The Energy of a Photon

Finds the energy of a single photon as E = hc ÷ wavelength. Shorter wavelengths carry more energy, which is why ultraviolet light burns skin and visible light does not.

Explanation

Light is a wave, and it is also a stream of particles called photons. The energy each photon carries depends on nothing but the wavelength.

E = \dfrac{hc}{\lambda}

$h$ — Planck's constant, $6.626 \times 10^{-34}$ J·s
$c$ — the speed of light, $2.998 \times 10^8$ m/s
$\lambda$ — the wavelength in metres

Shorter wavelength, more energy, since $\lambda$ sits in the denominator.

Example

Find the energy of one photon of green light, at 550 nm.

E = \dfrac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{550 \times 10^{-9}} = 3.61 \times 10^{-19}\ \text{J}

In electronvolts that is 2.25 eV.

A shortcut worth memorising

Working in electronvolts and nanometres, the whole formula collapses to

E\ [\text{eV}] \approx \dfrac{1240}{\lambda\ [\text{nm}]}

For 550 nm, $1240 \div 550 = 2.25$ eV, exactly as above. Visible light runs from 1.8 eV at the red end (700 nm) to 3.1 eV at the violet end (400 nm).

Why only ultraviolet burns you

Breaking a chemical bond takes a few electronvolts.

Radio waves — around $10^{-6}$ eV. Far too feeble to break anything
Visible light — 1.8 to 3.1 eV, just short of the mark
Ultraviolet — 3.1 to 12 eV. Enough to cut the bonds in DNA

The decisive point is that this depends on wavelength, not brightness. No intensity of red light will ever give you sunburn: each individual photon is too weak, and a flood of weak photons is still a flood of weak photons. Meanwhile faint ultraviolet does damage with every photon that lands.

This is precisely Einstein's explanation of the photoelectric effect, and it was the evidence that light comes in particles.