In the second half of the PES youtube video (https://www.youtube.com/watch?v=NRIqXeY1R_I), there is a discussion of
the PES of Kr, where the signal from the electrons in the 3d AO (atomic orbital) are at a higher energy level than the
electrons in the 4p AO, which implies that the 4p AO should be filled before the 3d AO, which is inconsistent with using
the aufbau principle to write the electron configuration of Kr. How to resolve this apparent contradiction ?
8/4/13 12:32 PM as a reply to Paul Matsumoto..
Aufbau rules are for the filling of empty orbitals. PES measures the amount of kinetic energy that a removed electron
maintains and then calculates the energy that was required to remove that electron from an already filled orbital. This is
seen frequently in the transition metals, where electrons fill the 4s before the 3d, but when ionizing, the 4s electrons
typically empty before the 3d. Likewise, if doing successive ionizations of Krypton, we would expect to
remove 4p electrons first, then 4s, and finally 3d. Hence, we state that Krypton has 8 valence electrons, and we do not
count the 3d electrons as valence.
Below, I've attached a diagram from Oxtoby, Gillis, Campion, Principles of Modern Chemistry, 6th ed., Thomson
Brooks/Cole, 2008, p. 197. You can see from the data, that upon the p subshell filling (or once the d subshell is
completely filled, however you want to think about it), the energy of that subshell drops significantly, so it is much more
bound to the nucleus. This makes some intuitive sense with Coulomb's law (more electrons and more protons means
more attraction), but it also explains the data in the YouTube video. Once the 3d sublevel is filled (as it is for Krypton), the
energy requirements to remove 4p electrons is significantly less than the energy requirements to remove 3d
electrons. Note that the graph is on a logarithmic scale, so what looks like a small difference is actually almost an order of
magnitude.
Hope this clarifies,
Jamie Benigna
The Roeper School
Birmigham, MI
I thought that PES is based on the photoelectric effect, so PES relates to the work function term in Einstein's equation
describing the photoelectric effect, which studies atoms in the solid state, while ionization energy studies atoms in the gas
state. As such, is the energy to remove an electron in an atom in the solid and gas state the same ? That is, is "ionization
energy" identical to the "work function term" in Einstein's equation describng the photoelectric effect ? If it is not the
same, what is the relative values and its rationale ?
On a related note, based on a simulation of the photoelectric effect (http://phet.colorado.edu/en/simulation/photoelectric), I
could rationalize how to get the value on the x-axis (energy of electron in the atom) in the PES simulation
(http://www.chem.arizona.edu/chemt/Flash/photoelectron.html) based on Einstein's equation, but am uncertain how to get
the value on the y-axis (relative number of electrons) / area of the various peaks in the PES spectrum, since e.g. with an
incident photon of sufficient energy to emitt an electron in the 1s AO would also be sufficient to emit an electron in the 2s
AO, so how to isolate the signal from the 1s versus 2s electron ? That is, how to relate the simulation in the photoelectric
effect to the relative area of the peaks in the PES simulation ?
8/23/13 4:00 PM as a reply to Paul Matsumoto.
Paul,
You are correct that PES is built off of the photoelectric effect. The simulated spectra that you see on the arizona site
have modified the y-axis. To answer your other questions, a little bit of the instrumentation has to be explained. Since it
is the photoelectric effect, the frequency (energy) of the incident photons are known, and this results in a variety of
different electrons in the sample being ejected. As long as the energy of the incident light is high enough, electrons from
all sublevels in an atom will be ejected. The ejected electrons travel through a hemispherical energy analyzer where
voltage is applied to determine the kinetic energy of the electrons (i.e. how much voltage is needed to move the electrons
along the hemispherical pathway). The electrons leaving the hemisphere hit a photomultiplier so that individual electrons
can be counted. The output from the machine is usually "intensity" on the y-axis, and it is usually reported in units of
counts/sec or counts/min.
The work function is dealt with in the calibration of the instrument. The machine is designed so that the work function of
the machine is the bigger factor than the work function of the sample, and the machine is calibrated with reference
samples to know what the work function of the machine will be. This is removed from the data before the output is
done. The energy of the incident photon that causes electrons to be removed equals the sum of the ejected electron's
kinetic energy, the binding energy (i.e. which orbital it came from), and the work function; so the measured kinetic energy
of the electrons can be used to back calculate the binding energy. If one turns the frequency of incident light to a higher
level, the ejected electrons have more kinetic energy. Since binding energy does not depend on the energy of light used,
it is what is put on the x-axis. On the y-axis, the number of counts are plotted, so the higher the peak (really, the greater
the area under the curve), the more electrons that have that particular binding energy. The instrument can differentiate
between 1s and 2s electrons, for example, since the 1s electrons are more bound to the nucleus, so when hit with the
same wavelength of light, they will travel away with lower kinetic energy.
If you're dealing with a pure elemental sample, then the intensities of signals are proportionate to the number of
electrons. If you're dealing with argon, for example, you'll get way more 2p electrons than 2s electrons, so the peak will
be larger. As long as it is a pure sample, then peak size can be equated easily to "relative number of electrons" since
you'll get more electrons with the 2p kinetic energy hitting the analyzer. However, the output from most instruments just
plot "intensity" or "signal strength" versus "binding energy."
Now if you have a sample that is a mixture or a compound, you can actually get infromation about elemental abundances
if you select your peaks for analysis. For example, if looking at titanium dioxide, you could see that the 1s peak for
titanium is smaller than the 1s peak for oxygen, since there are more oxygen atoms, there will be more 1s electrons from
oxygen emitted.
Hope this clarifies,
Jamie Benigna
The Roeper School
Birmingham, MI
Email to various ap chem / physics teachers & past galileo student currently in PhD chem program
. . . I prepared an Excel file
http://chem-is-try.us/class/ap/supplement/various%20topics/Li%20PES%20simulation.xlsx
to relate a simulation of the photoelectric effect
http://phet.colorado.edu/en/simulation/photoelectric
to the PES spectrum simulation
http://www.chem.arizona.edu/chemt/Flash/photoelectron.html
since it is my understanding that the PES spectrum is based on the photoelectric effect.
As such, if you have the time, could you check if my Excel simulation correctly relates the two preceding
simulations (from colorado & arizona). . . .
no response from anyone