Photoelectric Effect

Equivalent forms

K_{max} = \frac{hc}{\lambda} - \phi

A linear equation that broke classical physics and birthed the photon.

Symbol	Name	SI unit	Dimension	Range
$K_max$	Max kinetic energyoutput Maximum kinetic energy of ejected electrons	J	M·L²·T⁻²	0 – 1e-17
$ν$	Frequency Frequency of incident light	Hz	T⁻¹	100000000000000 – 10000000000000000
$φ$	Work function Minimum energy to free an electron from the metal	J	M·L²·T⁻²	1.6e-19 – 1e-18

Unit systems

Symbol	SI	CGS	Imperial
$K_max$	J	erg	eV
$ν$	Hz	Hz	Hz
$φ$	J	erg	eV

Where it holds

Valid for single-photon absorption in clean metallic surfaces. Breaks down for very intense lasers (multi-photon ionization), and must be modified for semiconductors and insulators where band structure matters.

Dimensional analysis

$[K_\max ] = [h]\cdot [\nu ] = (M\cdot L^{2}\cdot T^{-1})\cdot (T^{-1}) = M\cdot L^{2}\cdot T^{-2} \checkmark$ (energy)

Discovery

Albert Einstein · 1905

Einstein proposed light quanta to explain Lenard's puzzling photoelectric data, earning him the 1921 Nobel Prize.

Try this

Why does dim blue light eject electrons but bright red light doesn't?

Light of wavelength 400 nm strikes a metal with work function 2.0 eV. What is the maximum kinetic energy of the ejected electrons?

Research status: stable

Real-world applications

Photomultiplier tubes and night-vision devices.
Solar cells (photovoltaic effect — extension to semiconductors).
X-ray photoelectron spectroscopy (XPS) for chemical analysis.
CCD/CMOS camera sensors converting photons to electrons.

Common misconceptions

Brighter light does NOT give electrons more energy — it only ejects more of them. K_max depends on frequency, not intensity.
There is no time delay even at very low intensity — the photon model predicts instantaneous emission, contradicting classical wave theory.
The work function depends on the metal and surface conditions, not the light source.

Experimental verification

Millikan's painstaking 1916 measurements (he was trying to *disprove* Einstein) yielded a perfectly linear

K_\max vs \nu

plot with slope h, measuring Planck's constant

to \sim 0.5

% accuracy. Lenard's earlier observations of frequency-dependent threshold and intensity-independent K_max were already inexplicable classically.

Derivation

Einstein postulated that light of frequency

\nu

consists of quanta of energy

E = h\nu

.

When a photon strikes a metal, it transfers all its energy to a single electron.

The electron must overcome the binding energy

\varphi

(work function) to escape; whatever is left becomes kinetic energy:

K_\max = h\nu - \varphi

.

Below

\nu _{0} = \varphi /h

, no escape is possible.

Limiting cases

h\nu

<

\varphi

⟶

K_\max = 0

(no emission)Below the threshold frequency, no electrons are ejected regardless of intensity — pure quantum behavior.

h\nu = \varphi

⟶

K_\max = 0

(threshold)Defines the cutoff frequency

\nu _{0} = \varphi /h

characteristic of the metal.

h\nu

≫

\varphi

⟶

K_\max \approx h\nu

At high photon energies the work function is negligible; electron KE tracks photon energy directly.

What if…

What if you double the light intensity?

Twice as many electrons are ejected per second, but each one still has the same K_max. Photoelectric current doubles; stopping voltage is unchanged.

What if the metal is replaced by one with double the work function?

The threshold frequency doubles. Light that previously worked may no longer eject electrons; remaining ejected electrons have lower K_max.

What if classical wave theory were right?

K_max would depend on intensity, there would be a measurable buildup time at low intensity (seconds to hours), and there would be no sharp threshold — none of which is observed.

1

400 nm light on a 2.0 eV metal

Given ·

λ:: 4e-7
φ eV:: 2

Find · K_max in eV

Steps

Use the shortcut E_photon $(eV) = 1240/\lambda (nm) \to E = 1240/400 = 3.10\,\mathrm{eV}$
$K_\max = h\nu - \varphi = 3.10 - 2.00 = 1.10\,\mathrm{eV}$
Convert if needed: $1.10 \times 1.602\times 10^{-19} \approx 1.76\times 10^{-19} J$

Result · Photon energy

= 1240/400 = 3.10\,\mathrm{eV}

;

K_\max = 3.10 - 2.0 = 1.10\,\mathrm{eV}

2

Threshold wavelength for cesium (φ = 2.1 eV)

Given ·

φ eV:: 2.1

Find ·

\lambda _{0}

Steps

At threshold $K_\max = 0 \Rightarrow h\nu _{0} = \varphi \Rightarrow \lambda _{0} = hc/\varphi$
$\lambda _{0} (nm) = 1240/\varphi (eV) = 1240/2.1 \approx 590\,\mathrm{nm}$
Cesium responds to visible light up to yellow — useful in photodetectors.

Result ·

\lambda _{0} = hc/\varphi = 1240/2.1 \approx 590\,\mathrm{nm}

(yellow-green)

De Broglie Wavelength→planck radiation