Question

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

Answer 1

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.

Explanation:

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

`E=hf`
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:

`E=hf`
`f=E/h`
`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.