How can we possibly know what elements are present in the stars? It’s not like we can nip off and get ourselves a sample …
The answer is spectroscopy – using light as an analytical technique. You should start by watching Brian Cox put it all into context.
We need to delve into the physics of light briefly before we can explain the techniques of atomic emission and absorption spectroscopy. Please don’t skip this!
Visible light forms a small part of the electromagnetic spectrum, with radiation in the frequency range 4 x 1014 – 8 x 1014 Hz. If we pass white light through diffraction grating or a glass prism, we can ‘split’ visible light into the familiar colour spectrum.
The frequency of radiation is directly proportional to its energy and inversely proportional to its wavelength.
c = 𝛌𝛎
where c = speed of light (m), 𝛌 = wavelength (m)* and
𝛎 = frequency (Hz)
* Note that in order to carry out calculations using this equation we must convert wavelength from nm to m (1nm = 1 x 10-9m)
However, it turns out that light can exist as particles as well as in the form of waves. Einstein suggested that in the case of light, a particle was more akin to a ‘packet’ of energy which was called a photon. The energy of a photon is directly proportional to the frequency of the light.
Atomic absorption spectroscopy
So we know that if we shine white light through diffraction grating we see the full spectrum of visible light, from red to indigo, but if we pass the light coming from a star through diffraction grating we see the full spectrum of visible light with distinct black lines present at specific frequencies.
These black lines correspond to specific frequencies of light that are missing – they have been absorbed by atoms and ions in the corona of the star.
In the Bohr model of the atom electrons exist in defined, ‘quantised’ energy levels, and these energy levels are unique to the atoms of an individual element. This means that the distance between two energy levels, say from n=1 to n=2, is identical in all hydrogen atoms, but it (ΔE) will not be the same in a lithium atom.
When we shine white light on an atom, the atom (not the electron) absorbs a specific frequencies of light and uses that energy to promote electrons to a higher energy level. The greater ΔE, the higher the frequency of light needed to promote the electron.
ΔE = h𝛎
where ΔE is the difference in energy between two energy levels, h = Planck’s constant and 𝛎 = frequency
If we look at the atomic absorption spectrum for a typical star, we might see something like this:
- there is more than one line for each element present in the star’s corona because there are a number of different promotions possible for an atom (n=1 to n=2, n=1 to n=3, n=2 to n=3 etc) and in each case ΔE will be different and hence the frequency of light absorbed will be different.
- the absorption spectrum is evidence that energy levels in an atom are quantised and that the actual energy associated with each level is unique to an element, as ΔE for, say, n=1 to n=2 for a calcium atom is clearly different to that of a magnesium atom (if you look at the spectrum above, the frequency of light absorbed by each element is different).
- the absorption spectrum for an emergent is described a black lies on a coloured background, with the lines getting closer together as the frequency increases – this is evidence that the energy levels get closer together, the further they are from the nucleus, and is predicted by the maths.
Atomic emission spectroscopy
I am assuming you are familiar with flame tests as an analytical technique for identifying certain metal elements in a compound …
Potassium characteristically colours the flame lilac, strontium red and sodium yellow.
So what is actually happening here?
Atoms absorb energy (from the flame in this case, but it could be by electrical spark) and promote electrons to higher energy levels. The electrons then fall back to a lower energy level and emit a photon of visible light with a specific frequency that its directly proportional ΔE, the difference between the energy levels in that element’s atoms because we know that ΔE = h𝛎.
We can use a spectroscope to view an emission spectrum and what we see is a black background with coloured lines that get closer together as the frequency increases.
Below you can see the atomic emission spectrum for hydrogen, with each coloured line on the spectrum corresponding to a specific transition of an electron from a higher energy level to a lower one:
Finally, you should be aware that each series of spectral lines for a hydrogen atom have been named.
- Lyman (emission) for electrons falling back to n=1
- Balmer (emission) for electrons falling back to n=2
- Paschen (emission) for electrons falling back to n=3
Practice questions
- The element Gallium was identified from its atomic emission spectrum in 1875. Explain why each element has its own unique emission spectrum.
2. Students carry out a flame test on an unknown salt. The flame turns a lilac colour and the lilac line occurs at 405nm in the emission spectrum.
(a) Describe the appearance of an atomic emission spectrum.
(b) Calculate the energy (in kJ mol-1) associated with the wavelength of the lilac line in the spectrum.
3. When strontium was first discovered it was mistaken for barium, but these two elements give different flame colours when heated. Explain why elements emit different colours of light when heated, and how analysis of this light could be used to prove that strontium and barium are different elements. You may use diagrams in your answer.
4. The emission spectrum of lithium shows an intense line at the red end of the spectrum.
(a) Explain the origin of this red line, and other lines, in the emission spectrum of lithium. You should annotate the diagram below as part of your answer.
(b) Label the diagram below with an arrow indicating increasing frequency.
(c) Explain why emission spectra can be used to identify the different elements in a firework.
5. Describe the differences and similarities between an absorption and an emission spectrum for the same element.
6. The atomic emission spectrum of sodium shows an intense yellow line which is due to the transition of an electron from its first excited state to its group state.
(a) State the subshells involved in this transition
(b) The wavelength of this yellow line is 589nm. Calculate the energy change associated with this transition.
Answers
- In emission spectroscopy electrons are promoted to higher energy levels ✔️when the atoms absorb energy. These electrons then fall back to lower levels, ✔️emitting a specific frequency of visible light that corresponds to a specific difference in the quantised energy levels ( ΔE = h𝜈 ) ✔️. Since ΔE is unique to each element, the emission spectrum will be unique ✔️.
- (a) Coloured lines on a black background ✔️, lines are closer together at higher frequencies ✔️
(b) c = 𝛌𝜈 so if we rearrange we get 𝜈 = c / 𝛌 (remember to convert nm into m; 405nm = 4.05 x 10-7m); 𝜈 = 3.00 x 10-8 / 4.05 x 10-7 = 7.407 x 1014 Hz ✔️
E = h𝜈 E = 6.63 x 10-34 x 7.407 x 1014 = 4.91 x 10-19 J (per photon / atom) ✔️
Multiply by Avogadro’s number to calculate energy per mole of photons / atoms, and divide by 1000 to convert answer to kJ
(4.91 x 10-19 x 6.02 x 1023 ) / 1000 = 296 kJ mol-1 ✔️
3. When atoms are heated they promote electrons to higher energy levels ✔️. These electrons then fall back to lower levels emitting specific frequencies of visible light ✔️as the energy is lost as photons ✔️. The energy levels in an atom are quantised and unique to each element ✔️. Since ΔE = h𝜈, the frequencies, and hence colours, of light emitted are unique to each element ✔️. A number of transitions is possible giving us an emission spectrum unique to each element ✔️- the emission spectrum of barium will not be the same as that of strontium (the wavelengths of the coloured lines will differ) so we can show that they are different elements .
4. (a) Excited electrons fall back to lower electronic energy levels emitting specific frequencies of visible light ✔️as the energy is lost as photons ✔️. The diagram should be labelled to show that the red line on the emission spectrum is z ✔️ (because red light has the lowest frequency of all visible light and so must correspond to the smallest ΔE and ΔE = h𝜈 ✔️).
(b) The arrow should point from red to blue and be labelled ‘increasing frequency’ ✔️
5. Emission spectrum consists of a black background with coloured lines whereas an absorptions spectrum has a coloured background with black lines ✔️. They both show the lines getting closer together as frequency increases ✔️.
6. (a) 3p subshell to 3s subshell
(b) 𝜈 = c / 𝛌 so 3.00 x 108 / 5.89 x 10-7 = 5.09 x 1014 Hz ✔️
E = h𝜈 ; E = (6.63 x 10-34) x (5.09 x 1014 ) = 3.38 x 10-19 J ✔️
Per kJ mol-1 ……. (3.38 x 10-19 x 6.02 x 1023) / 1000 = 203 kJ mol-1 ✔️