Answer to Question #123018 in Quantum Mechanics for Muhammad Adnan Sabir

Question

In the photoelectric effect, electromagnetic radiation incident on a metal
surface may eject electrons, but only if the frequency of the radiation exceeds
a threshold value.
Show that this frequency threshold can be understood if the mechanism
for the photoelectric effect involves an interaction between a photon and an
electron, and if the energy needed to eject the electron has a minimum value.
The minimum energy needed to eject an electron from the surface of
magnesium is 3.68 eV. Show that light with a frequency below 8:89 -
1014 Hz
cannot produce photoelectrons from magnesium, no matter how intense the
illumination may be.

Answer 1

To eject the electron from the metal we need a photon, that has enough energy to ionize the atom and to give the electron an amount of kinetic energy. Therefore, there is a threshold of minimal energy needed to eject the electron. So we may write the formula

$E_{\text{ph}} = E_{\text{ion}} + \dfrac{m_ev^2}{2}$ (see https://en.wikipedia.org/wiki/Photoelectric_effect#Mathematical_description). To obtain the minimal energy of photon we may consider the kinetic energy of electron equal to 0, so $E_{\text{ph}} = E_{\text{ion}}, \;\; h\nu = E_{\text{ion}}.$

So $\nu = \dfrac{E_{\text{ion}}}{h} = \dfrac{3.68\cdot1.6\cdot10^{-19}\,\mathrm{J}}{6.626\cdot10^{-34}\,\mathrm{J\cdot s}} = 8.89\cdot10^{14}\,\mathrm{Hz}.$ This is the minimal frequency needed to produce the ejection of electron. The intensity of light will influence only the number of ejected electrons, but the ejection is produced only with light of the enough frequency.