Electromagnetic radiation interacts with matter in different ways. One way is the Photoelectric Effect where an incident ray interacts with an atom and ejects an electron. The maximum kinectic energy of the electron is equal to energy of the incident ray less the binding enery of the electron (known as the Work Function. Recall that the energy of a photon is \(E=hf\).
\( KE_{max} = E_{\gamma} - \Phi \)
\( KE_{max} = hf - \Phi \)
The photoelectric effect is seen when light is shown on a metal. If the energy of the incident light is less than the Work Function, no electrons will be emitted. The cut-off frequency is the minimum frequency required to eject an electron.
\( KE_{max} = hf - \Phi \)
\( 0 = hf - \Phi \)
\( f_{c} = \frac{\Phi}{h} \)
Light of \(12\cdot10^{15} \; [Hz]\) is shown on a metal. What is the value of the work function for the metal if the electrons are observed to have a kinectic energy of \(40 \; [eV]\)?