Colour moves the energy, brightness moves the current

Emission requires a minimum frequency. Below it no electron leaves, however intense the light and however long it shines.

The reasoning that fails: a very bright red beam delivers enormous total energy, so surely it eventually frees an electron. It does not, because each electron interacts with one photon at a time. If $hf$ falls below $\phi$, adding photons only adds more photons that individually cannot do the job, and the current stays at exactly zero.

So compare $hf$ for one photon against the work function, and if it falls short, raise the frequency rather than the brightness.

This all-or-nothing dependence on colour is the reason the photon model was needed. A classical wave would deliver energy continuously, so a dim beam would simply take longer to accumulate enough. Experiment shows no such delay and no such accumulation.

$\phi$ is the minimum energy needed to free an electron from that material. So it changes when the plate changes and holds still when the lamp changes.

1. One metal, two beams: one $\phi$, two values of $hf$, so two values of $K_{\text{max}}$.
2. Two metals, one beam: one $hf$, two values of $\phi$, so two values of $K_{\text{max}}$.

Moving from green light to ultraviolet does not raise the work function. It raises $hf$ while $\phi$ holds, which is why $K_{\text{max}}$ rises.

A small piece of exam evidence supports this: work function values are supplied to you, which is itself a reminder that they are material data rather than something the experiment is measuring about the light.

Above threshold, the two experimental controls do entirely different things.

1. Frequency sets the maximum kinetic energy of each electron, and therefore the stopping voltage.
2. Intensity sets how many electrons come off per second, which is the photocurrent.

So doubling the brightness does not double the stopping voltage: $K_{\text{max}} = hf - \phi$ contains no intensity at all. What doubles is the plateau current, because twice as many photons eject twice as many electrons.

On a plot of current against voltage, more intensity lifts the saturation current and leaves the stopping voltage exactly where it was. Trying to overcome a stopping potential by brightening the lamp is the same error in the other direction.

Change the colour to move the stopping voltage; change the brightness to move the current. That split is the evidence for photons.

Rearranging $eV_{\text{stop}} = hf - \phi$ into the form of a straight line:

$$V_{\text{stop}} = \frac{h}{e}f - \frac{\phi}{e}.$$

Match it to $y = mx + b$ and every feature is named:

1. Slope $= h/e$. A slope in volt-seconds times $e$ gives $h$ in joule-seconds; read the vertical axis in electronvolts instead and the slope gives $h$ directly in eV$\cdot$s.
2. Vertical intercept $= -\phi/e$.
3. Horizontal intercept $= f_0 = \phi/h$, the threshold frequency.

The compact way to hold it: the slope carries a universal constant; the intercepts carry the metal. So the slope is the same for every plate, and switching plates slides the line without tilting it. Taking the slope as the work function, or the vertical intercept as $h$, gets both halves backwards, and forgetting the factor of $e$ when the axis is in volts is the quieter version.