How do you calculate the energy, in eV, off a photon of light of wavelength 490 nm?

1 Answer
Jan 29, 2017

You can do it like this:

Explanation:

To find the energy of a photon you use the Planck Expression:

sf(E=hf)

sf(h) is the Planck Constant which = sf(6.63xx10^(-34)color(white)(x)J.s)

sf(f) is the frequency

This becomes:

sf(E=(hc)/lambda)

sf(c) is the speed of light = sf( 3.00xx10^(8)color(white)(x)"m/s")

sf(lambda) is the wavelength

:.sf(E=(6.63xx10^(-34)xx3.00xx10^(8))/(490xx10^(-9))color(white)(x)J)

sf(E=4.00xx10^(-19)color(white)(x)J)

If 1 Coulomb of charge is moved through a potential difference of 1 Volt then 1 Joule of work is done.

The electron volt is the work done in moving 1 electron through a potential difference of 1 Volt.

The charge on the electron is taken here to be sf(1.60xx10^(-19)color(white)(x)C)

:.sf(1"eV"=1.60xx10^(-9)color(white)(x)J)

:. The energy of the electron in electron volts

sf(=(4.00xx10^(-19))/(1.60xx10^(-19))=2.5color(white)(x)"eV")