Photoelectric effect — Einstein's explanation, work function: Foundations

This page is the toolbox. The parent note freely uses symbols like $ν$ , $h$ , $ϕ$ , $E = h ν$ , "eV", $V_{0}$ . If you have never met these, the parent will feel like a foreign language. Here we build every one of them from zero, in the order they depend on each other, each with a picture and a reason it must exist.

0 — What a metal actually looks like inside

Before any light arrives, picture the metal itself.

Figure — Photoelectric effect — Einstein's explanation, work function

Look at the figure. The blue shaded region is the inside of the metal — a comfortable low-energy home for electrons (the orange dots). Above the dashed line is the vacuum outside. The vertical gap an electron must jump is the escape cost. We will name that cost $ϕ$ in a moment. Notice electrons sit at different depths: the one nearest the top (least tightly held) needs the smallest jump.

Why we need this picture: the entire topic is "how does light give an electron enough energy to make that jump?" Without seeing the well, none of the later symbols mean anything.

1 — Frequency $ν$ : the "colour" of a light wave

In the figure, both waves travel to the right at the same speed. The blue wave is stretched out — few wobbles per second — low $ν$ (redder light). The orange wave is squashed — many wobbles per second — high $ν$ (bluer light).

Wavelength $λ$ (Greek "lambda") is the partner of $ν$ : the distance between two wave crests, in metres. The marked arrow in the figure shows one $λ$ . Speed of light $c$ ties them together: $c = ν λ \Rightarrow ν = \frac{c}{λ} .$ Why this relation? In one second the wave advances a distance $c$ . If $ν$ crests pass and each is $λ$ apart, the total distance is $ν \times λ$ . That distance must equal $c$ . So high frequency ⇔ short wavelength — squashed waves. (This links to Wave-particle duality of light, where the same light is also treated as particles.)

2 — Energy, and the tiny unit we measure it in

Why introduce eV at all? Because work functions and photon energies land at "a few eV". Writing $2.3$ eV is friendlier than $3.68 \times 1 0^{- 19}$ J. This unit is exactly why the parent's shortcut $h c = 1240$ eV·nm exists.

3 — Planck's constant $h$ : the size of one energy bundle

Here is the pivot of the whole chapter.

Notice the shape of $E = h ν$ : energy is proportional to frequency. Double the frequency (bluer light) ⇒ double the energy per bundle. That is why blue light kicks electrons harder than red.

4 — Photon : one bundle of light

The figure makes the two dials visible. Top row: dim blue light — few photons, but each is a big (high-energy) packet. Bottom row: bright red light — many photons, but each is a small packet. This single picture is the heart of the whole topic: an electron meets one photon and asks "are you alone big enough to free me?" — the number of your neighbours is irrelevant.

Why we need the photon idea: a continuous wave would let an electron slowly accumulate energy from many weak waves. Photons forbid that — a below-size packet simply fails, no matter how many arrive.

5 — Work function $ϕ$ : the escape toll

Different metals have different $ϕ$ (sodium ≈ 2.3 eV, harder metals larger). It is a toll: a photon must pay at least $ϕ$ before anything escapes.

6 — Threshold frequency $ν_{0}$ and threshold wavelength $λ_{0}$

Combine the last two ideas. A photon worth $h ν$ must pay a toll $ϕ$ . The smallest frequency that can just afford the toll — leaving nothing over — defines the threshold.

Below $ν_{0}$ (light too red), no electron ever escapes, however bright. At or above $ν_{0}$ , escape happens. This sharp cutoff is exactly what classical waves could not explain.

7 — Kinetic energy $K E$ and $K E_{m a x}$ : the leftover

After the photon pays the toll $ϕ$ , whatever energy remains is handed to the electron as motion: $photon in h ν = toll ϕ + leftover K E .$ This is just book-keeping of energy — nothing is created or lost (see Energy conservation).

Why the subscript "max"? Electrons deeper in the well pay more than $ϕ$ to reach the surface, so they emerge slower. The surface electron (paying exactly $ϕ$ ) keeps the most leftover — the maximum: $K E_{m a x} = h ν - ϕ .$

8 — Charge $e$ , potential $V$ , and stopping potential $V_{0}$

The last symbols connect the invisible KE to something a voltmeter can read.

At the stopping point, all the electron's kinetic energy is spent climbing the hill: $e V_{0} = K E_{m a x} = h ν - ϕ \Rightarrow V_{0} = \frac{h}{e} ν - \frac{ϕ}{e} .$ That is a straight line of $V_{0}$ against $ν$ : slope $h / e$ (same for every metal!), and the line crosses zero at $ν_{0}$ . The parent note plots exactly this. In eV bookkeeping, $K E_{m a x}$ in eV equals $V_{0}$ in volts numerically, because the $e$ cancels — that is why " $0.80$ eV" gives " $0.80$ V".

Prerequisite map

Equipment checklist

I can state what frequency

ν

means and its unit

Number of wave wobbles per second, unit hertz (Hz) =

s^{- 1}

I know how

ν

λ

and

c

relate

c = ν λ

, so

ν = c / λ

— higher frequency means shorter wavelength.

I can convert between joules and electron-volts

1 eV = 1.6 \times 1 0^{- 19}

I know Planck's constant and what it does

h = 6.63 \times 1 0^{- 34}

J·s; it turns frequency into one photon's energy via

E = h ν

I can explain what a photon is

One indivisible bundle of light energy worth

h ν

; brightness = number of photons, colour = energy per photon.

I can define the work function

ϕ

The minimum energy to free the least tightly bound electron from the metal surface.

I know the threshold formulas

ν_{0} = ϕ / h

and

λ_{0} = h c / ϕ

; below

ν_{0}

no emission ever.

I understand why it is

K E_{m a x}

not

K E

Surface electrons keep the most leftover energy; deeper ones pay more than

ϕ

and emerge slower.

I can relate stopping potential to KE

e V_{0} = K E_{m a x} = h ν - ϕ

; plotting

V_{0}

ν

gives a line of slope

h / e

I know the size of the electron's charge

e = 1.6 \times 1 0^{- 19}