Einstein's equation and the stopping potential

How many photons does a torch send?

A small torch with a green LED looks as if it sends out a smooth stream of light. Einstein said it does not. It sends a stream of separate packets of energy called photons. How many? We can count them.

The energy of one photon is $E = h\nu = hc/\lambda$. With the energy in eV and the wavelength in nm, $hc = 1240$ eV nm.

A worked example. A torch sends out $0.50$ W of green light of wavelength $530$ nm. How many photons does it send each second?

1. Energy of one photon. $E = \dfrac{1240\ \text{eV nm}}{530\ \text{nm}} = 2.34\ \text{eV}$.
2. In joule. $2.34\ \text{eV} \times 1.6 \times 10^{-19}\ \text{J/eV} = 3.74 \times 10^{-19}\ \text{J}$.
3. The number each second. We want $N$, so $N = \dfrac{P}{E} = \dfrac{0.50\ \text{W}}{3.74 \times 10^{-19}\ \text{J}} = 1.34 \times 10^{18}\ \text{s}^{-1}$.
4. Check. More than a billion billion photons a second. That is why light looks smooth to the eye.

Slide the wavelength from violet to red at the same power. Does the number of photons go up or down? It goes up. Red photons carry less energy, so the same power needs more of them.

Einstein's picture in full

In 1905 Einstein proposed that light arrives in photons, each with energy $E = h\nu$. The intensity is the number of photons that cross a unit area each second, times the energy of one photon. One photon gives all its energy to one electron, or none of it.

An electron needs the work function $\phi$ of the metal to leave. The rest of the photon energy is kinetic energy:

Each law now follows.

• Intensity. More light means more photons, so more electrons and a bigger saturation current. Each photon has the same energy, so $K_{\max}$ does not change.
• Frequency. A higher frequency means more energy in each photon, so $K_{\max}$ and $V_0$ rise.
• Threshold. At $\nu_0$, a photon has just enough energy: $h\nu_0 = \phi$. Below it, no single photon can free an electron, and two photons cannot join forces.
• No delay. One photon gives its energy in one step.

Why "largest"? An electron deep in the metal loses some energy on its way to the surface. So the electrons leave with every energy from zero up to $K_{\max}$. That is why the current in the curves of the last lesson falls slowly as the collector becomes more negative.

Not every photon frees an electron. Many are reflected, and some only warm the metal. The fraction that frees an electron is the quantum efficiency. It is often a few per cent or less.

The graph of stopping potential against frequency

Put $K_{\max} = eV_0$ into Einstein's equation:

This is a straight line. Its slope is $h/e = 4.14 \times 10^{-15}$ V s. It cuts the frequency axis at $\nu_0 = \phi/h$. Its intercept on the $V_0$ axis is $-\phi/e$.

The slope is $h/e$ for every metal, so the lines of two metals are parallel. A metal with a larger work function has its line further to the right.

Millikan's test

Einstein's equation was bold. Robert Millikan did not believe it. Between 1914 and 1916 he built a vacuum tube in which a knife could shave the surface of sodium and other metals, so the surface stayed clean. He measured $V_0$ at many frequencies. The points fell on a straight line.

The slope gave $h = 6.57 \times 10^{-34}$ J s, within about half of one per cent of the best value known. Millikan had hoped to prove Einstein wrong. He proved him right. Einstein received the Nobel Prize in 1921 for this explanation, and Millikan received it in 1923.

Photocells are put to use in automatic doors, burglar alarms, the light meter of a camera and the sound track of old films.

A worked example. A student measures the stopping potential of a metal M. It is $0.55$ V at $7.5 \times 10^{14}$ Hz and $1.38$ V at $9.5 \times 10^{14}$ Hz. Find Planck's constant, the threshold frequency and the work function.

1. The slope. $\dfrac{1.38\ \text{V} - 0.55\ \text{V}}{2.0 \times 10^{14}\ \text{Hz}} = 4.15 \times 10^{-15}\ \text{V s}$.
2. Planck's constant. The slope is $h/e$, so $h = 1.6 \times 10^{-19}\ \text{C} \times 4.15 \times 10^{-15}\ \text{V s} = 6.64 \times 10^{-34}\ \text{J s}$.
3. The threshold. At the first point, $0.55\ \text{V} = 4.15 \times 10^{-15}\ \text{V s} \times (7.5 \times 10^{14}\ \text{Hz} - \nu_0)$. So $\nu_0 = 7.5 \times 10^{14}\ \text{Hz} - 1.33 \times 10^{14}\ \text{Hz} = 6.17 \times 10^{14}$ Hz.
4. The work function. $\phi = h\nu_0$, and in eV this is the slope times $\nu_0$: $4.15 \times 10^{-15}\ \text{V s} \times 6.17 \times 10^{14}\ \text{Hz} = 2.56$ eV.
5. Check. At the first point, $h\nu - eV_0$ in eV is $4.15 \times 10^{-15} \times 7.5 \times 10^{14} - 0.55 = 2.56$ eV. It agrees.

Practise with new numbers

Try it in Python

Real measurements never lie exactly on a line. The program fits the best straight line through six readings of the stopping potential, and reads off Planck's constant, the threshold frequency and the work function.

Common mistakes

• Taking the slope as $h$. The slope of $V_0$ against $\nu$ is $h/e$. Multiply it by $e$ to get $h$.
• Reading the wrong intercept. The intercept on the $V_0$ axis is $-\phi/e$, which is negative. The work function is its size, in eV.
• Mixing eV and joule. When you divide a power in watt by a photon energy, the energy must be in joule.
• Thinking different metals give lines of different slope. Every metal gives the slope $h/e$. Only the position of the line changes.

Remember these

• $E = h\nu = hc/\lambda$, with $hc = 1240$ eV nm. Photons each second: $N = P/E$.
• $K_{\max} = h\nu - \phi = eV_0$.
• The graph of $V_0$ against $\nu$ is a straight line. Slope $h/e$. Intercept on the $\nu$ axis $\nu_0$. Intercept on the $V_0$ axis $-\phi/e$.
• Photocurrent $I = \eta Ne$, with $\eta$ the fraction of photons that free an electron.

Check yourself

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

1. In Millikan's experiment, what does the slope of the graph of stopping potential against frequency give?

2. The stopping potential lines of metals P and Q are parallel. P cuts the frequency axis at $6.0 \times 10^{14}$ Hz and Q at $8.0 \times 10^{14}$ Hz. Which metal has the larger work function? Which has the larger stopping potential at $10 \times 10^{14}$ Hz, and by how much?

3. A source sends $4.0 \times 10^{17}$ photons each second, and each has an energy of $2.5$ eV. What is its power?

4. The intensity of the light is doubled, and twice as many photons arrive each second. Why does $K_{\max}$ stay the same?