Why hydrogen glows in exact colours: electron jumps in the Bohr atom

How to use this simulation

1. The electron starts in the first orbit, n = 1, closest to the nucleus. Leave the control on “Automatic” and watch it hop up and down a fixed list of orbits, or switch to “My choice”.
2. Pick a higher orbit, say n = 3, from the buttons. A coloured photon flies in from the edge, the electron absorbs it and jumps outward. The photon had to carry exactly 12.09 eV, no more and no less, because that is the gap between n = 1 and n = 3.
3. Now pick n = 2. The electron falls and a photon flies out. Read the readings: it carries 1.89 eV and has wavelength 656.3 nm, which is the red Balmer line.
4. Look at the energy diagram on the left. The arrow shows the last jump. The highlighted level is where the electron is. The levels crowd together as n grows, getting closer to 0 eV, where the electron is free.
5. Check the spectrum strip along the bottom. The white marker sits on the line your last jump produced. Lyman lines are ultraviolet, Balmer lines are mostly visible, and Paschen lines are infrared.
6. Click an orbit in the picture or a level in the diagram to send the electron there directly. Try going from n = 6 down to n = 2 in one jump, then in two or more smaller jumps, and compare the photons.

The puzzle Bohr had to solve

By 1913 two facts did not fit together. Rutherford had shown that an atom is a tiny nucleus with electrons somewhere around it. But an electron going round a nucleus is constantly accelerating, and a charge that accelerates should radiate energy away and spiral into the nucleus in a tiny fraction of a second. Atoms should not exist, yet they do.

And then there was the light. Glowing hydrogen gives out only certain wavelengths. If the electron could have any energy, the gas should glow with a smooth smear of every colour. A prism splits it into a few separate lines instead.

Bohr’s answer was to stop asking why and simply state the rules that match the observations. His atom follows three of them, and with them it gets the hydrogen spectrum right to a remarkable accuracy.

Bohr’s three rules

Here is the whole model in three sentences.

• The electron moves only in certain allowed circular orbits, called stationary orbits, and does not radiate while it stays in one.
• An orbit is allowed only if the electron’s angular momentum is a whole number n times h ÷ 2π. That whole number, n = 1, 2, 3 and so on, labels the orbit.
• The electron emits or absorbs radiation only when it jumps between two orbits. The photon carries exactly the difference of the two energies: hf = E(higher) − E(lower).

The orbits: radius grows as n²

Working out the allowed orbits gives a neat result. The radius of orbit n is n² times the radius of the first orbit, which is called the Bohr radius, a₀ = 0.0529 nm. So orbit 2 is four times as big as orbit 1, orbit 3 is nine times, and orbit 6 is thirty-six times.

The atom in the simulation is drawn to scale. That is why the first orbit is a tiny circle hugging the nucleus while orbit 6 fills the picture: the ratio of their radii is 36. The electron also moves slower in outer orbits, as its speed in orbit n is the first-orbit speed 2.19 × 10⁶ m/s divided by n.

The energy levels: negative, and crowding together

The energy of the electron in orbit n works out as Eₙ = −13.6 ÷ n² eV. The minus sign matters. Zero energy means the electron is free, far away from the nucleus, at rest. An electron bound to the nucleus has less energy than that, so its energy is negative. The more tightly bound, the more negative.

In the ground state (n = 1) the energy is -13.60 eV. To tear the electron completely away you must supply 13.6 eV. That number is the ionisation energy of hydrogen, and it matches experiment well.

Because of the 1 ÷ n² the levels get closer together as n grows. In the energy diagram on the left of the simulation, levels 4, 5 and 6 are almost on top of each other, squeezed against the 0 eV line. The big gaps are at the bottom, which is why the biggest photons come from jumps that end on n = 1.

Jumps, photons and colour

When the electron drops from a higher level to a lower one, the energy difference leaves as one photon. Going up works the same way in reverse: the atom swallows a photon, but only if the photon’s energy is exactly the gap. A photon with slightly too much or too little energy passes straight through, which is why a gas only absorbs the very same lines it emits.

A bigger gap means a higher energy photon and a shorter wavelength. The relationship is E = hc ÷ λ, which in handy units is λ (nm) = 1,239.84 ÷ E (eV). The jump from n = 2 to n = 1 releases 10.20 eV, so λ = 1,239.84 ÷ 10.20 = 121.5 nm, deep in the ultraviolet. The jump from n = 3 to n = 2 releases only 1.89 eV, so λ = 656.3 nm, which is red light.

The photon in the simulation is coloured from its wavelength. Visible light runs from about 380 nm (violet) to 750 nm (red). Photons outside that range are invisible to us, so the simulation draws ultraviolet in lavender and infrared in rose, just so you can follow them.

The Rydberg formula

Years before Bohr, spectroscopists had found, by trial and error, a formula for the wavelengths of the hydrogen lines. Bohr’s model produced exactly that formula from first principles, which is what made it so convincing.

For a jump between a lower orbit n₁ and a higher orbit n₂, the reciprocal of the wavelength is R times (1/n₁² − 1/n₂²). The Rydberg constant is R = 1.097 × 10⁷ m⁻¹.

The spectral series

All the jumps that end on the same lower orbit make a family called a series. Each series has its own named discoverer and its own region of the spectrum. Within a series the lines bunch closer and closer towards a limit, because the upper levels crowd together.

• Lyman: the biggest energy gaps, so the shortest wavelengths. They include 121.5 nm, 102.6 nm and 97.2 nm.
• Balmer: the only series with lines we can see. Hα (3 → 2) is red at 656.3 nm, Hβ (4 → 2) is blue-green at 486.2 nm, Hγ (5 → 2) is violet at 434.1 nm and Hδ (6 → 2) is violet at 410.2 nm.
• Paschen: smaller gaps, so longer wavelengths in the infrared, such as 1,875.2 nm, 1,281.9 nm and 1,093.9 nm.

Worked example 1: the blue-green Balmer line

Find the wavelength and photon energy when the hydrogen electron falls from n = 4 to n = 2.

Worked example 2: absorbing a photon

A hydrogen atom in its ground state absorbs a photon and ends in n = 3. What energy and wavelength must the photon have?

The energy needed is E₃ − E₁ = -1.51 − (-13.60) = 12.09 eV. Then λ = 1,239.84 ÷ 12.09 = 102.6 nm, which is ultraviolet (it is the Lyman line 3 → 1 run backwards). A photon of 10 eV is too weak and one of 13 eV would not stop at n = 3 either; the atom only accepts the exact gap, unless the photon is energetic enough to ionise it completely.

Worked example 3: ionisation from n = 2

How much energy frees an electron that is already in n = 2? The energy of that level is -3.40 eV, so you need 3.40 eV to reach 0 eV. The photon wavelength for that is 1,239.84 ÷ 3.40 = 364.7 nm, which is exactly the Balmer series limit.

From the ground state you need 13.6 eV, or a photon of 91.2 nm, which is the Lyman series limit. Notice that the series limit is just the ionisation energy of the lower level.

Worked example 4: how many lines can one atom make?

An electron is in n = 4. It can fall straight to 3, 2 or 1, or stop on the way. The number of different jumps is n(n − 1) ÷ 2, which is 6 for n = 4.

• From n = 5 there are 10 possible lines, and from n = 6 there are 15.
• A real gas has many atoms in many levels at once, so all those lines appear together. One atom makes only one jump at a time.

What the Bohr model gets wrong

The model is a stepping stone, not the final answer. It works for hydrogen and for hydrogen-like ions that have a single electron. For helium with one electron removed, He⁺, the energies simply scale by the square of the nuclear charge, so the ground state is 54.4 eV. It fails for atoms with two or more electrons.

It also cannot explain why some spectral lines are brighter than others, why lines split in a magnetic field, or why an electron should not radiate when it circles. Modern quantum mechanics replaces the sharp circular orbits with fuzzy clouds of probability called orbitals. The energy levels, though, come out right, and Bohr’s ladder is still the picture most people use for the first look at spectra.

Common mistakes

These cost marks every year.

• Dropping the minus sign. Eₙ is negative, and the energy of a jump is a difference of two negative numbers.
• Swapping n₁ and n₂ in the Rydberg formula and getting a negative wavelength. The smaller n goes first.
• Forgetting that R is in m⁻¹. Convert metres to nanometres at the end.
• Thinking a photon of any energy above the gap will do. For a jump between bound levels it must match exactly.
• Counting lines wrongly. One electron makes one jump at a time, but a gas shows them all.
• Saying all Balmer lines are visible. Hα to Hδ are, but the series limit at 364.6 nm is ultraviolet.

Revision list

Say these aloud before the exam.

• rₙ = n² a₀, with a₀ = 0.0529 nm. vₙ = v₁ ÷ n.
• Eₙ = −13.6 ÷ n² eV. Ground state -13.60 eV, ionisation energy 13.6 eV.
• Photon energy = E(higher) − E(lower) = hf = hc ÷ λ. λ (nm) = 1,239.84 ÷ E (eV).
• 1/λ = R (1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹.
• Lyman (n₁ = 1) ultraviolet; Balmer (2) visible; Paschen (3) infrared.
• From n there are n(n − 1) ÷ 2 possible lines.

Frequently asked questions