Photon energy

A photon of frequency $\nu$, angular frequency $\omega =2\pi \nu$, or vacuum wavelength $\lambda$ carries energy

$$E=h\nu =\mathrm{\hslash }\omega =\frac{hc}{\lambda }$$

where $h=6.62607\times {10}^{-34}$ J·s is Planck's constant, $\mathrm{\hslash }=h\mathrm{/}2\pi$, and $c=2.99792\times {10}^8$ m/s. In the units photonics actually uses, with the wavelength in nanometers,

$$E[\text{eV}]=\frac{1239.84}{\lambda [\text{nm}]}$$

The constant is $hc\mathrm{/}e$ in eV·nm, and its rounded form, 1.24 eV·µm, is the most used number in the field: it converts a bandgap in electron-volts to the absorption edge in micrometers and back. The same photon has a frequency $\nu [\text{THz}]=299792.458\mathrm{/}\lambda [\text{nm}]$, a wavenumber $\tilde{\nu }[{\text{cm}}^{-1}]={10}^7\mathrm{/}\lambda [\text{nm}]$, and an energy in joules of $1.98645\times {10}^{-16}\mathrm{/}\lambda [\text{nm}]$. A 1550 nm photon is therefore 0.7999 eV, 193.41 THz, 6452 cm⁻¹ and $1.2816\times {10}^{-19}$ J, four descriptions of one quantity that different communities prefer: energy for semiconductors, frequency for telecom channel plans, wavenumber for spectroscopy.

Photons per second

For optical power $P$ at wavelength $\lambda$ the photon flux is

$$\mathrm{\Phi }=\frac{P}{h\nu }=\frac{P\lambda }{hc}$$

so 1 mW at 1550 nm is $7.80\times {10}^{15}$ photons per second, 1 mW at 405 nm is $2.04\times {10}^{15}$, and a single photon per nanosecond at 1550 nm is 0.13 nW, the scale at which quantum key distribution receivers and single-photon detectors operate. The shot-noise floor of any measurement of that power is the statistical fluctuation of this count, $\sqrt{\mathrm{\Phi }\mathrm{\Delta }t}$ photons in a time $\mathrm{\Delta }t$, which is why a 1550 nm signal at a given power is quieter in relative terms than a 405 nm signal of the same power: it has 3.8 times as many photons.

What the photon energy decides

A semiconductor absorbs a photon only if the energy exceeds its bandgap, so the 1.24 eV·µm rule fixes which materials work where: silicon (1.12 eV) is opaque below 1.1 µm and transparent at 1310 and 1550 nm, germanium (0.66 eV) absorbs to 1.88 µm, InGaAs lattice-matched to InP (0.75 eV) to 1.65 µm, and GaN (3.4 eV) emits at 365 nm and, alloyed with indium, at 405 to 530 nm. A photodetector of quantum efficiency $\eta$ delivers one electron per absorbed photon, so its responsivity is

$$\mathcal{R}=\frac{\eta e}{h\nu }=\eta \frac{\lambda [\mu \text{m}]}{1.24}\text{ A/W}$$

which is why a perfect 1550 nm detector gives 1.25 A/W and a perfect 850 nm detector only 0.69 A/W, and why the ideal responsivity rises linearly with wavelength until the bandgap cuts it off. The same energy budget runs the other way in a laser: the photon energy divided by the junction voltage, $h\nu \mathrm{/}eV$, is the voltage-defect factor in the wall-plug efficiency, and a 1550 nm laser with a 1.2 V drop wastes a third of its electrical energy before any other loss is counted. Thermal energy is the natural comparison: $kT$ is 25.9 meV at 300 K, so a 1550 nm photon carries 31 times the thermal energy and a 10.6 µm photon only 4.5 times, which is why mid-infrared and terahertz detectors need cooling and visible ones do not, and why a photon of 0.8 eV cannot be excited thermally at room temperature but a phonon of 30 meV can.

Common questions

What is the energy of a 1550 nm photon?

0.800 eV, or $1.28\times {10}^{-19}$ J; its frequency is 193.4 THz. A 1310 nm photon is 0.946 eV, a 980 nm pump photon 1.265 eV, and a 405 nm photon 3.06 eV.

How do you convert wavelength to photon energy?

Divide 1239.84 eV·nm by the wavelength in nanometers, or 1.24 eV·µm by the wavelength in micrometers. For joules, divide $1.986\times {10}^{-16}$ J·nm by the wavelength in nanometers. The relation is exact for vacuum wavelength; inside a medium the wavelength shortens by the refractive index but the photon energy and frequency do not change.

Why does a shorter wavelength mean higher energy?

Because the energy is proportional to the frequency, $E=h\nu$, and frequency and vacuum wavelength are inversely related through $c=\lambda \nu$. Halving the wavelength doubles the frequency and the energy: a 775 nm photon carries exactly twice the energy of a 1550 nm photon, which is what second-harmonic generation does, combining two photons into one.

The Photonics Unit Converter converts between wavelength, frequency, photon energy and wavenumber, and gives the photon flux and the ideal-detector responsivity for a given power. The wavelength calculator gives the photon energy of any wavelength from radio to X-rays, with the region of the spectrum it falls in.

References: P. J. Mohr, D. B. Newell, and B. N. Taylor, "CODATA recommended values of the fundamental physical constants," Reviews of Modern Physics 88, 035009 (2016), for $h$, $c$, $e$ and $hc\mathrm{/}e$. B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed. (Wiley, 2019), chapter 13. S. M. Sze and K. K. Ng, Physics of Semiconductor Devices, 3rd ed. (Wiley, 2007), appendix, for the bandgaps quoted.