What complete energy (in $mathrmkJ$) is included in $1.0~mathrmmol$ the photons, all v a frequency of $2.75 cdot 10^14~mathrmHz$?

My price was:The power of photon in hydrogen atom is provided by the formula $E=h u$, wherein $h$ is Planck continuous and $ u$ is the frequency. ~ that ns got an answer different from the exactly answer i beg your pardon is $110~mathrmkJ$.

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Could anyone define why?



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

$$E=Nh u$$

where $N$ is the variety of photons, in this situation

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

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

If I use $E=nN_mathrm A h u$, I get the exactly answer you point out of $110~mathrmkJ$. Execute you?

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edited Nov 10 "16 at 14:48
answered Jul 2 "14 at 18:37

Ben NorrisBen Norris
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