Light as particles: the photon and the photoelectric effect

A glowing iron rod

Heat an iron rod in a fire. First it glows dull red. Then it becomes orange, yellow and finally white. The colour depends on the temperature. Scientists at the end of the 19th century tried to explain the amount of light of each colour that a hot object gives out. The wave theory gave a result that did not agree with the measurements at short wavelengths.

In $1900$, Max Planck found a way out, with a bold idea: energy is not given out in a smooth stream. It is given out in small packets, called quanta. The energy of one quantum of light of frequency $\nu$ is

Here $h = 6.626 \times 10^{-34}\ \text{J s}$ is Planck's constant. It is very small, so for large objects the packets are too small to notice. For an atom, they matter a great deal.

The photoelectric effect

When light falls on a clean metal surface, electrons can come out of it. This is the photoelectric effect. Hertz and Lenard found these facts:

1. A threshold frequency. For each metal there is a lowest frequency $\nu_0$. Below it, no electron comes out, however bright the light is, and for however long.
2. No delay. Above $\nu_0$, electrons come out at once, even in very dim light.
3. The number of electrons is proportional to the intensity (brightness) of the light.
4. The kinetic energy of the electrons depends on the frequency of the light, and not on its intensity.

The wave theory fails on the first, second and fourth points. A bright wave carries a lot of energy, so it should free electrons at any frequency, and a very dim wave should need time to build up enough energy.

Einstein's explanation

In $1905$, Einstein said that light itself is made of packets, which we call photons. A photon of frequency $\nu$ has the energy $h\nu$. One photon gives all of its energy to one electron, at once.

To leave the metal, the electron must pay a fixed amount of energy, called the work function $W$. The rest of the photon's energy stays with the electron as kinetic energy:

• If $h\nu < W$, the photon cannot free an electron. This is the threshold: $\nu_0 = \dfrac{W}{h}$.
• A brighter light has more photons, so more electrons come out. The energy of each photon does not change.
• A higher frequency gives each electron more energy.

The graph of $KE_{\max}$ against $\nu$ is a straight line. Its slope is $h$, the same for every metal. Its intercept on the frequency axis is $\nu_0$, and metals with a larger work function have a higher threshold.

The stopping potential $V_0$ is the voltage that just stops the fastest electrons: $eV_0 = KE_{\max}$.

A useful shortcut

Chemists measure photon energy in the electronvolt: $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$. Putting the numbers in, $E = \dfrac{hc}{\lambda}$ gives

Example. Light of $400\ \text{nm}$ falls on caesium, with $W = 2.14\ \text{eV}$.

• The photon has $E = \dfrac{1240}{400} = 3.10\ \text{eV}$.
• $KE_{\max} = 3.10 - 2.14 = 0.96\ \text{eV}$, so the stopping potential is $0.96\ \text{V}$.
• The threshold wavelength is $\lambda_0 = \dfrac{1240}{2.14} = 579\ \text{nm}$. Light of a longer wavelength, such as red, cannot free electrons from caesium.

You should find that no electrons leave until the photon energy passes the work function, however bright the light.

Counting photons

A beam of green light of $500\ \text{nm}$ and a power of $1\ \text{W}$ gives out $\dfrac{1}{3.98 \times 10^{-19}} = 2.5 \times 10^{18}$ photons every second. One photon is very small, and in everyday life we see so many of them together that light looks smooth.

A wave and a particle

Light shows interference, so it is a wave. Light also knocks out single electrons, one photon for one electron, so it also behaves as a stream of particles. This is the dual nature of light. No single picture is right at all times, and the experiment decides which behaviour we see.

Practise with new numbers

Try it in Python

This program finds the energy of a photon of green light, and the number of photons in one watt.

Remember these

• $E = h\nu = \dfrac{hc}{\lambda}$, with $h = 6.626 \times 10^{-34}\ \text{J s}$. In electronvolts, $E = \dfrac{1240}{\lambda\ (\text{nm})}$.
• Photoelectric effect: $h\nu = W + KE_{\max}$, and $W = h\nu_0$.
• No electron comes out below $\nu_0$. A brighter light gives more electrons, and a higher frequency gives faster electrons.
• Light behaves as a wave and as a stream of photons.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the energy, in eV, of a photon of $413\ \text{nm}$?

2. Why does red light free no electrons from zinc, however bright it is?

3. The stopping potential for some photoelectrons is $1.2\ \text{V}$. What is their greatest kinetic energy?

4. What happens to the energy of each photon when the frequency of light is doubled?