What total power (in $mathrmkJ$) is contained in $1.0~mathrmmol$ of pholoads, all through a frequency of $2.75 cdot 10^14~mathrmHz$?

My answer was:The energy of photon in hydrogen atom is offered by the formula $E=h u$, wright here $h$ is Planck continuous and $ u$ is the frequency. After that I acquired a solution different from the correct answer which is $110~mathrmkJ$.

You are watching: How much energy (in kj) do 3.0 moles of photons, all with a wavelength of 655 nm, contain?

Could anyone describe why?


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The formula $E=h u$ is for the energy of one photon. You have actually a mole of photons. You have to usage a slightly modified form:

$$E=Nh u$$

wright here $N$ is the variety of photons, in this case

$$eginalignN&=ncdot N_mathrm A\<6pt>&=1 mathrmmol imes6.02 imes10^23 mathrmmol^-1\<6pt>&=6.02 imes10^23endalign$$

Note that you are being asked to report energy in kilojoules not kilojoules per mole.

If I usage $E=nN_mathrm A h u$, I obtain the correct answer you cite of $110~mathrmkJ$. Do you?


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edited Nov 10 "16 at 14:48
user7951
answered Jul 2 "14 at 18:37
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Ben NorrisBen Norris
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