hv = IE + zmu
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7.129 A technique called photoelectron spectroscopy is used
to measure the ionization energy of atoms. A gaseous
sample is irradiated with UV light, and electrons are
ejected from the valence shell. The kinetic energies of
the ejected electrons are measured. Because the energy
of the UV photon and the kinetic energy of the ejected
electron are known, we can write where ν is the frequency of the UV light, and m and u
are the mass and velocity of the electron, respectively.
In one experiment the kinetic energy of the ejected
electron from potassium is found to be 5.34 × 10−19 J
using a UV source of wavelength 162 nm. Calculate the
ionization energy of potassium. How can you be sure
that this ionization energy corresponds to the electron in
the valence shell (i.e., the most loosely held electron)?
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