Bohr's model and the energy levels

From Rutherford to Bohr

The course "Modern Physics: Light, Atoms, and Quanta" gave Bohr's results in short form. In this lesson we derive them, and then we use them on ions and on collisions.

Bohr kept Rutherford's atom: one electron circling a nucleus of charge $+Ze$. Then he added three rules.

1. Stationary orbits. The electron can move only in certain circles. In them it does not radiate, although it is accelerating.
2. Quantised angular momentum. The angular momentum in an allowed orbit is a whole number of units of $\hbar = \dfrac{h}{2\pi}$:

3. Jumps. The atom radiates or takes in light only when the electron jumps from one orbit to another. The photon has the energy difference: $h\nu = E_i - E_f$.

Deriving the radius and the speed

For an electron of mass $m$ in the orbit of radius $r$, the pull of the nucleus provides the centripetal force:

Rule 2 gives $v = \dfrac{n\hbar}{mr}$. Put this in (1). The $r$ in the bottom line leaves, and we find

The length $a_0$ is the Bohr radius, the radius of the first orbit of hydrogen. The speed follows from rule 2:

The speed in the first orbit of hydrogen is about $\dfrac{1}{137}$ of the speed of light. So Bohr could use ordinary mechanics.

Deriving the energy

The kinetic energy comes from (1): $K = \tfrac12 mv^2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{2r}$. The potential energy of the electron and the nucleus is $U = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r}$. So $U = -2K$, and the total energy is

For the first orbit of hydrogen, $K = +13.6\ \text{eV}$, $U = -27.2\ \text{eV}$ and $E = -13.6\ \text{eV}$. The total energy is negative because the electron is bound. It takes $13.6\ \text{eV}$ to free it.

An example. The ion $\text{Be}^{3+}$ has one electron and $Z = 4$. Take the level $n = 3$.

1. Radius. $r_3 = \dfrac{0.0529\ \text{nm} \times 9}{4} = 0.119\ \text{nm}$.
2. Speed. $v_3 = \dfrac{4}{3} \times 2.19 \times 10^{6}\ \text{m/s} = 2.92 \times 10^{6}\ \text{m/s}$.
3. Energy. $E_3 = -\dfrac{13.6\ \text{eV} \times 16}{9} = -24.2\ \text{eV}$.

The charge $Z$ shrinks the orbits and speeds the electron up. It makes the levels $Z^2$ times deeper.

Set $Z = 1$ and $n_i = 3$, $n_f = 2$. Then raise $Z$ to 2. Does the photon get more energy or less? By what factor?

The time for one turn

In the orbit $n$, the electron goes round a circle of length $2\pi r_n$ at speed $v_n$. One turn takes

The outer orbits are slow. The time grows as $n^3$.

Exciting an atom by collisions

An atom can also be raised to a higher level by a collision. A bombarding electron can give any part of its energy, so long as the atom takes at least one gap. A photon is different. It must have exactly the energy of a gap, or the atom does not take it in.

So an electron beam of $12\ \text{eV}$ can lift hydrogen from $n = 1$ to $n = 2$, and leaves the rest of its energy as its own motion. This was shown by Franck and Hertz in 1914. It was a second proof that atoms have only certain energies.

De Broglie's explanation of the orbits

Why should the angular momentum be $n\hbar$? In 1923 Louis de Broglie gave a reason. An electron is a wave of wavelength $\lambda = \dfrac{h}{mv}$. A wave that goes round a circle meets itself again. It can form a steady pattern only if a whole number of wavelengths fits in the circle:

That is Bohr's rule. For the first orbit of hydrogen, $\lambda = \dfrac{6.63 \times 10^{-34}\ \text{J s}}{9.1 \times 10^{-31}\ \text{kg} \times 2.19 \times 10^{6}\ \text{m/s}} = 0.333\ \text{nm}$. The circumference, $2\pi \times 0.0529\ \text{nm}$, is $0.332\ \text{nm}$: one wavelength.

Where Bohr's model stops

• It gives the right levels for hydrogen and for ions with one electron. It fails for atoms with two or more.
• It cannot say how bright each line is.
• It cannot explain why the orbits do not radiate. It just says so.

Quantum mechanics took its place. It keeps Bohr's main idea: an atom has only certain energies.

Practise with new numbers

Try it in Python

The program prints the radius, speed and energy of the first five levels of an atom with one electron.

Common mistakes

• Putting $Z$ in the wrong place. The radius has $Z$ in the bottom line. The speed has it in the top line. The energy has $Z^2$ in the top line.
• Calling the energy positive. A bound electron has negative total energy.
• Mixing the orbit number and the charge. $n$ numbers the levels. $Z$ is the nuclear charge.
• Using Bohr's model for an atom with two electrons. It works only with one.

Remember these

• $r_n = \dfrac{0.0529\ \text{nm}\,n^2}{Z}$, $v_n = \dfrac{2.19 \times 10^6\ \text{m/s}\,Z}{n}$, $E_n = -\dfrac{13.6\ \text{eV}\,Z^2}{n^2}$.
• $K = -E$, $U = 2E$. Ionisation energy from the level $n$: $\dfrac{13.6\,Z^2}{n^2}$ eV.
• A photon is taken in only if its energy is exactly a gap. An electron in a collision can give part of its energy.
• de Broglie: $2\pi r = n\lambda$ gives $mvr = \dfrac{nh}{2\pi}$.
• Bohr's model is right for one-electron atoms only.

Check yourself

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

1. By what factor is the radius of the level $n = 3$ of $\text{Li}^{2+}$ larger than that of the first level of hydrogen?

2. In the first orbit of hydrogen, the kinetic energy is $13.6\ \text{eV}$. What are the potential energy and the total energy?

3. As $n$ grows, does the speed of the electron grow or fall?

4. How does de Broglie's idea explain that only certain orbits are allowed?