Photon energy

E=h f=\dfrac{hc}{\lambda}

Result unit: J (often eV)

What it describes

The energy of a light quantum is directly proportional to its frequency.

E

photon energy

[J]

h

Planck's constant (

\approx 6.63\cdot10^{-34}

)

[J·s]

f

frequency

[Hz]

c

speed of light

[m/s]

\lambda

wavelength

[m]

Light carries energy in portions (quanta). A photon of higher frequency (e.g. UV) carries more energy than a photon of red light.

Photoelectric effect, spectra, radiation energy. Often expressed in eV (

1

=1.6\cdot10^{-19}

J).

A photon with

f=5\cdot10^{14}

Hz. Calculate the energy (

h=6.63\cdot10^{-34}

Substitution

h=6.63\cdot10^{-34}

f=5\cdot10^{14}

Result

E=6.63\cdot10^{-34}\cdot5\cdot10^{14}\approx3.3\cdot10^{-19}

Found this formula useful? Share it.

It's free — send it to a friend who's revising.

Teaching? Embed this formula on your site.

Free widget — paste it on your school site, blog or notes. It links to the full version.