What is the energy of a photon emitted with a wavelength of 448 nm?

1 Answer
Jun 15, 2016

#4.44 * 10^-19J#

Explanation:

A photon is a particle of light that is released when an electron transitions from a high energy state to a low energy state.

The energy associated with the process is given by the following equation:

#E=h*f#

where #E# is the energy associated with the photon, #h# is Planck's constant (#6.63*10^-34J*s)#, and #f# is the characteristic frequency associated with the photon (expressed in Hertz, or #s^-1#).

With the information given in the problem, it might at first seem as if the problem cannot be solved. However, frequency, #f#, can be related to wavelength, #lambda#, in the following manner:

#f=c/lambda#

Since the speed of light, #c#, is relatively constant (#3.00*10^8m/s#) (this is true for a vacuum, but light slows down in other mediums) and wavelength is known, the second equation can be substituted into the first:

#E=(h*c)/lambda#
#E=((6.63*10^-34J*s)*(3.00*10^8*m/s))/(448*10^-9*m)#
#E=4.44*10^-19J#