Photoelectric Effect

Equivalent forms

eV_0 = hf - \phi

A linear equation that toppled the wave-only view of light.

Symbol	Name	SI unit	Dimension	Range
$K_max$	Max kinetic energyoutput Maximum KE of ejected photoelectrons	J	ML^2T^-2	0 – 1e-17
$h$	Planck's constant Fundamental quantum of action	J*s	ML^2T^-1	6.62607015e-34 – 6.62607015e-34
$f$	Photon frequency Frequency of incident light	Hz	T^-1	10000000000000 – 100000000000000000
$phi$	Work function Minimum energy to eject an electron from the metal	J	ML^2T^-2	1e-19 – 1e-18

Unit systems

Symbol	SI	CGS	Imperial
$K_max$	J	erg	ft*lbf
$h$	J*s	erg*s	ftlbfs
$f$	Hz	Hz	Hz
$phi$	J	erg	ft*lbf

Where it holds

Valid for single-photon absorption in metals and semiconductors. Multi-photon regime requires intense lasers (non-linear photoemission).

Dimensional analysis

K_\max \to [M*L^2*T^-2]

[M*L^2*T^-1][T^-1] - [M*L^2*T^-2] \to [M*L^2*T^-2]

Both sides are energy.

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Discovery

Albert Einstein · 1905

Einstein's explanation of Hertz/Lenard's observations won him the 1921 Nobel Prize and proved light quantization.

Try this

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

Electrons are only ejected from metal if the photon frequency exceeds a threshold. What is the max KE of photoelectrons from sodium (work function 2.28 eV) under 400 nm light?

Research status: stable

Real-world applications

Photomultiplier tubes
CCD and CMOS image sensors
Night vision devices
Solar cells and photodiodes
ARPES for condensed matter research

Common misconceptions

Intensity does NOT affect K_max, only photon count — a classical wave theory would predict otherwise
The emission is effectively instantaneous (< 10 fs), not delayed as a wave energy-buildup model would require
The work function is a property of the metal surface, not the bulk

Experimental verification

Millikan's 1916 photoelectric experiments measured h to within 0.5%. Modern angle-resolved photoemission spectroscopy (ARPES) uses this relation to map band structures of solids.

Derivation

Model the photon as a quantum of energy hf.

Energy conservation on absorption by a single electron:

hf = \varphi + K_\max

, where phi is the minimum energy to release the electron.

Solving gives

K_\max = hf - \varphi

.

K_max is measured as e*V_0 where V_0 is the stopping potential.

Limiting cases

hf < phi⟶

K_\max = 0

(no emission)Below threshold no electrons escape, no matter how intense the light.

hf = \varphi

⟶ f_threshold

= \varphi /h

Exactly at threshold electrons barely escape with zero KE.

hf >> phi⟶

K_\max \approx hf

At very high frequency the work function is a small correction.

What if…

What if you double the intensity?

Twice as many electrons come out per second, but K_max is unchanged.

What if you use a laser below threshold?

No emission from single-photon process. With extreme intensity two-photon absorption can kick in (non-linear regime).

What if the metal were perfectly clean?

Work function drops slightly (surface contamination raises phi); small but measurable shift in threshold.

1

Sodium under 400 nm light

Given ·

lambda:: 4e-7
phi eV:: 2.28

Find · K_max

Steps

$f = c/\lambda = 299792458 / 4e-7 = 7.4948e14\,\mathrm{Hz}$
$hf = 6.62607015e-34 * 7.4948e14 \approx 4.97e-19\,\mathrm{J} = 3.10\,\mathrm{eV}$
$K_\max = 3.10 - 2.28 = 0.82\,\mathrm{eV} = 1.31e-19\,\mathrm{J}$

Result ·

K_\max \approx 0.82\,\mathrm{eV} \approx 1.31e-19\,\mathrm{J}

2

Threshold frequency for cesium (phi = 2.1 eV)

Given ·

phi eV:: 2.1

Find · f_threshold

Steps

$\varphi = 2.1 * 1.602176634e-19 = 3.365e-19\,\mathrm{J}$
$f = \varphi /h = 3.365e-19 / 6.62607015e-34$
$f \approx 5.08e14\,\mathrm{Hz}$

Result ·

f \approx 5.08e14\,\mathrm{Hz}

(green light)

Photon Energy (Planck Relation)→de Broglie Wavelength→Bohr Hydrogen Energy Levels→