Welcome to AP Chemistry Topic 3.12. These revision notes build the full story of the photon, from the energy carried by a single packet of light to the photoelectric effect that proved light is quantized. Read through the notes and worked examples below, then check your understanding with the 18-question quiz.
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Light as Particles
In the photoelectric effect, light striking a clean metal surface ejects electrons, but only when the light’s frequency is high enough. This behavior cannot be explained by a wave model, in which energy arrives spread out and accumulates over time. It is explained by treating light as a stream of particles called photons, each carrying a discrete packet of energy.
A photon’s energy is set by its frequency alone: E = hν, where h = 6.626 × 10⁻³⁴ J·s. Raising the frequency raises the energy of every individual photon. Raising the intensity of the light does something completely different: it increases the number of photons arriving per second, while the energy of each photon stays the same. Frequency controls how hard each photon hits; intensity controls how many photons hit.
The Work Function and Threshold Frequency
An electron in a metal is held by attractive forces, and a minimum amount of energy is needed to free it. That minimum is the work function, Φ, a property of the specific metal. A photon can eject an electron only if its energy at least equals Φ.
The threshold frequency, ν₀, is the frequency at which the photon energy exactly equals the work function: Φ = hν₀. Below ν₀, no electrons are ejected at all, regardless of intensity or exposure time, because every single photon falls short. Above ν₀, electrons are ejected immediately, even by very dim light.
The corresponding threshold wavelength is λ₀ = hc/Φ. Light with a wavelength longer than λ₀ cannot eject electrons from that metal, whatever its color brightness. Metals differ: a metal with a large work function needs ultraviolet light, while one with a small work function responds to visible light.
Where the Extra Energy Goes
A photon transfers all of its energy to one electron in a single interaction. Energy conservation gives the central equation of this topic:
KE_max = hν − Φ
If hν is greater than Φ, the surplus becomes kinetic energy of the ejected electron. This is a maximum because electrons below the surface lose additional energy escaping; surface electrons, which lose nothing extra, leave with KE_max.
When frequency rises at constant intensity, each photon carries more energy, so KE_max rises in direct proportion. When intensity rises at constant frequency, more photons strike per second, so more electrons are ejected per second, but each leaves with the same KE_max as before.
Plotting KE_max against ν gives a straight line. Its slope is Planck’s constant, h, the same for every metal. Its intercept on the frequency axis is ν₀, which is different for each metal because Φ differs. Below ν₀ the line would predict negative kinetic energy, which is physically meaningless: ejection simply does not occur.
Worked Examples
Example 1: kinetic energy and threshold frequency
Light of frequency 1.20 × 10¹⁵ s⁻¹ strikes a metal with work function 4.5 × 10⁻¹⁹ J. Find the maximum kinetic energy of the ejected electrons and the threshold frequency.
Photon energy: E = hν = (6.626 × 10⁻³⁴)(1.20 × 10¹⁵) = 7.95 × 10⁻¹⁹ J.
KE_max = hν − Φ = 7.95 × 10⁻¹⁹ − 4.5 × 10⁻¹⁹ = 3.45 × 10⁻¹⁹ J.
Threshold frequency: ν₀ = Φ/h = (4.5 × 10⁻¹⁹)/(6.626 × 10⁻³⁴) = 6.79 × 10¹⁴ s⁻¹. The incoming frequency is well above this, which is consistent with electrons being ejected with energy to spare.
Example 2: energy of one mole of photons
A laser emits light of wavelength 600 nm. Find the energy of one mole of these photons.
Energy per photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(6.00 × 10⁻⁷) = 3.31 × 10⁻¹⁹ J.
Per mole: 3.31 × 10⁻¹⁹ × 6.022 × 10²³ = 1.99 × 10⁵ J/mol, about 199 kJ/mol.
Traps Specific to This Topic
Brighter light below the threshold frequency ejects nothing. Doubling intensity doubles the photon count, not the photon energy, so a red beam of any brightness fails where dim ultraviolet succeeds.
Do not say that increasing frequency increases the number of electrons ejected. Frequency sets KE_max; intensity sets the number ejected per second. The two variables are tested precisely because students swap them.
Doubling the frequency does not double KE_max. KE_max = hν − Φ, so doubling ν doubles hν but Φ is subtracted only once. KE_max more than doubles.
Work functions are quoted in joules or in electronvolts (1 eV = 1.602 × 10⁻¹⁹ J). KE_max = hν − Φ works in either unit, but both terms must use the same one before subtracting.
A KE_max versus ν graph that starts at the origin is wrong. The line must cut the frequency axis at ν₀, and the metal with the larger ν₀ has the larger work function.