The Photoelectric Effect

This was known about for quite a few years before Einstein explained what was going on. His explanation forced us to look at light in a completely new way.

A gold leaf electroscope with a zinc cap is charged negatively. When light from a U.V. lamp shines on the cap the electroscope loses its charge. When light from a normal lamp shines on the cap it does not, even if it is very intense.

Why doesn't light from a normal lamp discharge the electroscope?
How come the light from the very bright normal lamp doesn't have enough energy to enable the electrons to escape?
How come the light from the much dimmer U.V. lamp does?

Light had always been thought of as a continuous wave, like a water wave. If we increase the intensity then we are increasing the amplitude. Higher frequency radiation just had a smaller wavelength.

Let us now imagine that light comes in little packets called photons. ( Plank's Theory )

The photoelectric effect shows us that there is a direct link between the frequency of a photon and its energy.

The relationship is E = h f  where E is the photons energy, f is its frequency in Hz and h is Planck's constant. (6.6 x 10^-34 Js)

More intense light means more photons.

The energy required for an electron to escape from a particular metal is known as its work function,    ( thi ). In general, more reactive metals have smaller work functions.

According to Einstein
An individual photon gives its energy to an individual electron. Visible light does not have enough energy to release the electrons because the photons have less energy than the work function for zinc. U.V. photons do have enough energy and the electrons escape, the difference in energy becoming kinetic energy.

Using the apparatus below we can find the energy of the photoelectrons emitted when different frequencies of light fall on a reactive metal by measuring the stopping potential across a photodiode.

Can you show that the gradient of this graph = h / e ?