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There is a question, that says:

What total energy (in $\mathrm{kJ}$) is contained in $1.0~\mathrm{mol}$ of photons, all with a frequency of $2.75 \cdot 10^{14}~\mathrm{Hz}$?

My answer was:
The energy of photon in hydrogen atom is given by the formula $E=h\nu$, where $h$ is Planck constant and $\nu$ is the frequency. After that I got an answer different from the correct answer which is $110~\mathrm{kJ}$.

Could anyone explain why?

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  • $\begingroup$ what values did you use in the equation? $\endgroup$ – ron Jul 2 '14 at 17:40
  • $\begingroup$ $2.75 \times 10^{14}$ x 6.63e-34 = 1.82e-19 KJ ?? $\endgroup$ – Maher Jul 2 '14 at 17:46
  • $\begingroup$ I get the same answer as you, except your answer is in J/mol, you need to divide by 1000 to convert it to kJ/m $\endgroup$ – ron Jul 2 '14 at 17:56
  • $\begingroup$ I did, but still far away from the right answer ! $\endgroup$ – Maher Jul 2 '14 at 18:00
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The formula $E=h\nu$ is for the energy of one photon. You have a mole of photons. You need to use a slightly modified form:

$$E=Nh\nu$$

where $N$ is the number of photons, in this case

$$\begin{align} N&=n\cdot N_\mathrm A\\[6pt] &=1\ \mathrm{mol}\times6.02\times10^{23}\ \mathrm{mol^{-1}}\\[6pt] &=6.02\times10^{23} \end{align}$$

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

If I use $E=nN_\mathrm A h \nu$, I get the correct answer you cite of $110~\mathrm{kJ}$. Do you?

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  • $\begingroup$ Yes I do ! but I didn't know that avogadro number is used to convert mol to photons .. I know that it is used to convert mol to atoms or molecules.. thanks ! $\endgroup$ – Maher Jul 3 '14 at 11:44
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    $\begingroup$ The units on Avogadro's number are $\text{mol}^{-1}$, so you can convert between number of anything and moles of anything. A mole is just a counting aid for very large numbers of things. $\endgroup$ – Ben Norris Dec 3 '14 at 11:40

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