Photoelectric effect — Einstein's photon model: Visual walkthrough

This is a visual walkthrough of the parent topic.

Step 1 — The picture we are trying to explain

WHAT: We draw the experiment before we explain it — a plate, incoming light, an electron leaving.

WHY: Every symbol we invent later must point at something in this picture. No floating algebra.

PICTURE:

Figure — Photoelectric effect — Einstein's photon model

Look at the amber arrow: that is the incoming light. The cyan dot leaving the surface is the electron. The whole mystery is: when does that cyan dot fly off, and how fast?

Step 2 — Two dials on the light: colour and brightness

Before any physics, we must be precise about what we can change about light. There are exactly two independent dials.

WHAT: We separated one knob ("colour", $ν$ ) from another knob ("brightness", intensity).

WHY: The whole puzzle is that experiment says the colour knob decides whether electrons fly, while common sense screams it should be the brightness knob. Naming them apart lets us test each.

PICTURE:

The top row: same brightness, different colours (different $ν$ ). The bottom row: same colour, different brightness. We will turn each dial and watch the electron.

Step 3 — Light comes in packets (the photon idea)

Here is Einstein's leap, borrowed from Planck's Quantum Theory.

WHAT: We assigned each photon an energy that grows with frequency (colour), not with brightness.

WHY THIS TOOL — why multiply by $h$ ? We need a rule that ties energy to colour so that a redder photon is genuinely feebler than a bluer one, no matter how many arrive. A simple straight-line rule "energy $= h \times ν$ " does exactly that, and Planck had already found the constant $h$ from a completely different experiment (black-body glow). Reusing his number, not inventing a new one, is what made this convincing.

PICTURE:

Notice: a bluer photon (right) is drawn as a bigger coin than a redder one (left). Turning up brightness only makes more coins of the same size — it never makes a single coin bigger. Hold that image; it settles every confusion later.

Step 4 — The metal charges a toll to leave: the work function

An electron sits inside the metal, held there by attraction to the positive ions. To escape into open air it must be paid enough to break free.

WHAT: We named the "escape toll" $ϕ$ .

WHY: Energy is never free to leave — the electron is bound. Any honest energy budget must subtract this cost first.

PICTURE:

The cyan "wall" is the barrier of height $ϕ$ . An electron climbing out must spend at least $ϕ$ worth of energy just to reach the top of the wall.

Step 5 — One photon, one electron: the energy bookkeeping

Now the key rule of the model: one photon gives all its energy to exactly one electron — all or nothing, in a single instant.

WHAT: We wrote "energy in = energy spent + energy left over" for a single photon–electron meeting.

WHY: Energy cannot appear or vanish (conservation of energy). The photon's energy has only two places to go: pay the toll, or become motion. That splits $h ν$ cleanly into $ϕ + K E$ .

PICTURE:

The amber bar (total $h ν$ ) is chopped into a fixed cyan piece ( $ϕ$ , the toll) and a white leftover ( $K E_{m a x}$ ). Make the amber bar taller (bluer light) and the leftover grows; the toll never changes.

Step 6 — Rearranging into Einstein's equation

WHAT: Solve the Step-5 balance for the leftover $K E_{m a x}$ by moving $ϕ$ to the other side.

WHY: We want the observable — how fast electrons come out — alone on the left, so we can predict and measure it.

$K E_{m a x} = h ν - ϕ$

Why " $max$ "? The equation assumes the electron paid only the minimum toll $ϕ$ . Electrons starting deeper in the metal lose extra energy scraping their way out, so they emerge slower. The fastest electron is a surface one that paid exactly $ϕ$ — hence the leftover is the maximum possible KE.

Step 7 — The threshold: where the leftover hits zero

Now the make-or-break case. Turn the colour dial toward red (lower $ν$ ). The amber bar shrinks. What happens when it is exactly as tall as the toll?

Setting $K E_{m a x} = 0$ in Step 6: $0 = h ν_{0} - ϕ \Rightarrow ϕ = h ν_{0}$

So we may rewrite Einstein's equation entirely in terms of frequencies: $K E_{m a x} = h ν - h ν_{0} = h (ν - ν_{0})$

$ν$ — colour of the light you shine.
$ν_{0}$ — the metal's built-in threshold colour.
$h (ν - ν_{0})$ — leftover grows in proportion to how far above threshold you are.

PICTURE:

Three cases side by side:

Left ( $ν < ν_{0}$ ): amber bar shorter than the toll wall. The photon cannot even pay to escape → no electron, ever, no matter how many such photons (how bright). This is the degenerate case classical physics got wrong.
Middle ( $ν = ν_{0}$ ): bar exactly equals toll → electron just escapes with zero speed.
Right ( $ν > ν_{0}$ ): bar taller → electron leaves with leftover $K E_{m a x}$ .

Step 8 — Measuring the leftover: the stopping potential

We cannot read an electron's speed with a ruler. We measure it with a voltage.

An electron of charge $e$ climbing a voltage $V_{0}$ spends energy $e V_{0}$ . To stop the fastest electron it must spend all its $K E_{m a x}$ :

Combine with Step 6: $e V_{0} = h ν - ϕ \Rightarrow V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$

WHY this is beautiful: Read as $V_{0}$ (up) versus $ν$ (across), this is a straight line. Its slope is $h / e$ — so a simple graph measures Planck's constant.

PICTURE:

The straight line crosses zero at $ν = ν_{0}$ (below which there is no line — no electrons to stop). The slope triangle is labelled $h / e$ ; the negative intercept on the vertical axis is $- ϕ / e$ .

Worked check (numbers matching the parent)

The one-picture summary

This single diagram compresses all eight steps: a photon coin of size $h ν$ arrives, spends $ϕ$ climbing the wall, and the leftover white slab is $K E_{m a x}$ , which a reverse voltage $V_{0}$ exactly cancels.

Recall Feynman retelling — the whole walkthrough in plain words

Light comes in coins (photons); a bluer coin is worth more than a redder one ( $E = h ν$ ). To leave the metal an electron must pay a fixed entry toll $ϕ$ . Rule of the game: one coin goes to one electron, all at once. So the electron's leftover pocket money is coin minus toll: $K E_{m a x} = h ν - ϕ$ . If the coin is too small to cover the toll (light too red, $ν < ν_{0}$ ), nothing happens — and stacking up more small coins never helps, because coins can't combine. If the coin is big enough, the electron leaves instantly with the change as speed. To measure that change we push back with a voltage until the electron just fails to arrive; that voltage $V_{0}$ times the charge $e$ equals the leftover. Plot $V_{0}$ against colour $ν$ and you get a straight line whose steepness hands you Planck's constant.

Recall

What splits the photon's energy $h ν$ into two parts? ::: The escape toll $ϕ$ (work function) and the leftover $K E_{m a x}$ : $h ν = ϕ + K E_{m a x}$ . Why is it $K E_{m a x}$ and not just $K E$ ? ::: Only surface electrons pay the minimum toll $ϕ$ ; deeper ones lose extra energy and come out slower, so $K E_{m a x}$ is the fastest case. What does a negative value of $h (ν - ν_{0})$ mean physically? ::: No emission at all — the single photon couldn't pay the toll. What is the slope of $V_{0}$ versus $ν$ ? ::: $h / e$ , which lets you measure Planck's constant. Why doesn't extra brightness give faster electrons? ::: Brightness = more coins of the same size; each electron still meets one photon of energy $h ν$ , so its leftover is unchanged.

Connections

Planck's Quantum Theory — source of $E = h ν$ and the constant $h$ reused in Step 3.
Wave-Particle Duality — the photon as light's particle side.
Work Function and Binding Energy — the toll $ϕ$ of Step 4; same idea in Photoelectron Spectroscopy.
Bohr Model of the Atom — quantized light–matter energy exchange.
Compton Effect — photons also carry momentum.
Electromagnetic Spectrum — why UV clears the threshold but red often can't.