The photoelectric effect and Einstein's light quanta

The previous section ended at a boundary line: Planck held that “one unit at a time” was a property of the oscillators in the cavity wall, while light travelling through space remained a continuous electromagnetic wave.

This section is about someone crossing that line.

Simple experiment, awkward results

The apparatus has three parts: a metal plate, a collector and an ammeter. Shine light on the plate; if electrons are knocked out and reach the collector, the ammeter reads.

By 1902 Lenard had the measurements nailed down. Four facts:

Estimate: how long classical theory says you must waitbasic~5 minexpand

Putting a number on “accumulating energy” shows how fatal fact 4 is.

Take a faint beam with power density $I=10^{-6}\ \text{W}/\text{m}^2$ (roughly a candle a few metres away). An atom has a cross-section of about $\sigma\approx10^{-20}\ \text{m}^2$.

The power intercepted by one atom is then

$$P=I\sigma=10^{-26}\ \text{W}\tag{1.3.1}$$

Ejecting an electron requires overcoming the work function, typically $W\approx2\ \text{eV}=3.2\times10^{-19}\ \text{J}$. The time to accumulate that is

$$t=\frac{W}{P}=\frac{3.2\times10^{-19}}{10^{-26}}\approx3\times10^{7}\ \text{s}\tag{1.3.2}$$

About a year.

The measured delay is under a nanosecond. Sixteen orders of magnitude apart — this is not a theory that lacks precision, it is a picture that is simply wrong.

Einstein changed one sentence

In 1905 the 26-year-old Einstein, working in a patent office, wrote a paper titled “On a heuristic point of view concerning the production and transformation of light”. He said:

That is the whole change. Now go through the four facts again:

$$h\nu = W + E_{k,\text{max}}\tag{1.3.3}$$

$W$ is the work function, the minimum cost of pulling an electron out of the metal. $E_{k,\text{max}}$ is what the electron has left over.

• Threshold frequency: if $h\nu<W$, a single photon cannot pay for the ticket and no electron emerges. More photons do not help, because pooling is not allowed — an electron absorbs one at a time. Hence $\nu_0=W/h$.
• What intensity does: intensity equals the number of photons. More photons knock out more electrons (larger current), but each still receives $h\nu$, so the energy ceiling is unchanged.
• Linear in frequency: $E_{k,\text{max}}=h\nu-W$, a straight line of slope exactly $h$ and intercept $-W$.
• Instant response: nothing needs accumulating. A photon arrives, hits an electron, and the transaction is done.

Millikan came to refute it and nailed it down instead

Einstein’s paper was not popular. Even Planck did not accept it — putting $h$ inside light in flight struck him as going too far.

Work it out: the threshold wavelength of sodiumbasic~4 minexpand

Sodium has a work function $W=2.28\ \text{eV}$. The corresponding threshold wavelength is

$$\lambda_0=\frac{hc}{W}=\frac{1240\ \text{eV}\cdot\text{nm}}{2.28\ \text{eV}}=544\ \text{nm}\tag{1.3.4}$$

(Remember $hc=1240\ \text{eV}\cdot\text{nm}$ — one of the most useful constants there is.)

544 nm is green. So green, blue and violet light eject electrons from sodium; yellow and red never will, however long you shine them.

One more: with 400 nm violet light the maximum kinetic energy is

$$E_k=\frac{1240}{400}-2.28=3.10-2.28=0.82\ \text{eV}\tag{1.3.5}$$

Stopping those electrons requires 0.82 V of reverse bias on the collector — and this stopping voltage is precisely what the experiment measures directly.

∎Key formulas

• Photon energy

  $$E=h\nu=\frac{hc}{\lambda}$$

  hc = 1240 eV·nm, the workhorse for photon-energy estimates

• Einstein's photoelectric equation

  $$h\nu=W+E_{k,\text{max}}$$

  One electron absorbs one photon — that sentence is the crucial one

• Threshold frequency

  $$\nu_0=\frac{W}{h},\qquad \lambda_0=\frac{hc}{W}$$

  Below it no intensity suffices

• Stopping voltage

  $$eU_0=E_{k,\text{max}}=h\nu-W$$

  Plot U₀ against ν: slope h/e, intercept gives the work function

?Self-check4 questions

1. 1.

   Which assumption actually forbids "bright enough low-frequency light will eject electrons"?

   • An electron absorbs only one photon at a time; the energies of several photons cannot be pooled
   • Photons are massless
   • A metal contains a finite number of electrons
   • Photons travel at c

   Submit
2. 2.

   Double the intensity of the incident light at fixed frequency. Which quantities change? (Select all that apply.)

   Select all that apply

   • The number of electrons emitted per second
   • The photocurrent
   • The maximum kinetic energy of the electrons
   • The stopping voltage

   Submit
3. 3.

   What was Millikan trying to do with his precision photoelectric experiments, and what happened?

   • He set out to refute Einstein's equation, and measured a perfect straight line whose slope gave an h matching the black-body value
   • He set out to confirm Einstein and succeeded
   • He was measuring the electron charge and the photoelectric effect was a by-product
   • He set out to prove light is a continuous wave

   Submit
4. 4.

   Ultraviolet light of 250 nm strikes tungsten, work function W = 4.3 eV. What is the maximum kinetic energy of the photoelectrons, in eV? (hc = 1240 eV·nm)

   eV15% relative tolerance

   Submit

What comes next

By now two of the three dead ends from section 1.1 have been undone by the same sentence: energy is exchanged in whole units.

The third remains: why atoms do not collapse, and why spectra are sets of sharp lines.

In 1913 a young Dane working under Rutherford in Manchester decided to try attaching “one unit at a time” directly to the electron’s orbit, and see what came out.