed through a prism, the “rainbow” spectrum is a continuous spectrum: light of all frequencies and wavelengths are present.

The light emitted by hot gases, however, have discrete spectra consisting of a set of discrete spectral line: only a few colours are present.

For example, the visible spectrum from hydrogen[1] contains only four wavelengths: 410 nm, 434 nm, 486 nm and 656 nm. What’s so special about these wavelengths?

Amazingly, Balmer (who was a secondary school mathematics teacher) was able to come up with a formula for those wavelengths!

\displaystyle \frac{1}{{{{\lambda }_{n}}}}=10973731.57(\frac{1}{{{{2}^{2}}}}-\frac{1}{{{{n}^{2}}}}) for \displaystyle n=3,4,5,6,...

Impressive as the formula may look, it doesn’t explain why hydrogen atoms are emitting light only at these wavelengths and not others.

In 1913, Niels Bohr, after thinking about the instability of Rutherford’s atom, Balmer’s hydrogen spectrum and Einstein’s photon, proposed the Bohr model, which asserts the electron in the hydrogen atom is only allowed to orbit at certain radii.

Just like the Earth-satellite system has different amounts of energy depending on the altitude of the satellite’s orbit, the proton-electron system has different amounts of energy depending on the radius of the electron’s orbit. If the electron can only have certain orbital radii, then the atom can only have certain amounts of energy. We say that the energy levels of a hydrogen atom are quantized.

According to Bohr, the hydrogen atom can make a transition from a higher energy level EH to a lower energy level EL by emitting a photon with energy Ephoton equal to the energy difference between those two energy levels.

\displaystyle {{E}_{{photon}}}={{E}_{H}}-{{E}_{L}}

If the atom is allowed only certain energy levels, it would imply that it cannot emit photons of just any energy. Instead it can only emit photons whose energies correspond to the energy difference between a pair of allowed energy levels. This may explain why the spectrum is discrete and not continuous!

Using Balmer’s formula, Bohr “worked backward” [1]and deduced that the allowed energy levels for the hydrogen atom is given by the formula

\displaystyle {{E}_{n}}=-\frac{{13.6}}{{{{n}^{2}}}}\text{ eV} for \displaystyle n=1,2,3,...

Note that

  1. \displaystyle n=1 represents the lowest possible energy, aka ground level of the hydrogen atom. If a hydrogen atom is already at the ground level, it cannot lose any more energy (hence the Bohr’s atom is saved from the Rutherford atom’s death spiral).
  2. All higher levels are called excited levels.
  3. The energy levels are all negative in values. This is because the atom is held together by the attraction between the proton and the electron, making the electric potential energy of the system negative. Bohr basically took the total energy to be the summation of this EPE and the KE of the orbiting electron. So the total energy is zero only when the electron is infinitely far (i.e. free) from the proton.
  4. 13.6 eV is thus the energy required to remove the electron from the hydrogen atom. This is also called ionization energy of the hydrogen atom.

Let’s take Bohr’s model out for a test drive. Suppose the electron makes a transition from \displaystyle n=3 to \displaystyle n=2. The emitted photon thus has energy

\displaystyle {{E}_{{photon}}}=(-1.51)-(-3.40)=1.89\text{ eV}

Since \displaystyle {{E}_{{photon}}}=\frac{{hc}}{\lambda }, the emitted photon’s wavelength can be calculated as follow.

\displaystyle \displaystyle \begin{aligned}\frac{{(6.626\times {{{10}}^{{-34}}})(2.998\times {{{10}}^{8}})}}{\lambda }&=1.89(1.602\times {{10}^{{-19}}})\\\lambda &=656\text{ nm}\end{aligned}

This explains why the hydrogen spectrum contains a spectral line at 656 nm, and not say at 646 nm or 666 nm! In fact, all the spectral lines of hydrogen can now be “explained”: The transitions that end at \displaystyle n=2 (which happen to produce photons in the visible spectrum) match up to the Balmer series. In addition, the transitions that end at \displaystyle n=1 and \displaystyle n=3 produce photons in the ultraviolet region and infrared region respectively. The corresponding spectral lines have been confirmed experimentally, and are known as the Lyman and Paschen series.

As you can see, the Bohr model was strongly supported by the spectral lines of hydrogen. Even though there were still many unresolved issues, the Bohr model represented an important milestone in our understanding of the structure of the atom.

Concept Test



Continuous Spectrum Violin

Bohr’s School Rule

Bohr’s Air Con

[1] Ok. Bohr actually started from the argument that the angular momentum L of the orbiting electron must be quantized (as in \displaystyle L=n\frac{h}{{2\pi }}), but you don’t have to know such details for the H2 syllabus.

[1] The hydrogen spectrum was intensely studied because hydrogen is the simplest atom: if you can’t explain even the hydrogen atom, you have no chance with other atoms.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s