Bohr's Model of the Hydrogen Atom

Coloured lines

Heat hydrogen gas, and look at its light through a prism. You do not see a smooth rainbow. You see a few sharp coloured lines, and nothing in between.

Why only these colours? In 1913, Niels Bohr gave an answer.

Bohr's three ideas

1. The electron goes round the nucleus in a circle. Only certain orbits are allowed. In these orbits, the electron does not radiate energy.
2. The allowed orbits are those where the angular momentum is a whole number multiple of $\hbar$. The symbol $\hbar$ is read "h bar", and it means $\dfrac{h}{2\pi}$.

3. The atom gives out or takes in light only when the electron jumps from one orbit to another. The photon has exactly the energy difference between the two orbits.

What the model gives for hydrogen

If we put idea 2 together with the pull of the nucleus, we get the energy and the radius of every orbit.

• $n = 1$ is the ground state. Its energy is $-13.6\ \text{eV}$.
• The energy is negative because the electron is bound. Energy zero means that the electron has just escaped. So it takes $13.6\ \text{eV}$ to ionise a hydrogen atom from the ground state.
• When $n$ gets bigger, the levels come closer together, and they approach zero.

Atoms with one electron

The model works for any atom that has only one electron. Examples are $\text{He}^+$ and $\text{Li}^{2+}$. If the nucleus has $Z$ protons, then

For $\text{He}^+$, $Z = 2$. Its ground state is $-13.6\ \text{eV} \times 4 = -54.4\ \text{eV}$.

Spectral lines

When the electron falls from a level $n_i$ to a lower level $n_f$, the atom gives out a photon.

The lines are grouped by the level where the electron lands.

• Lyman series, $n_f = 1$: ultraviolet.
• Balmer series, $n_f = 2$: visible light. It includes the red line of $656\ \text{nm}$.
• Paschen series, $n_f = 3$: infrared.

An atom in level $n$ can come down in many ways. A gas with many atoms can give up to $\dfrac{n(n-1)}{2}$ different lines.

You should find that jumps that land on $n = 1$ give ultraviolet photons, below about $122\ \text{nm}$, while jumps that land on $n = 2$ give visible light.

The atom can also take in a photon. But the photon must have exactly the energy of the gap between two levels. A photon of $10.2\ \text{eV}$ raises the electron from $n = 1$ to $n = 2$. A photon of $10\ \text{eV}$ is not taken in.

Where the model stops

Bohr's model gives the right answer for hydrogen and for ions with one electron. It fails for atoms with more electrons. It also cannot tell how bright each line is. Quantum mechanics took its place. But it keeps Bohr's main idea: an atom can have only certain energies.

Practise with new numbers

Try it in Python

This program finds the energy and the wavelength of the lines that end on $n = 2$.

Remember these

• $E_n = -\dfrac{13.6}{n^2}$ eV for hydrogen, and $-\dfrac{13.6\,Z^2}{n^2}$ eV for an ion with one electron.
• A jump gives a photon with energy $E_{\text{upper}} - E_{\text{lower}}$. Then $\lambda = \dfrac{1240}{E}$ nm.
• Lyman: ultraviolet. Balmer: visible. Paschen: infrared.
• To ionise an atom in level $n$, we need $\dfrac{13.6}{n^2}$ eV.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the wavelength of the photon from the jump $n = 3$ to $n = 1$?

2. How much energy ionises hydrogen from the level $n = 3$?

3. What is the energy of the ground state of $\text{He}^+$?

4. Why does hot hydrogen give only certain colours, and not a smooth rainbow?