Photon Energy (Planck Relation)

Equivalent forms

E = \hbar\omega

E = hc/\lambda

The equation that ended classical physics.

Symbol	Name	SI unit	Dimension	Range
$E$	Photon energyoutput Energy of a single photon	J	ML^2T^-2	1e-25 – 1e-13
$h$	Planck's constant Fundamental quantum of action	J*s	ML^2T^-1	6.62607015e-34 – 6.62607015e-34
$f$	Frequency Frequency of the electromagnetic wave	Hz	T^-1	100000000 – 100000000000000000000

Unit systems

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

Where it holds

Valid for all electromagnetic radiation, any frequency. E and f are single-photon quantities; beam intensity corresponds to photon number.

Dimensional analysis

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

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

Both sides are energy.

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Discovery

Max Planck · 1900

Planck introduced energy quanta to fit blackbody radiation data; Einstein extended it in 1905 to explain the photoelectric effect.

Try this

Why does UV light burn your skin but red light doesn't?

Energy of light comes in discrete packets proportional to frequency. What is the energy of a single UV photon at 1e15 Hz?

Research status: stable

Real-world applications

Solar cells and photovoltaics
X-ray imaging and radiation therapy
LED and laser design
Photon energy thresholds for chemical reactions

Common misconceptions

Brightness does not change single-photon energy — only photon count
$E = hf$ applies to the photon, not to a 'wave packet' of finite duration
Frequency here is the linear frequency f, not angular frequency $\omega = 2*pi*f$

Experimental verification

Photoelectric effect (Millikan 1916), Compton scattering (1923), single-photon interference experiments, and photon-counting detectors in astronomy confirm

E = hf

to parts-per-billion accuracy.

Derivation

Planck modeled blackbody oscillators as only emitting/absorbing energy in integer multiples of hf.

Fitting the blackbody spectrum required this quantization.

Einstein (1905) extended it: each electromagnetic mode of frequency f carries energy in packets of hf, which are localized photons.

Limiting cases

f = 5e14\,\mathrm{Hz}

(visible green)⟶

E \approx 3.31e-19\,\mathrm{J} \approx 2.07\,\mathrm{eV}

Visible photon — enough to drive photosynthesis.

f = 1e15\,\mathrm{Hz} (UV)

⟶

E \approx 6.63e-19\,\mathrm{J} \approx 4.14\,\mathrm{eV}

Above many chemical bond energies — hence ionization and sunburn.

f \to 0

(radio)⟶

E \to 0

Radio photons carry vanishingly small energy per photon, but coherent waves still transmit information.

What if…

What if h were larger?

Single photons would carry more energy — visible light would be ionizing and life as we know it impossible.

What if you halved the frequency?

Photon energy halves — red light instead of UV, and bond-breaking chemistry turns off.

What if you increase intensity but keep f fixed?

More photons per second, same E per photon — no change to photoelectric threshold.

1

UV photon energy

Given ·

f:: 1000000000000000

Find · E

Steps

$E = h * f$
$E = 6.62607015e-34 * 1e15$
$E \approx 6.626e-19\,\mathrm{J}$ , divide by 1.602176634e-19 to convert $\to 4.136\,\mathrm{eV}$

Result ·

E = hf = 6.62607015e-34 * 1e15 = 6.626e-19\,\mathrm{J} \approx 4.14\,\mathrm{eV}

2

Red laser (650 nm)

Given ·

lambda:: 6.5e-7

Find · E

Steps

$f = c/\lambda = 299792458 / 6.5e-7 \approx 4.61e14\,\mathrm{Hz}$
$E = hf = 6.62607015e-34 * 4.61e14$
$E \approx 3.06e-19\,\mathrm{J}$

Result ·

E = hc/\lambda \approx 3.06e-19\,\mathrm{J} \approx 1.91\,\mathrm{eV}

de Broglie Wavelength→Photoelectric Effect→Bohr Hydrogen Energy Levels→