A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A 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
$hv = IE + \frac{1}{2}mu^2$
where $v$ is the frequency of the UV light, $m$ and $u$ are the mass and velocity of the electron, respectively, and $h$ is the Planck constant.
In one experiment, the kinetic energy of the ejected electron from a metal is found to be $4.86 \times 10^{-19}$ J using a UV source of wavelength 127. nm. Calculate the ionization energy of the metal. Be sure your answer has the correct number of significant digits.