I understand that distinct absorption/emission spectra occur due to electrons requiring specific amounts of energy, and therefore specific wavelengths of light, to move between specific excited states. However, when you are completely removing an electron, as occurs in the photoelectric effect with metals or when ionizing an atom, do you still need a specific amount of energy (and therefore specific wavelength)? Or would the "leftover" energy from the light (hf - ionization energy in the context of ionization or hf - the work function in the context of the photoelectric effect) just become the electron's kinetic energy? If light removed an electron and had a quantity of energy exactly equal to the ionization energy or exactly equal to the work function, would the electron's kinetic energy be 0?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The latter. You simply need enough energy to ionize the atom. For example, to ionize hydrogen in the ground state (ionization potential 13.6 eV), any photon with energy greater than 13.6 eV (wavelength less than 91.2 nm) is capable of ionizing the atom; the higher the energy of the photon, the more kinetic energy will be injected into the system by the ionization process.
If the energy of the ionizing photon were exactly the ionization energy, you are correct that no energy would be injected by the photon. (That does not mean that the electron's kinetic energy will be zero, though.)
-
$\begingroup$ Would the ionizing wavelengths still appear in the absorption spectrum, even though an electron could absorb them to leave the atom? $\endgroup$– AkashJan 31, 2021 at 15:15
-
$\begingroup$ Yes. In astronomy, you see this commonly as a jump in the intensity in the spectrum of a star, galaxy, or quasar: the intensity will be much lower at wavelengths less than the ionization threshold. These jumps look very different than absorption lines, which are only at one wavelength. See, for example, Balmer jump (which corresponds to the ionization of a hydrogen atom in the n=2 state). en.wikipedia.org/wiki/Balmer_jump $\endgroup$ Jan 31, 2021 at 18:09