The Bohr model of hydrogen

Bohr’s three postulates

Narrated video. Transcript:

Rutherford’s planetary atom has a serious problem. An electron going round in a circle is accelerating, towards the centre.

Classical electromagnetism says an accelerating charge radiates electromagnetic waves. So the orbiting electron should keep losing energy.

Watch what would happen. As it loses energy, the orbit shrinks, and the electron spirals inward until it falls into the nucleus. No atom would be stable.

Worse, as it spirals in, it goes round faster and faster, so the frequency of the light it gives out would change smoothly. We would see a continuous spectrum, not sharp lines.

Real atoms are stable, and they give line spectra. In 1913 Niels Bohr kept Rutherford’s nucleus, but added three new rules, his postulates.

First postulate. The electron can revolve only in certain stable orbits without radiating energy. These are called stationary states, and each has a definite total energy.

Second postulate. The allowed orbits are those for which the angular momentum is a whole-number multiple of h over 2π: L equals n h over 2π, where n is 1, 2, 3 and so on. n is the principal quantum number.

Third postulate. An electron can jump from one allowed orbit to a lower one. It then emits a photon whose energy equals the energy difference: h ν equals E i minus E f.

So energy goes in or out only in definite packets. That is the seed of the line spectrum. Next, let us find the allowed orbits.

Radius and speed of the orbits

Narrated video. Transcript:

We need two equations. The Coulomb force supplies the centripetal force: m v² over r equals 1 over 4πε₀, times e² over r².

And Bohr’s quantum condition: m v r equals n h over 2π.

Divide the first equation by the second, and r cancels. The speed in the nth orbit is v n equals e² over 4πε₀, divided by n h over 2π. Simplified, v n equals e² over 2 ε₀ h n.

Put in the numbers for n equal to 1: v₁ is about 2.18 × 10⁶ metres per second. That is about one 137th of the speed of light. In higher orbits the electron is slower: v is proportional to 1 over n.

Now put v n back into m v r equals n h over 2π. The radius of the nth orbit is r n equals n² h² ε₀ over π m e².

For n equal to 1 this is the Bohr radius, a₀ equals h² ε₀ over π m e². It works out to 5.29 × 10⁻¹¹ metre, about half an ångström.

So r n equals n² a₀. The second orbit is 4 times as big as the first, the third 9 times, the fourth 16 times. The orbits spread out quickly.

Let us check n equal to 3: r₃ equals 9 × 0.53 ångström, which is 4.77 ångström, and v₃ equals 2.18 × 10⁶ over 3, about 7.3 × 10⁵ metres per second.

Energy levels

Narrated video. Transcript:

The total energy of an electron in an orbit of radius r is E equals minus e² over 8πε₀ r. Now put in r n.

The result is E n equals minus m e⁴ over 8 n² ε₀² h². Putting in the constants, E n equals minus 13.6 over n² electron volt.

Draw these energies as horizontal lines: an energy level diagram. n equal to 1 is at minus 13.6 eV. n equal to 2 is at minus 3.4 eV. n equal to 3 at minus 1.51, n equal to 4 at minus 0.85.

The levels crowd together as n grows, and approach zero. At n equal to infinity, E is zero: the electron is free of the nucleus.

The lowest level, n equal to 1, is the ground state. The atom normally sits there. The levels above it are excited states.

To lift the electron from the ground state to n equal to 2, the atom needs exactly minus 3.4 minus minus 13.6, which is 10.2 eV. This is the first excitation energy.

To remove the electron completely, from n equal to 1 to infinity, takes 13.6 eV. This is the ionisation energy of hydrogen, and it matches the value measured in experiments.

Notice the signs. Kinetic energy K equals minus E n, a positive number. Potential energy U equals 2 E n. For the ground state, K is 13.6 eV and U is minus 27.2 eV.

So the Bohr model gives definite radii, speeds and energies, all fixed by one whole number, n.

Try it: move the electron between levels

Interactive simulation. Things to try:

• Tap levels 1 to 6 in turn. Watch the orbit radius grow as n² and the speed fall as 1/n.
• Check the readouts: kinetic energy is always −E_n and potential energy 2E_n.
• Go from n = 1 up to n = 2: the atom must absorb 10.2 eV. Press ∞ from n = 1: the ionisation energy, 13.6 eV.
• Notice how the levels crowd together just below 0 eV.

The third orbit

Compared with the first Bohr orbit of hydrogen, the electron in the third orbit has

1. radius 3 times, speed 1/3
2. radius 9 times, speed 1/3
3. radius 9 times, speed 1/9
4. radius 9 times, speed 3 times

Ionising an excited atom

A hydrogen atom is in its first excited state (n = 2). How much energy is needed to ionise it?

1. 13.6 eV
2. 10.2 eV
3. 3.4 eV
4. 6.8 eV

Summary

• Quantum condition: L = mvr = nh/2π, n = 1, 2, 3, …
• Radius and speed: r_n = n² a₀ (a₀ = 5.29 × 10⁻¹¹ m), v_n = v₁/n
• Energy levels: E_n = −13.6/n² eV
• Photon emitted: hν = E_i − E_f

• Classically, an orbiting electron would radiate and spiral into the nucleus, giving a continuous spectrum. Bohr’s postulates avoid this.
• Electrons move without radiating in stationary states whose angular momentum is a multiple of h/2π.
• A jump between states emits or absorbs one photon with hν equal to the energy difference.
• Ground state −13.6 eV; ionisation energy 13.6 eV; first excitation energy 10.2 eV. K = −E_n, U = 2E_n.