How would you explain the photoelectric effect in terms of energy?

1 Answer
Mar 29, 2016


A minimum amount of energy (called the work function ) is required to dislodge a photon from a surface. That energy must be delivered by a single packet of light energy, otherwise known as a photon.


Photons are "packets" or particles of light, each containing as specific amount of energy, which is the product of the frequency of the light.

The energy of a single photon is

where #f# is the frequency in Hz, and #h# is Planck's constant, equal to #6.626*10^-34 J*s# (#h# can also be expressed in terms of eV, as #4.136*10^-15 eV*s#).

What this means is if, for example, the work function of an electron on a surface is 2 eV (which is #3.204*10^-19J# which is an uglier number, and the reason we use eV for small energies), then we need to hit it with a photon of at least 2 eV to dislodge it. Interestingly, the energy has to be delivered at once, so we can't hit it with, say three photons of 1 eV.

We can calculate the minimum frequency of light required to dislodge the electron:

#f=(2eV)/(4.136*10^-15 eV*s)#
#f=4.836*10^14 "Hz"#

Light with a lower frequency will never dislodge that electron.
Light with a higher frequency can give the photon extra kinetic energy to fling it away from the surface faster.