Bohr's model of the hydrogen atom

Saving the atom, and explaining the lines

Rutherford's atom has a problem: an electron that circles the nucleus should spiral inwards. And hydrogen gives out light of only certain wavelengths. In $1913$, the Danish physicist Niels Bohr made a model that answered both questions, for hydrogen and for atoms like it with a single electron, such as $\text{He}^+$ and $\text{Li}^{2+}$.

Bohr's postulates

1. Fixed orbits. The electron moves in a circular orbit around the nucleus. Only certain orbits are allowed. In such an orbit, the electron has a fixed energy, and it does not radiate. These are the stationary states.
2. Quantised angular momentum. The angular momentum of the electron in an allowed orbit is a whole number of $\hbar$:

Here $\hbar$ is read "h bar", and $n$ is the principal quantum number.

3. Jumps. The electron can jump from one allowed orbit to another. It gives out light when it falls to a lower energy, and takes in light when it rises. The photon has the energy that is the difference of the two levels:

What the model gives

For an electron around a nucleus of charge $+Ze$, the postulates and the laws of electricity give the following results for orbit number $n$.

• The radius is larger for a larger $n$, and it grows as $n^2$. The value $52.9\ \text{pm}$ is the radius of the first orbit of hydrogen, the Bohr radius.
• The energy is negative. Zero is the energy of an electron that is free and at rest, far from the nucleus. An electron held by the nucleus has less energy than this.
• The levels get closer together as $n$ grows.

The energy of the ground state

For hydrogen, $n = 1$ is the ground state, with $E_1 = -2.18 \times 10^{-18}\ \text{J} = -13.6\ \text{eV}$. To remove the electron completely, we must give it $13.6\ \text{eV}$. This is the ionisation energy of hydrogen. For one mole of atoms it is $2.18 \times 10^{-18} \times 6.022 \times 10^{23} = 1.31 \times 10^{6}\ \text{J mol}^{-1}$, which is $1312\ \text{kJ mol}^{-1}$. This agrees with the measured value.

The higher levels, $n = 2, 3, \ldots$, are excited states.

You should find that the levels crowd together near zero, and that every level of $\text{He}^+$ lies four times lower than the same level of hydrogen.

Explaining the lines

If the electron falls from level $n_2$ to level $n_1$,

The wavenumber is $\tilde{\nu} = \dfrac{\Delta E}{hc}$. Here $hc = 6.626 \times 10^{-34} \times 3.00 \times 10^{8} = 1.988 \times 10^{-25}\ \text{J m}$, and $\dfrac{2.18 \times 10^{-18}\ \text{J}}{1.988 \times 10^{-25}\ \text{J m}} = 1.097 \times 10^{7}\ \text{m}^{-1}$, so

This is exactly the Rydberg formula, and the number in front is the measured value of $R_H$. Now we see what the series are. Lyman: the electron falls to $n = 1$. Balmer: to $n = 2$. Paschen: to $n = 3$. The lines come from the jumps between levels, and each jump gives one wavelength.

Atoms like hydrogen

The same formulae work for $\text{He}^+$ ($Z = 2$) and $\text{Li}^{2+}$ ($Z = 3$), which also have one electron.

• The energy of the ground state is $Z^2$ times larger. For $\text{He}^+$ it is $-13.6 \times 4 = -54.4\ \text{eV}$. For $\text{Li}^{2+}$ it is $-13.6 \times 9 = -122.4\ \text{eV}$.
• The orbit is $Z$ times smaller. For $\text{Li}^{2+}$, the first orbit has a radius of $\dfrac{52.9}{3} = 17.6\ \text{pm}$.

Where the model fails

• It works only for atoms with one electron. For helium, with two electrons, it gives wrong values.
• Lines that look single are really made of several very close lines (the fine structure). A magnetic field splits them further (the Zeeman effect). Bohr's model cannot explain this.
• It cannot explain how atoms form bonds, or the shapes of molecules.
• It says that we know both the position and the speed of the electron exactly, which we learn in the next course is impossible.
• It does not explain the wave nature of matter.

Bohr's idea of energy levels survives in the better theory. His circular orbits do not.

Practise with new numbers

Try it in Python

This program lists the radius and the energy of the first five orbits, and the wavelength of the first Balmer line.

Remember these

• Allowed orbits have $m_e v r = n\hbar$. In these orbits the electron does not radiate.
• $r_n = 52.9\ \text{pm} \times \dfrac{n^2}{Z}$ and $E_n = -13.6\ \text{eV} \times \dfrac{Z^2}{n^2}$.
• A jump between levels gives or takes a photon: $\Delta E = h\nu$.
• The model gives the Rydberg formula, and the ionisation energy of hydrogen, $13.6\ \text{eV}$.
• It works only for one-electron species, and it cannot explain bonding or the wave nature of the electron.

Check yourself

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

1. How much energy removes the electron from $\text{He}^+$ in its ground state?

2. Which jump in hydrogen gives the red Balmer line of $656\ \text{nm}$?

3. How many times larger is Bohr's third orbit than the first, for the same atom?

4. What does Bohr's second postulate say about light?