Transcript PES ppt

Photoelectron Spectroscopy

Straightforward evidence for the validity of orbital diagram

Konsler 2013

Electron Configuration Is there any direct evidence that this diagram is accurately showing potential energy of electrons on the atom?

Some Available Evidence • Atomic Emission Spectra • Successive Ionization • Photoelectron Spectroscopy

The topic is probably best introduced in1 st year Chemistry under Atomic Electronic Structure

Atomic Emission Spectra

Atomic emission spectra are evidence that there are specific allowed PE values by showing the energy difference between some of them.


Differences between PE values, not the values themselves.

Ambiguous origin and destination orbitals.

Impossible to visually compare PE of orbitals on multiple atoms; large data sets will require many calculations.

Successive Ionization Using successive ionization it is usually possible to determine the number of valence electrons an atom has. This is evidence that our PE diagram is an accurate representation

Limitations Each ionization causes a reorganization of the remaining electrons, meaning the successive ionization is not a measurement of the characteristics of the original atom. We infer there is a relationship between the ions and the parent atom.

Limitations It’s not feasible to do more than 5 successive ionizations, making core electrons (and sometimes even valence electrons) impossible to obtain values for.

Limitations It’s not possible to determine if electrons at the same energy level on the atom have the same PE


, because we remove them one at a time. As a result, the method is really a support only for n, not the other 3 quantum numbers.

Photoelectron Spectroscopy



= hv

Atom Each event happens event. Since E photon moment of the event.


single atom. This is a quantum > IE electron all the electrons on the atom, the electron removed is random and the KE of the electron is a it exists on the atom at the for a for characteristic of that electron as Monochromatic Beam of X-Rays IE electron = E photon - KE

KE = mv

2 e -


Negatively Charged Hemisphere (Constant V) Retarding Voltage is directly proportional to KE electron Dependent Variable (y)


Negatively Charged Hemisphere (Constant V) Independent Variable (x)

Retarding Voltage of Lens

Electron Detector Beam of Atoms Electrons scan past detector Electrostatic Lens (Focuses and slows electrons) The intensity is measured as the retarding voltage on the lens is reduced at a constant rate X-Rays (monochromatic) Electrons ejected

Photoelectron Spectroscopy is still limited by E photon . If IE electron is too large, the electron will not be detected. This means that electrons with low values of n on atoms with high value of Z may be off the left hand side of the chart.

High Ionization Energy Low electron PE Low Ionization Energy High electron PE 20 MJ/mol 10 MJ/mol IE electron = E photon - KE electron 0 MJ/mol

However, a fluorescence source emits X-Rays with energy of over 1000 eV, several times greater than for successive ionization instrumentation. In addition, core electrons are removed from a neutral atom, requiring less energy.

High Ionization Energy Low electron PE Low Ionization Energy High electron PE 20 MJ/mol 10 MJ/mol IE electron = E photon - KE electron 0 MJ/mol

Photoelectron Spectrum

Valence Helium possesses the valence electron with the lowest PE of all elements. The measured value for He is 2.37 MJ/mol. Any values observed with greater values (to the left of this line) must be core electrons. Note, d-subshell core electrons will be to the right of this line.

20 MJ/mol 10 MJ/mol 0 MJ/mol “Valence Line”: He(1s) = 2.37 MJ/mol

Photoelectron Spectrum

Relative Intensity = 2 Since each event removes a random electron from a separate atom, the relative intensity shows the proportion of electrons at that PE.

Relative Intensity = 1 20 MJ/mol 10 MJ/mol 0 MJ/mol

Photoelectron Spectrum

19.3 MJ/mol RI = 2 Valence 1.36 MJ/mol RI = 2 0.80 MJ/mol RI = 1 20 MJ/mol 10 MJ/mol 0 MJ/mol

Photoelectron Spectrum

19.3 MJ/mol RI = 2



Boron (Z=5)

Analysis: 1) Valence has 2 values: 2) RI is 2 to 1 in valence: 3) Closest core has RI 2 not 6: 4) s 2 s 2 p 1 must be 1s 2 2s 2 2p 1 sp s 2 p 1 not p x s 2 p 1


2 1.36 MJ/mol RI = 2 0.80 MJ/mol RI = 1


1 20 MJ/mol 10 MJ/mol 0 MJ/mol

Photoelectron Spectrum 2p

6 3.67 MJ/mol

Sodium (Z=11) 1s

2 104 MJ/mol


2 6.84 MJ/mol


1 0.50 MJ/mol { } 8 MJ/mol 4 MJ/mol 0 MJ/mol

But there aren’t even any instruments!

• All PES spectra will probably be “simulated” for the following reasons: • PES post-dates atomic theory.

• Most PES is a solid-phase technique. Having so many nearby atoms will cause interference. For example, an electron promoted out of a low lying orbital will leave a vacancy which can be occupied by an outer orbital electron with concomitant production of a photon. If that photon is of high enough energy, it might eject a more easily ionized electron in the same atom or a nearby atom. This will result in the production of background interference and even spurious peaks. As a result, PES spectra derived from solids would be impossible for a non-expert to interpret.

• • • •

Why do it?

Although it requires “simulated spectra” it is superior to other methods we describe in General Chemistry because, even in principle, atomic emission and successive ionization do not provide data which could be used to identify an atom in the context of an exam. At best they would allow a student to choose the most likely among a set of possibilities.

Photoelectron spectroscopy allows questions to be written using visual as opposed to tabular data (as is required for successive ionization).

Students can be asked to interpret a spectrum but they can also legitimately be asked to draw one. It’s a decently easy grade if you give the axes and requires essentially no “give away” information in the stimulus.

Therefore, while the instruments aren’t accessible, understanding the principle allows more thoughtful questions to be asked about electronic structure than were previously possible.