The energy per mole of photon of electromagnetic radiation of wavelength 4000A° is ?

1 Answer
Truong-Son N.
Jun 5, 2018

About #"300 kJ/mol"#... How many #"kJ"# of energy would #"2 mols"# of photons with wavelength #4000 Å# have?

Well, there are #10^(-10) "m"# in one angstrom, so #4000 Å = 4 xx 10^(-7) "m"#. Each photon has energy

#E = hnu = (hc)/lambda#

where:

  • #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant.
  • #c = 2.998 xx 10^8 "m/s"# is the speed of light.
  • #lambda# is the wavelength in #"m"#.
  • #nu# is the frequency in #"s"^(-1)#.

Hence, the energy PER PHOTON is simply:

#E = (6.626 xx 10^(-34) "J"cdotcancel"s" cdot 2.998 xx 10^8 cancel"m"//cancel"s")/(4 xx 10^(-7) cancel"m")#

#= 4.97 xx 10^(-19) "J/photon"#

But this number is clunky... since we want the MOLAR photon energy,

#(4.97 xx 10^(-19) cancel"J")/cancel("photon") xx (6.0221413 xx 10^(23) cancel"photons")/("1 mol") xx ("1 kJ")/(1000 cancel"J")#

#=# #"299.07 kJ/mol"#

But since you only gave 1 sig fig, I can only report #color(blue)("300 kJ/mol")#.

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