Physics 6 Tutorial 2  The Bohr Model of the Atom
Contents 
The electron layout of an atom is usually presented as electrons orbiting a nucleus. It gives us an easy to understand arrangement of the nucleus and the electrons. Consider the lithium atom:
Here we see the nucleus in the centre, with the electrons arranged in shells. This was the structure of the atom as proposed by Neils Bohr (1885  1962). He linked together the spectral behaviour of the hydrogen atom with his model of the atom. While his explanation was not complete, it is still a good representation and useful for introducing quantum theory to students.
Bohr's Model gives more than the simple structure of an atom. It can be used to explain how excited atoms give out spectral lines. You can look this up in Quantum 4 for the hydrogen atom.
The hydrogen atom is the simplest atom, consisting of a single proton and a single electron.
Bohr's model went on to show how an excited electron could occupy higher orbits like this:
The electron normally occupies the ground state (n = 1) but when excited, it occupies a higher level. We can show the different states as higher orbits in this model. If the electron has sufficient energy, it can go to one of these higher orbits. It almost immediately falls back to the ground state, which it can do by dropping the whole way back to the ground state in one leap, or by dropping down the orbits one (or two) at a time. So if we have an electron at n = 3, the photons emitted are:
Red as the electron falls from n = 3 to n = 2;
UV as the electron falls from n = 2 to n = 1;
Shorter wavelength UV as the electron falls straight from n = 3 to n = 1.
We can quantify this in an equation. If an electron is at an excited level (E_{1}) and makes a transition to a lower level (E_{2}), then the energy DE of the photon given out can be worked out with the equation:
DE = E_{1} – E_{2}
The strange looking symbol D is Delta, a Greek capital letter 'D'. It is physics code for 'change in' or 'difference in'.
Since DE = hf, we can rewrite this as:
hf = E_{1} – E_{2}
Energy of an Electron in Orbit
Bohr used classical physics to describe the behaviour of the electron in an orbit.
Consider an electron of mass m and charge e orbiting a nucleus of charge of +Ze at a linear speed of v and a radius of r. We know that the centripetal force is:
Strictly speaking, we should write:
This is because centripetal force is an attractive force.
We also know from Fields 4 that for the force in an electric field:
Where Q is the charge.
In the hydrogen atom, the proton number Z = 1. So we have one electron and one proton, both of which carry the charge of magnitude 1e. However the proton has a positive charge, and the electron has a negative charge. The equation becomes:
The minus sign tells us that the force is attractive.
We can equate these two equations and write:
This is handy because we can get rid of the minus signs (which can be a damned nuisance at times). So we write:
The r term on the left cancels with the r^{2} term on the right to give:
The kinetic energy of the orbiting electron can also be found:
This gives us:
Since there is an electric field, there is potential energy The potential energy in an electric field is discussed in Fields 5. We can work out the potential energy using:
The potential energy is negative, because work is obtained by bringing in the negative charge from infinity to the radius r. The total energy is the sum of the kinetic and potential energies. Therefore:
If there are Z protons, we can rewrite this as:
This is the energy of a single electron orbiting a bare nucleus. We would achieve this with a hydrogen atom, an He^{+} ion, or an Li^{2+} ion. So this treatment is limited.
The radius of the ground state electron orbit in a hydrogen atom is 5.29 × 10^{11} m. This is called the Bohr radius.
(a) What is the total potential energy of the electron in the ground state in a hydrogen atom? (b) What is this in eV? (c) Comment on what this represents.
Electronic charge = 1.6 × 10^{19} C Permittivity of free space = 8.85 × 10^{12} F m^{1}.

The minus sign is important. It means that when an electron comes in from infinity, work is got out. Similarly, work has to be put in to raise the electron to infinity, i.e. to ionise the atom. In the case of hydrogen, this is 13.6 eV. We use the idea of an energy well, in which work has to be done to remove the electron.
Angular Momentum in an Orbit
In Quantum 6, we saw that the de Broglie relationship was:
When we discussed the de Broglie relationship, we did so in the context of an electron travelling in a straight line. However any particle with mass moving in any path with a velocity is going to have momentum.
An electron making a circular orbit will have an angular momentum:
L = Iw
Since the electron is a single particle, we can write an expression for the moment of inertia.
I = mr^{2}
So we can write an expression for angular momentum:
L = wmr^{2}
We also know from circular motion that:
v = wr
Therefore:
So we can substitute for for w:
This is true for any circular orbit. In particle physics, this is called orbital spin.
de Broglie Wavelength and the Bohr Model
An electron orbiting a nucleus forms a de Broglie standing wave. The de Broglie wavelength of an electron wave at its fundamental frequency is the circumference of its orbit.
The orbiting electron forms a standing wave. For a standing wave to be formed under these circumstances, there needs to be a whole number, n, of waves. At the fundamental frequency (or first harmonic) , n = 1.
We know that for each orbit, the electron will travel a distance of 2pr. So we can write:
nl = 2pr
From the de Broglie equation:
This further rearranges to:
The radius of the electron orbit in a hydrogen atom is 5.29 × 10^{11} m. This is called the Bohr radius.
An electron is orbiting a hydrogen nucleus at the ground state. Calculate: (a) the angular momentum and give the correct unit; (b) the linear speed of the electron; (c) the de Broglie wavelength of the electron, to 2 significant figures; (d) the angular velocity of the electron; (e) the frequency of electron orbit.
Planck constant = 6.63 × 10^{34} J s Mass of electron = 9.11 × 10^{31} kg 
We can use the photon energy formula to work out the wavelength of the photon emitted when the electron drops from the ionised state to the ground state.
The ionisation energy of the hydrogen atom is 13.6 eV. (a) Calculate the wavelength of photons emitted when the electron falls from the ionised state to the ground state. (b) Is this the same as the de Broglie wavelength?
Planck constant = 6.63 × 10^{34} J s 
The de Broglie wavelength is NOT the wavelength of emitted photons. 
The orbits which are occupied by an electron raised to higher energy levels are a whole number multiple of the de Broglie Wavelength.
Application of the Bohr Model to Excited Atoms
In Quantum 4, we saw that the energy ladder for the different energy levels was not even. The jump from the ground state (n = 1) to (n = 2) is 3.41 eV  13.6 eV = 10.19 eV. When it's excited, the diameter of the atom will change. Let's see how:
Worked Example At the n = 2 level, the potential energy of the excited electron of a Hydrogen atom is 3.41 eV. What is the diameter of the excited atom?
How does this compare with the diameter of the atom at ground state?
Electronic charge = 1.6 × 10^{19} C Permittivity of free space = 8.85 × 10^{12} F m^{1}.

Answer Formula first:
E_{tot} = 3.41 eV × 1.60 × 10^{19} J eV^{1} = 5.456 × 10^{19} J
r = (1 × (1.6 × 10^{19} C)^{2}) ÷ (8 × p × 8.85 × 10^{12} F m^{1} × 5.456 × 10^{19} J) = 2.11 × 10^{10} m
Ratio of the radii = 2.11 × 10^{10} m ÷ 5.29 × 10^{11} m = 3.99 times
The atom swells to about four times its original size. 
So what happens to the speed of the electron? To find out, we use:
Calculate the speed of the electron as it orbits the nucleus at level n = 2, if the radius of the excited atom is 2.11 × 10^{10} m. Compare your answer to Question 2 (b)
Planck constant = 6.63 × 10^{34} J s Mass of electron = 9.11 × 10^{31} kg

Drawbacks of the Bohr Model
Bohr's model works well for hydrogen, the simplest atom. It was worked out using classical physics. At the time, quantum physics was only just being worked out. Now we know that there are limitations:
It assumes that electrons orbit in welldefined orbits  and this contradicts Heisenberg's Uncertainty Principle;
The electrons do not orbit like satellites. They will crash into the nucleus;
It is possible for the electrons to occupy any orbit according to the potential energy. This does not comply with the quantum nature of energy levels in atoms, where an electron rises to a new energy when exactly the right energy is given to the electron. If the exact energy is not given, the electron will not absorb it.
It does not explain the Zeeman or Stark effects where spectral lines get split into components in magnetic fields or electric fields respectively.
We will look at the models of the atom that superseded the Bohr Model in the next tutorial.
Niels Bohr is said to have an impish sense of humour that enraged the Professor of Physics at Copenhagen University during his final oral exam. However it failed him when Richard Feynman introduced his Feynman diagrams  he was thunderously angry and was ready to take a swing at Feynman. As well as being a brilliant physicist, Bohr was a good footy player, and was the goalkeeper for a League Football team in Copenhagen. Bohr's brother, Harald, played in the Danish Olympic Football team in 1902, and played Hockey in the London Olympics in 1948.