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

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

$E = h f$
where $f$ is the frequency in Hz, and $h$ is Planck's constant, equal to $6.626 \cdot {10}^{-} 34 J \cdot s$ ($h$ can also be expressed in terms of eV, as $4.136 \cdot {10}^{-} 15 e V \cdot s$).

What this means is if, for example, the work function of an electron on a surface is 2 eV (which is $3.204 \cdot {10}^{-} 19 J$ 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:

$E = h f$
$f = \frac{E}{h}$
$f = \frac{2 e V}{4.136 \cdot {10}^{-} 15 e V \cdot s}$
$f = 4.836 \cdot {10}^{14} \text{Hz}$

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.