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

1 Answer
Ed Hitchcock
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.

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⋅10−34J⋅s (h can also be expressed in terms of eV, as 4.136⋅10−15eV⋅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=Eh
f=2eV4.136⋅10−15eV⋅s
f=4.836⋅1014Hz

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.

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