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
$$
h v=I E+\frac{1}{2} m u^{2}
$$
where \(v\) 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 \(5.34 \times 10^{-19}
\mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm}\). 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)?