The Photoelectric Effect

A dim lamp that works instantly

Put a clean metal plate in a vacuum tube and shine light on it. Electrons come off. That much is not surprising: light carries energy, the electrons are loosely held, energy goes in and an electron comes out.

Everything else about the experiment is surprising. Turn the lamp down until it is barely visible and electrons still come off within a nanosecond of the light arriving: there is no waiting while the energy piles up. Turn the lamp up bright and you get more electrons, but not faster ones. Then switch to red light and nothing comes off at all, no matter how bright you make it, no matter how long you wait.

A wave carries energy spread smoothly over the whole wavefront, and a brighter wave carries more of it, so a wave picture predicts the opposite of all three results. The last lesson pushed classical mechanics until it broke at high speed and you replaced it with relativistic energy and momentum. This is the other place classical physics breaks, and the repair is stranger.

Light comes in lumps

Suppose the beam is not a smooth wave of energy but a stream of particles, each carrying a fixed amount of energy set by the frequency $f$ of the light:

$E = hf$

The constant $h$ is Planck's constant, $6.63 \times 10^{-34}$ J·s, or $4.14 \times 10^{-15}$ eV·s if you want the answer in electronvolts. One lump is called a photon. Brightness is how many photons per second; frequency is how much each one carries. Those are two separate knobs, and that separation is the whole explanation.

Since $f = c/\lambda$, you can also write $E = hc/\lambda$, and for this topic you should memorise one number:

$hc = 1240\ \text{eV}\cdot\text{nm}$

With it, photon energy in electronvolts is 1240 divided by the wavelength in nanometres, in one step. Green light at 500 nm carries $1240/500 = 2.48$ eV per photon. That is the arithmetic behind half the problems in this unit.

Now the three puzzles dissolve. Emission is instant because an electron never had to accumulate anything: it absorbed one photon, all at once, or it did not. Brightness gives more electrons but not faster ones because each electron still only ever meets one photon. And red light does nothing at any brightness because every photon in the beam is individually too weak, and a million weak photons do not add up inside one electron.

The work function, and what is left over

An electron inside a metal sits in a well. It takes energy to lift it out through the surface, and the minimum energy required is called the work function $\phi$ of that metal. Sodium is 2.30 eV, tungsten 4.50 eV, cesium 2.14 eV: the lowest of the common metals, which is why cesium photocells respond to visible light while most metals need ultraviolet.

An electron absorbs one photon of energy $hf$. It spends $\phi$ getting out. Whatever survives is kinetic energy. An electron sitting right at the surface that loses nothing else on the way keeps the most, so

$K_{max} = hf - \phi = \frac{hc}{\lambda} - \phi$

The subscript is doing real work. Electrons from deeper in the metal bump into things on the way out and emerge with less, so the emitted electrons come out with a spread of energies from zero up to $K_{max}$. Problems almost always ask about the maximum.

Set $K_{max} = 0$ and you get the cutoff. Below the threshold frequency $f_0 = \phi/h$, or equivalently above the threshold wavelength $\lambda_0 = hc/\phi$, nothing is emitted. For tungsten, $\lambda_0 = 1240/4.50 = 276$ nm: solidly ultraviolet.

Measuring $K_{max}$ with a voltmeter

You cannot weigh an electron's kinetic energy directly, so you make it climb a hill. Face the cathode with a collecting anode and put a voltage across the gap that pushes the electrons back. An electron with kinetic energy $K$ can climb a potential difference $V$ only if $K > eV$. Raise $V$ until even the fastest electron turns around and the current reads zero. That value is the stopping potential $V_s$, and at it

$eV_s = K_{max} = hf - \phi$

Measured in electronvolts and volts this is friendly: an electron with 1.20 eV of kinetic energy is stopped by exactly 1.20 V. The numbers are the same, the units differ.

Rearranging gives the equation of a straight line:

$V_s = \frac{h}{e} f - \frac{\phi}{e}$

Plot the stopping potential against frequency and you get a line whose slope is $h/e = 4.14 \times 10^{-15}$ V·s. The striking part is that the slope is the same for every metal. Change the cathode and the line slides sideways without tilting: the horizontal intercept moves to the new $f_0$, and the vertical intercept sits at $-\phi/e$. Millikan measured that slope carefully, got Planck's constant out of a voltmeter and a prism, and ended the argument.

Example 1: plug in

Sodium, work function 2.30 eV, lit with 400 nm light. Find $K_{max}$ and the speed of the fastest electron.

Photon energy first: $E = 1240/400 = 3.10$ eV. Then $K_{max} = 3.10 - 2.30 = 0.80$ eV.

For the speed you have to leave electronvolts. Convert: $K = 0.80 \times 1.602 \times 10^{-19} = 1.28 \times 10^{-19}$ J. Then $v = \sqrt{2K/m_e} = \sqrt{2(1.28\times10^{-19})/(9.11\times10^{-31})} = \sqrt{2.81 \times 10^{11}} = 5.3 \times 10^{5}$ m/s. That is under 0.2% of $c$, so the non-relativistic formula was safe. Always check that.

Example 2: set up before you plug in

A cathode gives a stopping potential of 1.40 V at 400 nm and 2.43 V at 300 nm. Find the work function.

You have two unknowns hiding in one equation, so write it twice and subtract. The photon energies are $1240/400 = 3.10$ eV and $1240/300 = 4.133$ eV. Then

$1.40 = 3.10 - \phi \quad\text{and}\quad 2.43 = 4.133 - \phi$

The first gives $\phi = 1.70$ eV, the second gives $\phi = 1.703$ eV. They agree, which is the check that the data is consistent with one photon per electron. If a problem gives you two data points and asks for $h$ instead, subtract the two equations to kill $\phi$: the difference in stopping potentials divided by the difference in frequencies is $h/e$.

Example 3: the one that needs an idea

An isolated metal sphere of radius 2.00 cm and work function 3.00 eV sits in vacuum and is lit with 250 nm light. It charges up positively as electrons leave, until no more can escape. How many electrons left?

Nothing in the photoelectric equation mentions charge, so the idea has to come from elsewhere: a positively charged sphere pulls the departing electrons back. An electron leaving a sphere at potential $V$ must pay an extra $eV$ on the way out. Emission stops when that extra toll eats all of $K_{max}$.

Start with the toll-free case. $E = 1240/250 = 4.96$ eV, so $K_{max} = 4.96 - 3.00 = 1.96$ eV. Emission therefore stops once the sphere reaches $V = 1.96$ V.

Now use electrostatics, from much earlier in the course: an isolated sphere of radius $R$ carrying charge $Q$ sits at potential $V = kQ/R$. So

and the number of electrons is that charge divided by $e$:

$N = \frac{4.36 \times 10^{-12}}{1.602 \times 10^{-19}} = 2.72 \times 10^{7}$

Twenty-seven million electrons, and the sphere has charged up to about two volts. Both numbers feel small, which is the point: it takes very little charge to make a potential that matters on the scale of a single electron.

A photon also carries momentum

A photon has no mass, but relativity gives $E^2 = (pc)^2 + (mc^2)^2$, and with $m = 0$ that collapses to $E = pc$. So

$p = \frac{E}{c} = \frac{h}{\lambda}$

A 2.48 eV photon carries $p = (2.48)(1.602\times10^{-19})/(3.00\times10^{8}) = 1.32 \times 10^{-27}$ kg·m/s. Tiny, but momentum is conserved whether or not you find it convenient. An atom that emits a photon recoils; a mirror that reflects a beam feels a push.

For a beam of intensity $I$ (watts per square metre) striking a surface head-on, energy $I$ arrives per second per square metre, so momentum $I/c$ arrives per second per square metre. Force per area is pressure, so a perfectly absorbing surface feels $P = I/c$ and a perfectly reflecting one feels $P = 2I/c$, because reflecting reverses the photon's momentum instead of merely removing it. Sunlight at 1000 W/m² on a mirror gives $2(1000)/(3.00\times10^{8}) = 6.67 \times 10^{-6}$ Pa. That is a ten-billionth of atmospheric pressure, and it is enough to sail a spacecraft on if you make the sail big enough and wait.

Compton scattering: the photon proved

The photoelectric effect shows photons deliver energy in lumps. Compton showed they collide like billiard balls. Fire an X-ray at a nearly free electron and some of the scattered radiation comes back at a longer wavelength, with the shift depending only on the scattering angle $\theta$:

$\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)$

The constant $h/m_ec = 0.002426$ nm is the Compton wavelength of the electron. Conserve relativistic energy and momentum in two dimensions and this drops out; the derivation is worth doing once, but for solving problems the formula plus one observation is enough. The observation: $1 - \cos\theta$ runs from 0 at $\theta = 0$ to 2 at $\theta = 180°$, so the shift is largest for a photon that comes straight back, and the electron gets the most energy in that backscattered case.

Notice the shift is an absolute shift in wavelength, not a fractional one. For visible light at 500 nm a shift of 0.0024 nm is nothing. For an X-ray at 0.05 nm it is a 5% change, plainly visible. That is why the effect belongs to X-rays and gammas.

To find the electron's energy, do not look for a separate formula. Energy is conserved, so the electron gets whatever the photon lost:

$K = \frac{hc}{\lambda} - \frac{hc}{\lambda'}$

Try a 0.0500 nm X-ray scattering at 60.0°. The shift is $\Delta\lambda = 0.002426(1 - \cos 60°) = 0.002426(0.500) = 0.001213$ nm, so $\lambda' = 0.051213$ nm. The photon energies are $1240/0.0500 = 24800$ eV and $1240/0.051213 = 24213$ eV, and the difference is 587 eV to the electron. Less than 3% of what the photon carried, which matches the small fractional change in wavelength.

Recognising which tool a problem wants

Four techniques live on this page, and problems signal which one they want quite openly.

If you see a metal surface, a work function, or a stopping potential, it is $K_{max} = hc/\lambda - \phi$. Convert wavelength to photon energy with 1240 first, always. If the problem gives two wavelengths or two metals, write the equation twice and subtract to eliminate whichever unknown you do not want.

If the problem gives a power, an intensity, or a current and asks "how many", you are counting photons. Divide a total energy by the energy of one photon, or a total charge by $e$. Take a 1.00 mW laser at 632.8 nm. One photon carries $1240/632.8 = 1.96$ eV, which is $3.14 \times 10^{-19}$ J. Divide $1.00 \times 10^{-3}$ J per second by that and you get $3.19 \times 10^{15}$ photons per second.

If something is pushed, recoils, or is held up by light, use $p = h/\lambda$ for a single photon or $P = I/c$ (doubled for a mirror) for a beam. A laser levitating a 1.00 mg mirror needs $mgc/2 = 1470$ W, which tells you how weak radiation pressure is.

If the problem says free electron and a scattering angle, it is Compton, and the answer runs through $\Delta\lambda$ even when the question asks for an energy.

One thing this lesson deliberately did not explain: why a given metal has the work function it does, or why atoms emit only certain wavelengths in the first place. For that you need to know what the electron is doing inside the atom, which is where the Bohr model starts.