We can find the energy of a photon, given frequency, with this equation:
#E = h nu#, where #E# is the energy of the photon in #J#, #h# is Planck's constant of #6.626*10^-34 Js#, and #nu# is the frequency in #"Hz"#.
The problem gives us wavelength instead of frequency, but we can actually solve this problem by combining #E = h nu# with another equation:
#c = lambda nu#, where #c# is the speed of light at #3.0*10^8 "m/s"# and #lambda# is the wavelength in #m#
After combining them, we'll be able to find the energy of a photon given wavelength.
#c = lambda nu#
#nu = c/lambda#
#E = h nu#
#E = hc/lambda#
Now, before we plug in the wavelength of #555nm#, we need to convert #555nm# into #m#, which would be #5.55*10^-7m#.
#E = hc/lambda#
#E = 6.626*10^-34 Js xx (3.0*10^8 m"/"s)/(5.55*10^-7m)#
#E = 6.626*10^-34 Jcancel(s) xx (3.0*10^8 cancel(m)"/"cancel(s))/(5.55*10^-7cancel(m))#
#E = 3.58 * 10^-19 J#
This is the amount of energy for #1# photon. However, we want it for #1# mole of photons, which means that there's #6.022*10^23# photons.
#E = 3.58 * 10^-19 J xx 6.022*10^23 = "216000J/mol"#
After converting it into #kJ#, this would be #"216 kJ/mol"#.
