Question #1d33e

1 Answer
Dylan K.
Dec 6, 2016

#E# = #3.146*10^6J#

Explanation:

#E#photon = #(hc)/(lambda)#

#E#photon = #(6.626*10^-34J*s*2.998*10^8m*s^-1)/(3*10^-8m#

#E#photon = #5.227*10^-18J#

#(5.227*10^-18J)# per photon #*(6.02*10^23)# photon =

#E# = #3.146*10^6J#

Related questions
See all questions in Calculations with wavelength and frequency
Impact of this question
1949 views around the world
You can reuse this answer
Creative Commons License