Einstein’s equation and the photon

Einstein’s photoelectric equation

Narrated video. Transcript:

In 1905, Albert Einstein made a bold proposal. Light energy is not spread out continuously. It comes in discrete packets, called quanta, each of energy h ν, where ν is the frequency and h is Planck’s constant, 6.63 × 10⁻³⁴ joule second.

Today we call these packets photons. In the photoelectric effect, one electron absorbs one whole photon, all at once, and receives its entire energy h ν.

The electron spends part of this energy getting out of the metal. Electrons right at the surface need the least, just φ₀. Whatever is left becomes kinetic energy.

So the fastest photoelectrons have K max equals h ν minus φ₀. This is Einstein’s photoelectric equation.

Now every fact falls into place. K max depends on ν, and grows in a straight line with it. Intensity does not appear in the equation, so K max does not depend on intensity.

K max cannot be negative. So electrons come out only if h ν is greater than φ₀, that is, ν greater than ν₀ equals φ₀ over h. That is the threshold frequency. A bigger work function means a higher threshold.

Brighter light means more photons per second, not bigger photons. More photons free more electrons, so the photocurrent is proportional to the intensity.

And since each electron takes in its photon in a single event, there is no waiting. Emission is instantaneous, even for the dimmest light.

Write K max as e V₀: e V₀ equals h ν minus φ₀. So V₀ equals h over e, times ν, minus φ₀ over e. A straight line of V₀ against ν, with slope h over e, the same for every metal.

Between 1906 and 1916, Robert Millikan tested this with careful experiments. The lines were straight, and their slope gave a value of h in close agreement with Planck’s. Einstein received the Nobel Prize in 1921 for this work.

Worked example: caesium

Narrated video. Transcript:

The work function of caesium is 2.14 electron volt. Let us find its threshold frequency, and the light needed for a given stopping potential.

Threshold: h ν₀ equals φ₀. So ν₀ equals 2.14 × 1.6 × 10⁻¹⁹ joule, divided by 6.63 × 10⁻³⁴ joule second. That gives 5.16 × 10¹⁴ hertz.

The matching wavelength is c over ν₀: 3 × 10⁸ divided by 5.16 × 10¹⁴, about 581 nanometres. That is yellow light. Any wavelength longer than this, like red light, cannot free electrons from caesium.

Now light falls on the caesium, and the photocurrent is stopped by a stopping potential of 0.60 volt. What is the wavelength of the light?

From Einstein’s equation, h ν equals φ₀ plus e V₀: 2.14 plus 0.60, which is 2.74 electron volt.

Then λ equals h c over the photon energy: 6.63 × 10⁻³⁴ times 3 × 10⁸, divided by 2.74 × 1.6 × 10⁻¹⁹. That is 4.54 × 10⁻⁷ metre, or 454 nanometres: blue light.

A useful shortcut: h c is about 1240 electron volt nanometre. So λ in nanometres is 1240 divided by the energy in electron volt. 1240 divided by 2.74 gives about 453 nanometres, the same answer.

The same shortcut tells us a photon of 400 nanometre violet light carries 3.1 electron volt, and one of 700 nanometre red light only about 1.8.

The photon

Narrated video. Transcript:

So in its interaction with matter, light behaves as if it is made of particles: photons. Let us list what we know about them.

One. Each photon has energy E equals h ν, and momentum p equals h ν over c, which is h over λ. It travels at the speed of light, c.

Two. All photons of a given frequency have the same energy and momentum, whatever the intensity. Brighter light simply means more photons crossing each square metre every second.

Three. Photons are electrically neutral. Electric and magnetic fields do not deflect them.

Four. When a photon collides with a particle such as an electron, total energy and total momentum are conserved. But the number of photons need not be: a photon can be absorbed, or a new one created.

Now an example. A laser gives light of frequency 6.0 × 10¹⁴ hertz, with a power of 2.0 milliwatt. What is the energy of each photon?

E equals h ν: 6.63 × 10⁻³⁴ times 6.0 × 10¹⁴, which is 3.98 × 10⁻¹⁹ joule, about 2.5 electron volt.

How many photons per second? Power divided by the energy of one photon: 2.0 × 10⁻³ divided by 3.98 × 10⁻¹⁹. That is about 5.0 × 10¹⁵ photons every second.

Five thousand million million photons each second, from a small laser pointer. Each one is so tiny that light looks perfectly smooth and continuous to our eyes.

Try it: measure h

Interactive simulation. Things to try:

• Sodium already has two points. For caesium, press “Record (ν, V₀)” at a few different wavelengths.
• The slope of each line is h/e ≈ 4.14 × 10⁻¹⁵ V s. Multiply by e = 1.6 × 10⁻¹⁹ C to get h ≈ 6.6 × 10⁻³⁴ J s.
• Both lines are parallel: the same h for every metal. Only the intercept ν₀ = φ₀/h changes.
• Switch to the I–V graph and press “Find V₀”: the stopping potential equals K_max = hν − φ₀, in volts.

Violet light on potassium

Light of wavelength 400 nm falls on potassium (φ₀ = 2.30 eV). What is the stopping potential? (Take hc = 1240 eV nm.)

1. 0.80 V
2. 3.10 V
3. 5.40 V
4. 2.30 V

Counting photons

A laser of power 2.0 mW emits light of frequency 6.0 × 10¹⁴ Hz. How many photons does it emit each second?

1. 3.98 × 10⁻¹⁹
2. 1.2 × 10¹²
3. 5.0 × 10¹⁵
4. 5.0 × 10¹⁸

Summary

• Einstein’s photoelectric equation: K_max = hν − φ₀ = eV₀
• Threshold: ν₀ = φ₀ / h, λ₀ = hc / φ₀
• Stopping potential line: V₀ = (h/e) ν − φ₀/e
• Photon: E = hν, p = hν/c = h/λ

• Light energy comes in photons of energy hν. One electron absorbs one whole photon, at once.
• The equation explains everything: K_max rises linearly with ν and does not depend on intensity; there is a threshold ν₀; more intensity means more photons and so more current; there is no time lag.
• The V₀–ν graph is a straight line of slope h/e, the same for all metals. Millikan’s measurements confirmed it.
• All photons of a given frequency have the same energy and momentum. They are neutral, travel at c, and conserve energy and momentum in collisions, though their number can change.
• Handy: hc ≈ 1240 eV nm, so E (in eV) = 1240 / λ (in nm).