LocusBlog

Photoelectric effect

Shine short-wavelength light on a clean metal surface and electrons come off it. That is the photoelectric effect, and four measured features of it refused to fit the wave theory of light.

Emission starts the moment the light arrives, with no measurable delay, even when the source is faint. Below a certain frequency, which depends on the metal, no electrons come off at all, no matter how bright the light or how long you leave it on. Above that frequency, turning up the brightness produces more electrons but not faster ones. And the energy of the fastest electrons, measured as the reverse voltage needed to stop them reaching the collector, rises along a straight line in frequency, with the same slope for every metal anyone tried.

A wave delivers energy continuously, at a rate set by its intensity and spread over every electron it washes across. All four results say the opposite: energy arrives in separate lumps, the size of a lump is set by frequency alone, and one lump goes to one electron. Write the lump down and the list collapses to a single line.

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

$h$ is Planck’s constant, $f$ is the frequency of the light, $\phi$ is the work function of the metal - the energy needed to lift one electron out of the surface

What the wave picture actually predicts

The lag time is the easiest of the four to argue about in numbers, so start there. Treat the light as a wave of intensity $I$ washing over the surface, and let one electron collect energy from the patch of wavefront directly above it. An atom is roughly 0.1 nm across, so that patch has area $\pi r^2 \approx 3.1\times10^{-20}$ m$^2$.

Take a dim source, $I = 10^{-6}$ W/m$^2$. The power reaching one electron is

$$P = IA = 10^{-6} \times 3.1\times10^{-20} = 3.1\times10^{-26}\ \text{W}$$

Sodium needs $\phi = 2.46$ eV, which is $2.46 \times 1.602\times10^{-19} = 3.94\times10^{-19}$ J. Dividing,

$$t = \frac{3.94\times10^{-19}}{3.1\times10^{-26}} = 1.3\times10^{7}\ \text{s}$$

That is about 150 days of patient accumulation before the first electron should appear. The experiment shows emission with no delay the apparatus can resolve. The estimate is crude, and it does not need to be good. Argue the collecting area up by a factor of ten and the intensity up by a factor of a thousand, which is generous on both counts, and the wave picture still needs $1.3\times10^{7}/10^{4} = 1300$ s, over twenty minutes, to assemble one electron’s worth of energy. The measurement finds no delay at all.

Where 1240 eV.nm comes from

Photon energies in eV and wavelengths in nm are the natural units for this problem, and the conversion between them is one constant. Planck’s constant is $h = 4.135667\times10^{-15}$ eV s and the speed of light is $c = 2.99792458\times10^{8}$ m/s, so

$$hc = 4.135667\times10^{-15} \times 2.99792458\times10^{8} = 1.2398\times10^{-6}\ \text{eV m} = 1239.8\ \text{eV nm}$$

which everybody rounds to 1240. Then $E = hc/\lambda$ becomes $E(\text{eV}) = 1240/\lambda(\text{nm})$, and the threshold wavelength is $\lambda_c = 1240/\phi$ with $\phi$ in eV. The shortcut is worth memorising because it removes two powers of ten from every line of arithmetic.

The same constant explains the fourth experimental fact. Since $K_{max} = eV_s$,

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

a straight line in $f$ with slope $h/e = 4.136\times10^{-15}$ V s and intercept $-\phi/e$. The slope contains no property of the metal, which is why every metal gives a parallel line, and the intercept contains nothing but the work function, which is why the lines are offset from each other. Plotting stopping potential against frequency and reading off the slope is a direct measurement of Planck’s constant.

Sodium at 400 nm

Violet light at 400 nm on a sodium surface, work function 2.46 eV. The photon energy is

$$E = \frac{1240}{400} = 3.10\ \text{eV}$$

$$K_{max} = 3.10 - 2.46 = 0.64\ \text{eV}$$

so the fastest photoelectrons carry 0.64 eV and the photocurrent stops at a reverse bias of 0.64 V.

Check the photon energy in SI, where none of the shortcuts apply. $h = 6.626\times10^{-34}$ J s, $\lambda = 4.00\times10^{-7}$ m:

$$E = \frac{6.626\times10^{-34} \times 2.998\times10^{8}}{4.00\times10^{-7}} = 4.966\times10^{-19}\ \text{J}$$

Divide by $1.602\times10^{-19}$ J/eV and you get 3.100 eV. Same number, which it has to be, since the 1240 shortcut is that division done once in advance.

The threshold for sodium is $\lambda_c = 1240/2.46 = 504$ nm, which is green. Sodium responds to visible light, and that is unusual: most metals have work functions above 4 eV, which puts their thresholds shorter than $1240/4 = 310$ nm and squarely in the ultraviolet. Get the threshold the other way as a check: $f_c = \phi/h = 2.46/(4.1357\times10^{-15}) = 5.948\times10^{14}$ Hz, and $c/\lambda_c = 2.998\times10^{8}/(504\times10^{-9}) = 5.948\times10^{14}$ Hz. Two different constants, two different routes, same threshold.

Below the threshold, where the arithmetic keeps going

Put 600 nm orange light on that same sodium surface. The formula still returns a number:

$$K_{max} = \frac{1240}{600} - 2.46 = 2.07 - 2.46 = -0.39\ \text{eV}$$

A negative maximum kinetic energy is not a slow electron. It is no electron. The correct answer to “what is the stopping potential for sodium at 600 nm” is that there is no photocurrent to stop, and reporting $-0.39$ eV means the formula was trusted past the point where it describes anything: $K_{max} = hf - \phi$ is arithmetic that keeps working long after the physics has stopped. Compare $\lambda$ to $\lambda_c = 1240/\phi$ first, and remember the inequality runs backwards in wavelength: shorter wavelength means higher frequency means emission, so the condition is $\lambda < \lambda_c$, not greater.

Two smaller traps sit nearby. $K_{max}$ is a maximum, not the energy of every photoelectron: electrons liberated below the surface lose energy on the way out, so the real photocurrent carries everything from zero up to $K_{max}$, and the stopping potential only measures the top of that spread. And $h$ has two values in common use, $6.626\times10^{-34}$ J s and $4.136\times10^{-15}$ eV s. Pairing one of them with a work function quoted in the other’s units throws the answer out by a factor of $1.6\times10^{-19}$, which at least has the courtesy to be obvious.

The equation was published by Einstein in 1905 and was not immediately believed. Robert Millikan spent years building the apparatus that could test it cleanly, and by 1916 had the straight line of stopping potential against frequency for several metals, with the slope giving a value for $h$ from the photoelectric effect alone. Einstein’s 1921 Nobel Prize in Physics names the achievement as “his discovery of the law of the photoelectric effect”, not relativity. Millikan received the prize in 1923, cited for the elementary charge and for his photoelectric work.

Practice this until the threshold check is automatic rather than an afterthought: the photoelectric effect problems mix threshold, stopping potential and work function so that the first thing you have to decide is which of the three the question is actually asking for. The physics simulations let you sweep the frequency and watch the photocurrent switch off at $f_c$, and the rest of the practice catalogue carries the photon energy and matter wave problems that sit on either side of this one.

References