The Photoelectric Effect

A surprise

Shine light on a clean metal surface. Under the right conditions, electrons leave the metal. This is the photoelectric effect.

Physicists tried to explain it with the wave picture of light. It did not work. Einstein explained it in 1905 with a new idea, and this idea won him a Nobel Prize.

What the experiments show

1. Below a certain frequency $\nu_0$, called the threshold frequency, no electrons leave the metal at all. Even very bright light does not help.
2. Above the threshold, electrons leave at once, even with very dim light.
3. The maximum kinetic energy of the electrons depends only on the frequency of the light. It does not depend on the brightness.
4. A brighter light gives more electrons every second, so the current is bigger. The electrons are not faster.

The wave picture predicts the opposite of the first three results. A bright wave of any frequency should free the electrons after some time. And brighter light should give the electrons more energy.

Einstein's idea: photons

Light comes in packets. Each packet is a photon, and its energy is

Here $h = 6.63 \times 10^{-34}\ \text{J s}$ is Planck's constant. The Greek letter $\nu$ (read "nu") is the frequency of the light. In modern physics, we use $\nu$ for frequency, and in the lessons on waves and sound, we use $f$.

One photon gives all its energy to one electron. The electron needs some energy just to leave the metal. This is the work function $\phi$. The rest of the energy becomes the kinetic energy of the electron.

Let us see what this says.

• If $h\nu$ is smaller than $\phi$, no electron can leave. This is the threshold. At the threshold, $h\nu_0 = \phi$, and the threshold wavelength is $\lambda_0 = \dfrac{hc}{\phi}$.
• A brighter light has more photons. It does not change the energy of each photon. So it gives more electrons, and the same $KE_{\max}$.

A shortcut for the numbers

If we use electronvolts (eV) for energy and nanometres (nm) for wavelength, then $hc \approx 1240\ \text{eV nm}$.

An example. A metal has a work function of $2.3\ \text{eV}$. Light of $400\ \text{nm}$ falls on it. The photon energy is $\dfrac{1240\ \text{eV nm}}{400\ \text{nm}} = 3.1\ \text{eV}$. So $KE_{\max} = 3.1\ \text{eV} - 2.3\ \text{eV} = 0.8\ \text{eV}$.

The stopping potential

We make the collecting plate negative, and we slowly increase the voltage. At some voltage, even the fastest electrons cannot reach the plate. This voltage is the stopping potential $V_0$.

So a stopping potential of $0.8\ \text{V}$ means that $KE_{\max} = 0.8\ \text{eV}$. If we draw $V_0$ against the frequency $\nu$, we get a straight line.

The slope of the line is $\dfrac{h}{e}$. It is the same for every metal, so the lines for two metals are parallel. The metal with the bigger work function crosses the frequency axis farther to the right, at a higher threshold frequency.

You should find that no electrons leave below the threshold, and that above it the point slides up the line as the wavelength gets shorter.

Practise with new numbers

Try it in Python

This program shines light of six wavelengths on a metal that has a work function of $2.3\ \text{eV}$.

Now we use the line that we saw in the picture. This program has five readings of the stopping potential. It fits a straight line, and it finds Planck's constant and the work function.

Remember these

• $E = h\nu = \dfrac{hc}{\lambda}$, and $hc \approx 1240\ \text{eV}$ nm.
• $KE_{\max} = h\nu - \phi$. There are no electrons if $h\nu < \phi$.
• Brightness changes the number of electrons. Frequency changes their energy.
• $eV_0 = KE_{\max}$. The graph of $V_0$ against $\nu$ is a straight line with the slope $h/e$.

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 of a photon of wavelength $500\ \text{nm}$?

2. A sodium surface has a work function of $2.3\ \text{eV}$. Light of $300\ \text{nm}$ falls on it. What is the largest kinetic energy of the electrons?

3. A bright red beam cannot eject electrons from a metal, but a dim blue beam can. Why?

4. The stopping potential is $1.1\ \text{V}$. What is the largest kinetic energy of the electrons, in joules?