# Even if i keep on increasing the frequency,why does the no of photo-electrons emitted from a metal surface reach a maximum and remain constant at that value?

Jun 20, 2018

The number of incident photons (and hence electron emitted if any) per unit time is dependent on the intensity of the incoming radiation but not its wavelength or frequency.

#### Explanation:

An incident photon will ionize a single electron upon collision if the energy $E$ of the photon exceeds the work function $\phi$ of the electron. Otherwise, the photon will be reflected with no electron emission.

• Emission of an electron if $h \cdot \textcolor{n a v y}{f} = E > \phi$
• No emission if $h \cdot \textcolor{n a v y}{f} = E < \phi$

It is worth noting that despite the absorption of the incident photon, none of its energy would remain on the metallic surface or go to a second electron. Part of the photon energy is converted to electrical potential energy as the photoelectric electron leaves the potential well; the rest is converted to the kinetic energy of the electron. That is:

$h \cdot \textcolor{n a v y}{f} = E = \phi + \text{KE}$

Meaning that it is not possible for the metal surface to retain any of the photon's energy in a way that would allow the emission of another electron in the absence of incident photons, or for the electron emitted to release another photon that will ionize the second electron following the absorption of the initial photon. Neither is it possible for the photon to hit two electrons simultaneously since by the Pauli Exclusion principle the two electrons cannot occupy the identical quantum state.

Thus the frequency- or equivalently the energy- of the photon only decides whether there is an emission along with the energy of the electron emitted in cases of emission.