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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