Physics 6 Tutorial 16  The Quantum Atom
Contents 
In Quantum Physics Tutorial 4, we saw how electrons in excited atoms go up to certain energy levels. About a microsecond later, they fall back to lower energy levels releasing photons of very specific wavelengths. We also saw how a photon that excited an electron had to have exactly the right energy. A fraction less, the photon is not absorbed; a fraction more, it is not absorbed either. This explains how we see specific wavelengths in an emission spectrum.
We developed the theory further in Physics 6 Tutorial 2 to look at the Bohr Model of the hydrogen atom.
You may wish to review these tutorials before going on.
The Hydrogen Line Spectrum
The visible line spectrum of hydrogen consists of just four lines of different colours, like this:
However there are many more spectral lines that are caused by transitions between different energy levels:
Image by OrangeDog, Wikimedia Commons
The transitions give photons of the shortest wavelength in the ultraviolet region (100 nm). These are the most energetic. Wavelengths of 1000 nm are in the infra red region, while wavelengths of 10 000 nm (1 × 10^{5} m or 10 mm) are in the far infrared region, getting on towards the microwave region. These transitions result in photons of the least energy.
When electrons are excited, they will go to the level of the exciting radiation, provided that the exciting energy is of exactly the right energy. Within 1 ms, they will fall back to lower energy levels, emitting photons.
If the exciting radiation is 13.6 eV or above, the electrons will be removed, and the hydrogen atom will be ionised. However electrons will be attracted back to the ion. These electrons can go from the 0 eV level (the ionisation level, where the electron can escape) all the way down to the ground state (n = 1, 13.6 eV) in one big leap, or in many different steps. We also know that the rungs of the energy ladder are not equally spaced. Therefore we can see that many photons of different wavelengths can be observed. An electron dropping down can pause on any rung of the energy ladder on their way to the ground state. The lower the rung, the higher the energy of the photons emitted. The photons of the highest energy result from transitions from higher energy levels all the way down to the ground state. Therefore a series of wavelengths in the UV region is observed.
When observing ionised hydrogen, we observe all the emitted lines at the same time. This is because there are may thousands of millions of atoms being ionised. Also the way electrons fall down the electron levels can be in any combination.
If the electrons pause at the level above the ground state (n = 2), photons in the visible region are observed. These form another series of wavelengths. The idea is shown in this diagram:
Note that the levels in the diagram are regularly spaced for clarity. In reality the levels are not evenly spaced.
The transitions ending at n = 1 are called the Lyman series, after the American physicist, Theodore Lyman (1874  1954), who discovered them. Transitions that end at n = 2 are the Balmer series, named after Johann Balmer (1825  1898), a Swiss physicist. Transitions that end at n = 3 are called the Paschen series, after the German physicist, Friedrich Paschen (1865  1948).
Note that when the electron reaches its end level, it will still fall to the ground state, contributing to the Lyman series.
All the photons emitted in the Lyman series are in the UV region. This is because the transition from n = 2 to n = 1 is from 3.41 to  13.6 eV, a leap of 10.19 eV. A transition from n = 3 to n = 1 would give a photon of more energy, hence shorter wavelength. The main Lyman lines are shown below:
The lines are marked Ly (for Lyman  really?), with a Greek letter, for example Lya. In the energy ladder, the lines are the result of these transitions:
Worked example Calculate the wavelength of a photon emitted when an electron falls from level n = 6 to n = 1. 
Answer We need to work out the value of the energy change as the electron falls: DE = E_{1} – E_{2}
DE =  0.38 eV  13.59 eV = 13.21 eV
We need to convert this to joules: DE = 13.21 × 1.60 × 10^{19} J eV^{1} = 2.11 × 10^{18} J
We now need to use:
Therefore: l = (6.63 × 10^{34} J s × 3.00 × 10^{8} m s^{1}) ÷ 2.11 × 10^{18} J = 9.41 × 10^{8} m = 94.1 nm

The photons emitted in the Balmer series are in or around the visible light region. All the transitions end at n = 2. The lines are marked with Ba with a Greek letter. Their place on the energy ladders are shown:
The grey background is to make the bluegreen line Bad stand out a bit more.
The line Baa has a wavelength of 367 nm. Violet light has a minimum wavelength of 380 nm, so the shorter wavelength indicates that the Baa line is in the UV region. This is consistent with the observation of just four visible Balmer lines.
As with the two series above, the lines are marked with the letters Pa, with a Greek letter. The transitions end up at n = 3. They are in the infrared region. The idea is shown below:
There are series that end at n = 4 (Brackett Series), n = 5 (Pfund Series), n = 6 (Humphreys Series). The lines get fainter, as these transition are increasingly rare events. Transitions at n = 7 have been found in the far infrared region.
Equation for the Energy Levels
The electron energy levels in the excited hydrogen atom do not form a ladder with equally spaced rungs. They are very irregularly spaced. The energy at a certain level can be predicted using a simple equation:
E_{n} is the energy at a certain energy level (eV);
E_{0} is the ionisation energy (eV);
n is the energy level.
The spacing gets smaller and smaller, as would be expected with the energy being proportional to 1/n^{2}.
In the hydrogen atom, E_{0} is 13.6 eV. So we can rewrite the equation:
We can model this on a spreadsheet:
Energy Level  Energy level / eV 
1  13.60 
2  3.40 
3  1.51 
4  0.85 
5  0.54 
6  0.38 
7  0.28 
8  0.21 
9  0.17 
10  0.14 
We can plot these data a bargraph. Note that a line graph would not be appropriate here due to the quantum nature of the energy levels. The energy levels are integers. You could, of course, plug numbers into the equation where n = 1.25, and get a number out of it (E_{n} = 8.704 eV), but that number would be meaningless.
Notice that the signs are negative. This means that work is got out when the electron falls from the excited state to the ground state. The equation above is positive, showing the work required to raise an electron from an energy level to the ionised state.
Electron Standing Waves (*)
This will be examined only in the extension questions. It is difficult.
Electron behaviour is impossible to describe using classical physics. An orbiting electron can be in two places at once. It can appear anywhere within a shell without appearing at any point in between. There are certain points in which electrons have a high probability of appearing, and where the probability of them appearing is very low. We can model this as a standing wave.
We can use much of what we know about standing waves to explain the standing wave behaviour of electrons. Standing wave behaviour is covered in Waves Tutorial 4. You may wish to revise it before continuing.
In a resonating string, the line represents the position of the string in the standing wave. With electrons, the area enclosed by the line presents the probability of finding the electron. We will use the same ideas from standing waves in strings to help us to understand the way electron standing waves work. The classic way that we represent a standing wave in a string is a standing wave in one dimension.
We know that:
l = 2L
We also know the de Broglie relationship that links de Broglie wavelength and linear momentum:
Rearranging:
So we can write:
or
We can also write a relationship between kinetic energy and momentum:
and momentum is given by:
p = mv
So we can write:
If we combine:
and
we get:
So what?
We know that all the energy in an electron is kinetic, so we can work out what the energy is for the fundamental standing wave.
Let us assume that the hydrogen atom has a diameter of 1.06 × 10^{10} m. The electron standing wave looks like this:
Worked example Calculate the energy of the electron standing wave in eV shown above. (Mass of an electron = 9.1 × 10^{31} kg; Planck's constant = 6.6 × 10^{34} J s) 
Answer Use:
E_{k} = (6.6 × 10^{34} J s)^{2} ÷ (8 × 9.1 × 10^{31} kg × (1.6 × 10^{10} m)^{2}) = 2.34 × 10^{18} J
E_{k} = 2.34 × 10^{18} J ÷ 1.6 × 10^{19} J eV^{1} = 15 eV.
This is very approximate to the real ionisation energy of the hydrogen. The correct answer is 13.6 eV 
Let us suppose that we have two waves in the box:
By a similar argument to that above, we can write an equation for the energy of the two standing wave loops (one wavelength):
This is because we get one whole wavelength into the box instead of 1/2 wavelength. If the wavelength has a lower energy, it has a longer wavelength, so the length of the wave in the box increases.
Worked Example An excited electron in an atom falls from the ionised state to the n = 2 state (0 eV to 3.4 eV). (a) Calculate the energy in the photon emitted. (b) Use the equation above to calculate the size of the box needed to accommodate the electron in its excited state, assuming the box is one whole wavelength in size.. (Mass of an electron = 9.1 × 10^{31} kg; Planck's constant = 6.6 × 10^{34} J s) 
Answer (a) DE = 0 eV  3.4 eV = 3.4 eV (work is got out)
Convert the energy to joules: E = 3.4 eV × 1.6 × 10^{19} J eV^{1} = 5.44 × 10^{19} J
(b) Rearrange the equation above to:
L^{2} = (6.6 × 10^{34} J s)^{2} ÷ (2 × 9.1 × 10^{31} kg × 5.44 × 10^{18} J) = 4.40 × 10^{20} m^{2}
L = (4.40 × 10^{19} m^{2})^{0.5} = 2.1 × 10^{10} m 
The box swells as the the electron gets excited. This is because it accommodates two longer standing wave loops of lower energy.
The difficulty is that this model explains the idea of electron standing waves just like those on a string. However the numerical analyses above do not give convincing consistencies with the wavelengths and energy. In reality the waves loops are formed on the circumference of the electron shell, not across its diameter. Another problem is that the nucleus is at the same point as where the probability of the electron being found is the highest.
This could even result in an electron capture event.
The Bohr Model and Electron Standing Waves(*)
This helps us to understand the idea of electron standing waves, but it only shows what happens in onedimension. A more sophisticated model is one that shows us what happens in when we see two dimensions. This uses the Bohr Model:
The n = 1 resonance is circular, and represents a single standing wave loop (i.e. half a wavelength). The n = 2 resonance represents two standing wave loops (i.e. one whole wavelength). The n = 3 resonance has three standing wave loops (not very well drawn, but it will have to do). It shows 3/2 wavelengths.
For the Bohr model electronic resonance patterns, the radius is the Bohr Radius, a_{0}, where:
a_{0} = 0.0529 nm = 5.29 × 10^{11} m
The wavelength of the first resonance, n = 1, is given by:
The wavelength of the second resonance, n = 2, is given by:
If n doubles, the radius according to the Bohr Model increases four times. We have before that the energy goes down four times as n goes from 1 to 2. Therefore the wavelength increases by four times. So we can say:
Therefore:
The wavelength of the second resonance, n = 3, is given by:
Since the wavelength goes up as n^{2}, we can write:
Therefore:
Angular Momentum in Electron Orbits (*)
In the Bohr model, standing waves occur around the circumference of the electron shell. For a standing wave to form, the circumference has to be the same distance as a whole number of standing wave loops. When we looked at how electrons occurred in 1dimension, we used the linear momentum in the de Broglie wavelength equation, i.e.:
We know that in the quantum atom, the wavelengths are quantised, meaning that they have to have a very specific value. The same is true for the standing wave loops. Therefore the circumferences of the electron shells have to be quantised as well. The wavelength of the standing wave at any energy level is given by:
To work with a de Broglie wavelength of an electron orbiting the nucleus of an atom, we need to use the angular momentum for the circular motion:
L = mvr
L = Angular momentum (kg m^{2} s^{1});
v = velocity (m s^{1});
r = radius (m).
Do not use: L = Iw for angular momentum as this refers to rotational motion, not circular motion.
Note that L is used as the physics code for both angular momentum, and the length of a standing wave loop. Be careful what you are referring to. 
So we can combine the equation for angular momentum and the de Broglie equation:
From the equation:
we can write:
We can then substitute into the equation for the angular momentum:
We will use this relationship in the Bohr Orbit.
Classical Electron Orbit (*)
In Physics 4, we studied centripetal force in Further Mechanics Tutorial 1. We also looked at the forces between charges in Fields Tutorial Tutorial 4. You may wish to review these.
Consider an electron of charge e orbiting a nucleus of proton number Z. As the electron is travelling in a circular path of radius r, it is subject to a centripetal force F. Since the charges are opposite, the force will be negative, i.e., attractive. However we would have negative signs on both sides, which cancel out.
From circular motion, we can write:
From electric fields, we can write:
And we can combine the two to write:
Cancelling:
The electron in orbit has two kinds of energy, kinetic (because it's moving) and potential energy (the work that can be got out as a result of the centripetal force). We can work out the kinetic energy first:
We can work out the potential energy from the attraction between the two charges:
The minus sign is important to the argument, so don't miss it out.
The total energy of the orbiting electron around the hydrogen nucleus is:
The Bohr orbit combines the energy from the classical argument above with the quantisation of the angular momentum, i.e.:
and
Kinetic energy is related to momentum by:
By a similar argument we can related the kinetic energy to the angular momentum:
Now we can substitute for mvr:
This gives us an expression for the quantisation for the angular momenta for the energy levels. We can now equate this last equation to:
and write:
Cancelling:
Now we make r the subject:
Now we substitute this for r in the equation:
which gives us:
This looks quite a difficult result to take in until we realise that many of these terms are constants:
e  electronic charge = 1.60 × 10^{19} C;
m  mass of an electron = 9.11 × 10^{31} kg;
e_{0}  permittivity of free space = 8.85 × 10^{12} F m^{1};
h  Planck's constant = 6.63 × 10^{34} J s.
Hydrogen has a proton number of 1. Lets consider the first level, n = 1. So we can put these numbers in:
E_{k} = (1 × 9.11 × 10^{31} kg × (1.60 × 10^{19} C)^{4}) ÷ (8 × (8.85 × 10^{12 } F m^{1})^{2} × 1^{2} × (6.63 × 10^{34} J s)^{2})
= (9.11 × 10^{31} kg × 6.55 × 10^{76} C^{4}) ÷ (8 × 7.82 × 10^{23} F^{2} m^{2} × 4.40 × 10^{67} J^{2} s^{2})
= (59.67 × 10^{107} kg C^{4}) ÷ (275 × 10^{90} F^{2} J^{2 }s^{2} m^{2})
= 0.217 × 10^{17} J = 21.7 × 10^{19} J
Now we can convert this to eV
E = 21.7 × 10^{19} J ÷ 1.60 × 10^{19} J eV^{1} = 13.6 eV
Notice that there were a power of ten that was greater than 100. This will not work in a calculator. This is where the brain is quite useful...
In this calculation we used n = 1. We can easily modify this for any value of n, to give this relationship:
We can rearrange the equation:
to:
The right hand side of this equation consists of constants. The constants work out to the Rydberg Constant, which has a value of 21.7 × 10^{19} J or 13.6 eV. If Z = 1, we can rearrange to have E_{n} on the left hand side to give:
This relationship works well in hydrogen. The idea was that it should work for any proton number. But it doesn't.