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Electron configurations

In this blog, I discussed the “electron configuration” nomenclature which is so loved by chemists (strange people that they are….). Just to remind you, the noble gas neon, which is at number 10 in the periodic table, may be written as $1s^{2} \; 2s^{2} \; 2p^{6}$. If you add together the superscripts you get $2+2+6=10$, the number of electrons in neutral Helium. Titanium, which is at number 22 in the periodic table may be written as $1s^{2} \; 2s^{2} \; 2p^{6} \; 3s^{2} \; 3p^{6} \; 3d^{2} \; 4s^{2}$. Again, if you add together the superscripts you get $2+2+6+2+6+2+2=22$, the number of electrons in neutral Titanium. I explained in the blog that the letters s,p,d and f refer to “sharp, principal, diffuse” and “fine“, as this was how the spectral lines appeared in the 1870s when spectroscopists first started identifying them.

But, what I didn’t address in that blog on the electron configuration nomenclature is why do electrons occupy different shells in atoms? In hydrogen, the simplest atom, the 1 electron orbits the nucleus in the ground state, the n=1 energy level. If it is excited it will go into a higher energy level, n=2 or 3 etc. But, with a more complicated atom like neon, which has 10 electrons, the 10 do not all sit in the n=1 level. The n=1 level can only contain up to 2 electrons, and the n=2 level can only contain up to 8 electrons, the n=3 level can only contain up to 18 electrons, and so on. This leads to neon having a “filled” n=1 level (2 electrons), and a filled n=2 level (8 electrons), which means it does not seek additional electrons. This is why it is a noble gas.

Titanium on the other hand, with 22 electrons, has a filled n=1 level (2 electrons), a filled n=2 level (8 electrons), a partially filled n=3 level (8 electrons out of a possible 18), and a partially filled n=4 level (2 electrons out of a possible 32). Because it has partially filled n=3 and n=4 levels, and it wants them to be full, it will seek additional electrons by chemically combining with other elements.

What is the reason each energy level has a maximum number of allowed electrons?

It is all due to something called the Pauli exclusion principle.

Wolfgang Pauli, after whom the Pauli exclusion principle is named. He came up with the idea in 1925. In addition to this principle, he also came up with the idea of the neutrino.

The energy level n

Niels Bohr suggested in 1913 that electrons could only occupy certain orbits. I go into the details of his argument in this blog, but to summarise it briefly here, he suggested that something called the orbital angular momentum of the electron had to be divisible by $\hbar \text{ where } \hbar = h/2\pi, \text{ } h$ being Planck’s constant. We now call these the energy levels of an atom, and we use the letter n to denote the energy level. So, an electron in the second energy level will have $n=2$, in the third energy level it will have $n=3$ etc.

As quantum mechanics developed over the next 15-20 years it was realised that an electron is fully described by a total of four (4) quantum numbers, not just its energy level. The energy level $n$ came to be known as the princpical quantum number. The other three quantum numbers needed to fully describe the state of an electron are

• its orbital angular momentum, $l$
• its magnetic moment, $m_{l}$ and
• its spin, $m_{s}$

The orbital angular momentum $l$ quantum number

As I mentioned above, spectroscopists noticed that atomic lines could be visually categorised into “sharp”, “principal”, “diffuse” and “fine“, or $s,p,d \text{ and } f$. It was found that the following correspondence existed between these visual classifications and the orbital angular momentum $l$. This is the second quantum number. $l$ can only take on certain values from $0 \text{ to } (n-1)$. So, for example, if $n=3, \; l \text{ can be } 0,1 \text{ or } 2$.

spectroscopic name and orbital angular momentum
Spectroscopic Name letter orbital angular momentum $l$
sharp s $l=0$
principal p $l=1$
diffuse d $l=2$
fine f $l=3$

As this table shows, the reason a line appears as a “sharp” (s) line is because its orbital angular momentum $l=0$. If it appears as a “principal” (p) line then its orbital angular momentum must be $l=1$, etc.

The magnetic moment quantum number $m_{l}$

The third quantum number is the magnetic moment $m_{l}$, which can only take on certain values. The magnetic moment only shows up if the electron is in a magnetic field, and is what causes the Zeeman effect, which is the splitting of an atom’s spectral lines when an atom is in a magnetic field. The rule is that the magnetic moment quantum number can take on any value from $-l \text{ to } +l$, so e.g. when $l=2, \text{ } m_{l}$ can take the values $-2, -1, 0, 1 \text{ and } 2$ (5 possible values in all). If $l=3 \text{ then } m_{l} \text{ can be } -3, -2, -1, 0, 1, 2, 3$ (7 possible values).

The spin quantum number $m_{s}$

The final quantum number is something called the spin. Although it is only an analogy (and not to be taken literally), one can think of this as the electron spinning on its axis as it orbits the nucleus, in the same way that the Earth spins on its axis as it orbits the Sun. The spin can, for an electron, take on two possible values, either $+1/2 \text{ or } -1/2$.

Putting all of this together

Let us first of all consider the $n=1$ energy level. The only allowed orbital angular momentum allowed in this level is $l=0$, which means the only allowed values of $m_{l}$ is also 0 and the allowed values of the spin are $+1/2 \text{ and } -1/2$. So, in the $n=1$ level, the only allowed state is $1s$, and this can have two configurations, with the electron spin up or down (+1/2 or -1/2), meaning the $n=1$ level is full when there are 2 electrons in it. That is why we see $1s^{2}$ for Helium and any element beyond it in the Periodic Table. But, what about the $n=2, n=3$ etc. levels?

The number of electrons in each electron shell
State Principal quantum number $n$ Orbital quantum number $l$ Magnetic quantum number $m_{l}$ Spin quantum number $m_{s}$ Maximum number of electrons
1s 1 0 0 +1/2, -1/2 2
n=1 level Total = 2
2s 2 0 0 +1/2, -1/2 2
2p 2 1 -1,0,1 +1/2, -1/2 6
n=2 level Total = 8
3s 3 0 0 +1/2, -1/2 2
3p 3 1 -1,0,1 +1/2, -1/2 6
3d 3 2 -2,-1,0,1,2 +1/2, -1/2 10
n=3 level Total = 18
4s 4 0 0 +1/2, -1/2 2
4p 4 1 -1,0,1 +1/2, -1/2 6
4d 4 2 -2,-1,0,1,2 +1/2, -1/2 10
4f 4 3 -3,-2,-1,0,1,2,3 +1/2, -1/2 14
n=4 level Total = 32
5s 5 0 0 +1/2, -1/2 2
etc.

The astute readers amongst you may have noticed that the electron configuration for Titanium, which was $1s^{2} \; 2s^{2} \; 2p^{6} \; 3s^{2} \; 3p^{6} \; 3d^{2} \; 4s^{2}$, suggests that the $n=4$ level starts being occupied before the $n=3$ level is full. After all, the $n=3$ level can have up to 18 electrons in it, with up to 10 electrons in the $n=3, l=2$ (d) state. In the $n=3$ level the (s) and (p) states are full, but not the (d) state. With only 2 electrons in the $n=3, l=2$ (d) state, the $4s$ state starts being populated, and has 2 electrons in it. Why is this?

I will explain the reason in a future blog, but it has to do with the “shape” of the orbits of the different states. They are different for different values of orbital angular momentum $l$.

The Bohr model of the atom

In my blog on Niels Bohr, number 2 in The Guardian’s list of the 10 best physicists, I mentioned the Bohr model of the atom. In this blog I will go into more detail about this model, and how it agreed with the experimental results (the Rydberg formula) for the hydrogen atom.

Quantised orbits

In 1911, Rutherford had proposed that atoms have positively charged nuclei, with the negatively charged electrons orbiting the nuclei. One of the problems with this idea was that an orbiting electron would be accelerating, by virtue of moving in a circle. The acceleration is directed towards the centre of the circle. It was well known that when an electron is accelerated it radiates electromagnetic waves. Calculations showed that the orbiting electrons on Rutherford’s model should radiate away their energy in a few microseconds (millionths of a second), and spiral towards the nucleus. They clearly were not doing this, but why?

Bohr suggested in a paper in 1913 that electrons would somehow not radiate away their energy if they were orbiting in certain “allowed orbits”. If they were in these special orbits, the normal laws of EM radiation would not apply. He suggested that these allowed orbits were when the orbital angular momentum $L$ could be written as

$L = \frac{ n h }{ 2 \pi } = n \hbar \text{ (Equ. 1) }$

($\hbar = \frac{ h }{ 2 \pi } \text{ where } h$ is Planck’s constant, and is given its own symbol in Physics at it crops up so often). What is orbital angular momentum? Well, it is the rotational equivalent of linear momentum. Linear momentum is defined as $\vec{p} = m\vec{v} \text{ where } m \text{ is the mass and } \vec{v} \text{ is the velocity}$. Notice, momentum is a vector quantity, this is important in doing calculations involving collisions, such as the ones I did in this blog.

By analogy, orbital angular momenutm is defined as

$\vec{L} = \vec{r} \times m \vec{v}$

where $\vec{r}$ is the radius vector of the orbit, which is defined as pointing from the centre of the orbit along the radius. For a circular orbit, where the radius vector is at right angles to the velocity vector, we can just write $L=mvr$ where $L$ is the magnitude (size) of the vector $\vec{L}$.

The force keeping the electron in orbit

From Classical Physics, Bohr argued that the force which was keeping the electron in orbit about the positively charged nucleus was the well known Coulomb force, given by

$F = - \frac{ Z k_{e} e^{2} }{ r^{2} }$

where $k_{e}$ is the Coulomb constant (which determines the force between two 1 Coulomb charges separated by 1 metre), $Z$ is the atomic number of the atom, $e$ is the charge on an electron and $r$ is the radius of the orbit. The minus sign is telling us that the force is directed towards the centre, whereas our definition of the radius vector is that it is away from the centre, so they are in opposite directions.

We can equate this to the formula for the centripetal force on any object moving in a circular orbit, so we can write

$\frac{ m_{e} v^{2} }{ r } = \frac{ Z k_{e} e^{2} }{ r^{2} } \text{ (Equ. 2) }$

where $m_{e}$ is the mass of the electron and $v$ is the speed of its orbit.

Re-arranging Equation 2 we can write

$v = \sqrt{ \frac{ Z k_{e} e^{2} m_{e} r }{ m_{e}^2 r^{2} } }$

which then allows us to write the angular momentum as

$m_{e} v r = \sqrt{ Z k_{e} e^{2} m_{e} r } \text{ which (from Equ. 1) } = n \hbar$

This allows us to write an expression for the “radius” of an electron’s orbit as

$r_{n} = \frac{ n^{2} \hbar^{2} }{ Z k_{e} e^{2} m_{e} }$

where $n$ is the energy level of the electron. The so-called “Bohr radius” is the radius of an electron in the $n=1$ energy level for hydrogen ($Z=1$) and can be written

$\boxed{ r_{1} = \frac{ \hbar^{2} }{ k_{e} e^{2} m_{e} } \approx 5.29 \times 10^{-11} \text{ metres } }$

This is, indeed, about the size of a hydrogen atom.

The total energy of the electron

The total energy of the electron in its orbit is given by the sum of its kinetic energy and its potential energy. The kinetic energy is just given by $1/2 \; (mv^{2})$. What about the potential energy? The potential energy can be found by using the relationship between work and force; back in this blog I said that work was defined as the force multiplied by the distance moved. Energy is the capacity to do work, and is measured in the same units, Joules. So we can derive the potential energy of an electron in orbit due to the Coulomb force as

$P.E. = \int_{r}^{\infty} { F} dr = - \int_{r}^{\infty} \frac{ Z k_{e} e^{2} }{ r^{2} } dr$

where $dr$ is an incremental change in the radius. If we do this integration we get

$P.E. = - \frac{ Z k_{e} e^{2} }{ r }$

where the negative sign is telling us that we have to do work on the electron to increase its radius, or to put it another way that the force acts towards the centre but the radius vector acts away from the centre of the electron’s orbit. This means that the total energy $E$ is given by

$E = \frac{ 1 }{ 2 } m_{e} v^{2} - \frac{ Z k_{e} e^{2} }{ r }$

But, from Equ. 2 we can write the kinetic energy as

$\frac{ 1 }{ 2 } m_{e} v^{2} = \frac{ Z k_{e} e^{2} }{ 2r }$

So then the total energy $E$ can be written

$E = \frac{ Z k_{e} e^{2} }{ 2r } - \frac{ Z k_{e} e^{2} }{ r } = - \frac{ Z k_{e} e^{2} }{ 2 r }$
So, in the Bohr model, the energy of the $n^{th}$ energy level is given by

$\boxed{ E_{n} = - \frac{ Z k_{e} e^{2} }{ 2 r_{n} } \text{ or } -\frac{ Z^{2} (k_{e} e^{2})^{2} m_{e} }{2 \hbar^{2} n^{2} } }$

In the case of hydrogen, where $Z=1$ we can write

$\boxed{ E_{n} = -\frac{ (k_{e} e^{2})^{2} m_{e} }{ 2 \hbar^{2} n^{2} } \approx -\frac{ 13.6 }{ n^{2} } \text{ eV} }$

This was in perfect agreement with the Rydberg formula for the energy levels of hydrogen, which had been experimentally derived by the Swedish physicist Johannes Rydberg in 1888. As I will show in a future blog, Bohr’s model was a “semi-empirical” model, in that it was a step along the way to the correct model. It was produced by using a mixture of classical physics and quantum mechanics, and Bohr did not understand why his condition that only orbits whose angular momentum were equal to $n \hbar$ was true. The explanation was produced with the full theory of Quantum Mechanics in 1926 as a solution to Schrödinger’s wave equation for hydrogen.

I have referred a few times to line spectra. In my blog about the Messier catalogue, I mentioned that M42 and M1 were both examples of objects which exhibited an emission line spectrum. And in my blog on the three kinds of spectra, of course emission line spectra are one of the three types. So, how do emission line spectra come about?

Electrons in orbit about the nucleus

In a future blog I will explain in more detail why electrons orbit the nucleus of atoms in only certain allowed energy levels. This was first proposed by Niels Bohr in 1913, and the details were then worked out until a full explanation was developed by Erwin Schrödinger, Werner Heisenberg and Paul Dirac in 1926-1928 in what we now call Quantum Mechanics.

As I mentioned in my blog on the different types of spectra, it was noticed by Kirchhoff and Bunsen in the 1850s that different salts produced different types of line spectra when they were burnt in a flame. This is because each element has an unique spectral signature. We will look at the simplest spectrum, that of the hydrogen atom. It is the easiest to understand, because the hydrogen atom only has one electron. Additional electrons complicate matters, because the electrons interact with each other in their orbits, but for hydrogen things are nice and simple.

The emission line spectrum of hydrogen

Because hydrogen is the simplest element, it is the most abundant in the Universe. About 75% of the “normal” matter in the Universe is in the form of hydrogen. In the visible part of the spectrum, the hydrogen emission line spectrum looks like this.

Notice that there are a series of lines over towards the blue (left) end of the spectrum, and a prominent line in the red, which has an arrow pointing towards it. This line is so common and important in astronomy that it even has a special name, it is called the h-alpha (hydrogen alpha) line.

How do these lines come about?

The explanation for these different lines is that the different lines are produced by electrons jumping down between different energy levels in the hydrogen atom. The electron wants to be in the ground state, the n=1 level, in hydrogen. But it can be excited into higher levels, either by absorbing a photon (a particle of light), or by another electron hitting it. Once it finds itself in a higher energy level (n=2 or higher), it wants to jump back down to the n=1 level as quickly as it can. It is in jumping back down that the photons which we see in an emission line spectrum are produced.

The figure below shows the simplified energy level diagram for hydrogen, with the n=1, n=2, n=3 etc. energy levels. The Balmer series, which are all in the visible part of the spectrum, are produced when electrons jump from a higher energy level into the n=2 level. So e.g. from n=3 to n=2, or n=4 to n=2, or n=5 to n=2. It is the Balmer spectral lines which are shown in the figure above of a hydrogen emission spectrum.

The figure also shows three other “series”, the Lyman series which are in the ultraviolet part of the spectrum, and the Paschen and Brackett series, which are in the infrared part of the spectrum. The Lyman series all end in the n=1 level (ground state), the Paschen series all end in the n=3 level, and the Brackett series all end in the n=4 level.

The energies of the different energy levels in hydrogen

These visible-light lines were well measured in the 1880s, and are referred to as the Balmer series. Balmer had even derived an empirical formula for the wavelengths. He did not know about energy levels, but by slightly adapting his formula we can write an empirical formula for the energy levels of a hydrogen atom [empirical means it was derived through trial and error, with no physical explanation for why the formula works]. It is

$\text{energy in eV is } 13.6 \left(1 - \frac{1}{n^{2}} \right)$

This formula gives the energy in eVs, and these values are over on the right of the diagram for each energy level [Note: Chemists will often label the n=1 level as -13.6eV, and work upwards from this, physicists tend to label it as 0eV. It doesn’t matter, because all that is important is the energy difference between different levels]. eV stands for “electron volt”, and is just a more convenient unit for measuring the small energies involved than using the more usual Joules. It’s just like using nanometres to measure things on the atomic scale because metres are too big.
An electron volt is defined as $1 \text{eV} = 1.6 \times 10^{-19} \text{ Joules }$, so is pretty small as you can see.

There is a very simple relationship between the difference in energy between two energy levels and the wavelength of the photon produced when the electron jumps down. The energy of the photon is given by $E=h\nu$, where $h$ is called Planck’s constant and $\nu$ is the frequency of the photon. If you prefer to think in terms of wavelength instead of frequency they are very simply related; the wavelength $\lambda$ is just given by $\lambda = c/\nu$ where $c$ is the speed of light. So, putting this together, we can write that the wavelength $\lambda$ of a photon is given by

$\lambda = \frac{hc}{\Delta E}$

where $\Delta E$ is the energy difference between the two levels the electron jumps between. To go through a couple of examples, if an electron jumps from the n=3 to the n=2 level, the energy difference is $12.09 - 10.2 = 1.89 \text{ eV}$. We need to convert this to Joules, so $1.89 \text{ eV } = 1.89 \times 1.6 \times 10^{-19} = 3.024 \times 10^{-19} \text{ Joules }$. To get the wavelength from this we write $\lambda = (\; (6.63 \times 10^{-34}) \times (3 \times 10^{8}) \; )/(3.024 \times 10^{-19}) = 6.577 \times 10^{-7} \text{ metres } = 657.7 \text{ nanometres or } 6577 \text{ Angstrom}$. This is in the red part of the visible spectrum, and is the hydrogen-alpha line we were referring to earlier.

To work through a second example, if we look at the transition between n=5 and n=3 (part of the Paschen series) we get $\Delta E = 13.06 - 12.09 = 0.97 \text{ eV } = 1.552 \times 10^{-19} \text{ Joules}$. So the photon will have a wavelength of $(\; (6.63 \times 10^{-34} ) \times (3 \times 10^{8}) \; )(1.522 \times 10^{-19}) = 1.2800 \times 10^{-6} \text{ metres} = 1280 \text{ nanometres} = 12800 \text{ Angstrom}$, which is in the infra-red part of the spectrum.

What is true for hydrogen is also true for the other elements, it is just that there is no simple formula for working out the energies of the different energy levels like there is for the energy levels in hydrogen. But, as we shall see in another future blog, even the n=2 and n=3 levels in hydrogen are not all at exactly the same level, it depends on the angular momentum of the electron in a particular energy level. This is where the s,p,d,f lines that I mentioned in this blog comes into play. So what I have explained above is a first approximation, but perfectly fine for most uses.