Photon Energy (Planck-Einstein Relation)

Equivalent forms

E = \hbar \omega

E = p c

A universal constant h converts vibration into energy — the first crack that opened the quantum world.

Symbol	Name	SI unit	Dimension	Range
$E$	Photon energyoutput Energy carried by a single photon	J	M·L²·T⁻²	1e-22 – 1e-14
$f$	Frequency Frequency of the electromagnetic wave	Hz	T⁻¹	100000000 – 100000000000000000000
$lambda$	Wavelength Wavelength in vacuum	m	L	1e-12 – 1

Unit systems

Symbol	SI	CGS	Imperial
$E$	J	erg	eV
$f$	Hz	Hz	Hz
$lambda$	m	cm	in

Where it holds

Universally valid for free-space photons. Inside media the relation uses phase velocity; for matter waves use de Broglie

\lambda = h/p

.

Dimensional analysis

E \to [M\cdot L^{2}\cdot T^{-2}]

h\cdot f \to [M\cdot L^{2}\cdot T^{-1}]\cdot [T^{-1}] = M\cdot L^{2}\cdot T^{-2}

Both sides are energy.

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Discovery

Max Planck and Albert Einstein · 1900

Planck introduced E=hf as a mathematical trick to explain blackbody radiation in 1900. Einstein elevated it to physical reality in 1905, using it to explain the photoelectric effect — earning him the 1921 Nobel Prize.

Try this

Why does ultraviolet light burn your skin but radio waves don't?

Find the energy of a single photon of green light (λ = 550 nm) in joules and electron-volts.

Research status: stable

Real-world applications

Solar cells: photons must have E > band gap to free electrons.
X-ray imaging: high-frequency photons penetrate matter with enough energy to expose film/detectors.
LED color: emitted photon energy equals semiconductor band gap.
Photosynthesis: chlorophyll absorbs specific photon energies to drive electron transport.

Common misconceptions

Brightness sets photon energy — intensity sets photon count; frequency alone sets per-photon energy.
Photons have mass equal to $E/c^{2}$ — they have momentum $p = E/c$ but zero rest mass.
h is the same in different unit systems numerically — only the dimensional form is invariant.

Experimental verification

Verified by Millikan's 1916 photoelectric experiments (which he originally hoped would disprove Einstein). Modern verification reaches parts-per-billion through atomic clock comparisons and photon-counting calorimetry.

Derivation

Planck assumed blackbody oscillators emit energy in discrete units

\varepsilon = hf

to fit the observed spectrum, deriving the Planck distribution.

Einstein argued this discreteness is intrinsic to light itself: each photon carries

E = hf

.

Combining with

c = f\lambda

gives

E = hc/\lambda

.

The photoelectric experiment confirmed it: kinetic energy of ejected electrons depends only on f, not intensity.

Limiting cases

f \to 0

⟶

E \to 0

Zero-frequency photons carry no energy (classical limit).

\lambda \to 0

⟶

E \to \infty

Gamma rays carry enormous per-photon energy; pair production becomes possible above

2\,\mathrm{m}_e c^{2}

.

h \to 0

(classical)⟶

E \to 0

Classical limit recovers continuous wave energy with no per-photon quanta.

What if…

What if light enters a medium with refractive index n?

Frequency is invariant; wavelength shortens $to \lambda /n$ . Photon energy doesn't change crossing interfaces.

What if photon energy exceeds

2\,\mathrm{m}_e c^{2} \approx 1.022\,\mathrm{MeV}

?

Pair production becomes possible: a gamma photon can convert into an electron–positron pair in a strong field.

What if frequency is so low it's a radio wave?

Per-photon energy is tiny $(\sim 1e-25\,\mathrm{J}$ at MHz), so even weak signals contain quadrillions of photons — wave behavior dominates.

1

Energy of a green photon

Given ·

lambda:: 5.5e-7

Find · E in J and eV

Steps

Apply $E = h c / \lambda$
Numerator: $hc = 6.62607015e-34 \cdot 299792458 = 1.986e-25 J\cdot m$
Divide $by \lambda$ : $E = 1.986e-25 / 5.5e-7 = 3.613e-19\,\mathrm{J}$
Convert: $3.613e-19 / 1.602176634e-19 = 2.255\,\mathrm{eV}$

Result ·

E = hc/\lambda = (6.62607015e-34)(299792458)/(5.5e-7) = 3.613e-19\,\mathrm{J} = 2.255\,\mathrm{eV}

2

Photons per second from a 1 mW laser pointer (red, 650 nm)

Given ·

lambda:: 6.5e-7
power:: 0.001

Find · photons/second

Steps

Compute E_photon $= (6.62607015e-34)(299792458)/(6.5e-7) = 3.057e-19\,\mathrm{J}$
Divide power by per-photon energy: 1e-3 / 3.057e-19
Rate $\approx 3.27e15$ photons per second

Result · E_photon

= hc/\lambda = 3.057e-19\,\mathrm{J}

; rate

= P / E = 1e-3 / 3.057e-19 = 3.27e15

photons/s

Wave Speed Equation→Doppler Effect→