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

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Answer 1 · 2017-01-29T17:44:50

You can do it like this:

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")#