This is new material for the AP
test this year
It is not in your book
 It is not in most books


It is a good bet there will be a couple of
multiple choice questions on this
material
Ionization energy
The energy required to remove an electron.
 It is easiest to remove electrons from the
valence shell. As we move closer, it becomes
more difficult.
 However, removing electrons makes the atom
more positive. Therefore that impacts the
data.
 What would the ionization energy be if we
didn’t first remove valence electrons?

Ionizing vs. Nonionizing Radiation
Ultraviolet, x-rays, and gamma is ionizing.
 Generally, anything of lower energy is
nonionizing
 High energy photons are absorbed by atoms.
 Normally this excites the atom, gets the electrons
to jump up.
 If they jump high enough, they will reach escape
velocity, they will be going so fast the
electromagnetic pull from the nucleus won’t be
enough to pull them back.

Photoelecton Spectroscopy
PES begins by shooting at atom with a high
energy photon (normally UV or x-ray) that is
absorbed by the atom.
 An electron is ejected carrying off the excess
energy.
 The ejected electron is measured in a detector.
 The energy of the photon absorbed (h), will be
equal to the ionization energy (IE) + kinetic
energy (KE) of the electron.
 Therefore
IE = h - KE

PES
Data from PES
We find that all electrons in a given shell
require the same energy to remove
them.
 That is to say all electrons in the 1s require
the same energy to remove, and all in the
2s, and 2p are the same, however, the
energy to remove the 2p is different than
the 1 s.
 This is different from the previous
determinations of ionization energy, with
IE1, IE2, IE3 etc.

Plotting the data
The data from PES experiments is
plotted as peeks.
 The height of the peeks is proportional
to the number of electrons of equivalent
energy ejected during the experiment.
 In other words, the height is related to
the number of electrons in that energy
level.

Peeks
1s
2s
1s