The energy of a photon is the frequency multiplied by the Planck constant; #E = hf#, and the wavelength of a wave is its speed divided by it's frequency; #v=flambda => f = \frac{v}{lambda}#, so it's energy, #E = \frac{hv}{lambda}#. This rearranges to #lambda = \frac{hv}{E}#, where #h# is the Planck constant, #v# is the speed of light, and #E# is the energy of the photon #:. lambda = \frac{6.63*10^{-34}*3.0*10^8}{3.083*10^12} = 6.4515*10^-38m#
#= 6.4515*10^-29nm#
I guess that you meant to write #3.083*10^{-12}# not #3.083*10^{12}#, seeing as gamma rays have a wavelength of #~10^-11m#, which is #~10^-18# times larger.
If #E = 3.083*10^-12J#, using the same method, it would give a wavelength of #6.4515*10^-5nm#, which is a reasonable value for electromagnetic waves.
