PHOTOELECTRON SPECTROSCOPY (PES) – a simplified introduction
PES uses high-energy electromagnetic radiation (X-ray or UV) to probe the electronic structure of an
atom. When sufficient energy is provided, electrons will be emitted.
𝑿 + 𝒉𝝂 → 𝑿+ + 𝒆−
The energy of the photon is consumed in two ways:
(1) Break the force of attraction between the electron and the protons in the nucleus.
a. This is called the binding energy (BE), measured in Joules (J), megajoules/mole (MJ/mole)
or electron volts (eV).
𝟏 𝑱 = 𝟔. 𝟐𝟒 𝒙 𝟏𝟎𝟏𝟖 𝒆𝑽
b. Remember coulomb’s law: As the distance between the electron and proton DECREASES,
the attraction INCREASES and the amount of energy necessary to break the attraction (the
binding energy) INCREASES
c. The BINDING ENERGY is thus related to the IONIZATION ENERGY. If the sample is in the
gaseous phase, they are typically considered as identical. For our purposes we will treat
them as synonyms.
(2) Increase the kinetic energy of the electron. The energy of the impinging photons is KNOWN the
kinetic energy values of emitted electrons are MEASURED, allowing the binding energy to be
GRAPHED.
𝐸𝑝ℎ𝑜𝑡𝑜𝑛 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 + 𝐵𝑖𝑛𝑑𝑖𝑛𝑔 𝑒𝑛𝑒𝑟𝑔𝑦
OR
𝐸𝑝ℎ𝑜𝑡𝑜𝑛 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 + 𝐼𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑒𝑛𝑒𝑟𝑔𝑦
We will be exploring the binding energy for the removal of electrons in different sublevels as the FIRST
electron removed. We will not be exploring the CHANGES in binding energy that occur with the removal
of SEQUENTIAL electrons. We will have that conversation when we explore the periodic trends in
ionization energy. Each peak in the spectrum will represent the binding energy for an electron when it is
the first electron removed. PES spectra will thus provide us with two key pieces of information:
(1) A spectrum will show the binding energy for each sublevel in an atom and hence links
experimental evidence to the electron configuration model. The peak height shows the relative
number of electrons in the sublevel. The spectrum provides evidence for the claim that a 2p
electron is at a little higher energy than the 2p electron. Below is a simplified spectrum for
fluorine.
Dena K. Leggett, PhD Allen High School, Allen, TX AP Chemistry
1
Note (1) the ratio of peak height
is 5:2 representing the electron
ratio of 5:2 (2) the IE for the “p”
electrons is less than the “s”
electrons
Note that the IE of the 1s electron is
higher than the 2s & 2p. It is closer to
the nucleus so experiences greater
attraction.
2p5
2s2
1s2
CAUTION: evaluate the x-axis carefully.
Sometimes energy is graphed low to high
and others high to low. (sorry – hate it
for you!)
(2) Comparison of spectra beautifully demonstrates how the actual energy of energy levels varies as
the nuclear charge, Z, increases. This in turn illustrates why each element has a unique atomic
spectrum!
𝑭𝟐𝒑𝟓
Note that the IE for a Fluorine 1s electron
is greater than for nitrogen. This is due to
the differences in nuclear charge, Z.
REMEMBER COULOMB’S LAW! ZF = 9
and ZN = 7
𝑵𝟐𝒑𝟑
𝑵𝟐𝒔𝟐 𝑭𝟐𝒔𝟐
𝑵𝟏𝒔𝟐 𝑭𝟏𝒔𝟐
INQUIRY ACTIVITY:
1. A PES spectrum is shown above.
a. Identify the element and write its electron configuration.
b. Sketch the expected spectrum of phosphorus on the graph, making sure to show the
relative changes in positions (not the exact energy) of the peaks and the relative intensity
of each peak.
2. How many peaks would you anticipate for copper (ignoring any splitting within a sublevel)? The
relative intensity of peaks for fluorine is 2:2:5. What is the relative intensity of peaks for copper?
3. A student makes the following claim regarding the PES spectra of Mg2+ and Ne. Is the statement
true or false? Justify your answer.
Since Mg2+ is isoelectronic with Ne, the PES spectra will be identical.
4. A sample of an unknown element was studies using photons with energy of 198.2 MJ/mole. The
kinetic energy of the ejected electrons was determined to be 38.6 MJ/mole. What is the binding
energy for these electrons?
Dena K. Leggett, PhD Allen High School, Allen, TX AP Chemistry
3
5. A portion of the PES spectra for phosphorus and sulfur depicting the 3p and 3s electrons is shown
above. Although the nuclear charge for sulfur is GREATER than phosphorus, the binding energy
for the 3p electrons is unexpectedly LOWER for sulfur. Justify this observation in terms of
repulsive and attractive forces.
REFERENCES:
 http://www.pes.arizona.edu/aboutPES.htm
 http://www.chem.arizona.edu/chemt/Flash/photoelectron.html
 http://www.chem.qmul.ac.uk/surfaces/scc/scat5_3.htm
 http://scidiv.bellevuecollege.edu/jm/c140/c140%20f06/POGIL/PES_140.pdf
 https://sites.google.com/site/brinnbelyeascience/home/story-centered-ap-chemistrycurriculum/units-of-study-scc-chemistry/unit-3-atomic-structure/photoelectronspectroscopy
 http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CD8QFjAB
&url=http%3A%2F%2Fwww.shodor.org%2Fteacherforum%2Fmedia%2Fdoc%2FPhotoel
ectron%2520Spectroscopy%2520Inquiry%2520F.doc&ei=5wjkUYSaPIHzygG18IGgCw&us
g=AFQjCNF3PWr2zz7uZov_9uB_bc_KncpJ6A&bvm=bv.48705608,d.aWc