How does intensity affect the photoelectric effect?

1 Answer
May 9, 2015

If the frequency of electromagnetic waves is higher than the extraction threshold of the metal and electrons are emitted from the metal surface, then an increase of light intensity will result in a proportional increase of electrical current of the electrical circuit where the emitted electrons are conveyed.

This is simply explained by the photon model of light. In that model an electromagnetic wave carries its energy not like a continuum, but as multitude of grains of energy which have the same individisible amount of energy. The indivisible grain of light energy, also called "light dart" by Einstein, and then electromagnetic energy quanta or photons is given by the Planck-Einstein's law:

#E_"photon" = h*f"#

where #f# is the frequency of light. For each single photon at the same frequency.

From the point of view of photons, a more intense light is not made of "higher waves", but of a higher number of photons.

Than is obvious that, if each photon is capable to expel an electron, the more intense is the light the more number of electrons will be expelled. If frequency is under the threshold, then even the higher intense light will be equally incapable to eject even a single electron.

It is impossible to explain these experimental outcomes with the wave model of light.