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I’m having trouble with calculating the energy of $\pu{5.00E10}$ photons of frequency $\pu{1.00E9 s-1}$.

I know that to calculate the energy $E = hf$ is to be used. However, I am unaware as to what to do with the $\pu{5.00E10}$ value and its meaning.

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Analogy: you have 5 bottles of coke 1 L each; what is the total volume?

The formula you shown $(E = hf)$ is derived for exactly one photon. Now, since you have 50 billion of photons $(N = \pu{5.00e10}),$ guess what the total energy would be?

$$E_\mathrm{tot} = Nhf = (\pu{5.00E10})(\pu{6.63E-34 J s})(\pu{1.00E9 s-1}) = \pu{3.32E-14 J}$$

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  • $\begingroup$ How about if I am given instead of a frequency value a wavelength value along with the photon value, how do I answer solve for the energy. $\endgroup$
    – Huda Alnusairi
    Mar 17, 2019 at 12:08
  • $\begingroup$ $$f = \frac{c}{λ},$$ where $c$ – speed of light; $λ$ – wavelength. $\endgroup$
    – andselisk ♦
    Mar 17, 2019 at 12:24
  • $\begingroup$ For the speed of light do I use c= 3.00E8 and if I do, do I use the frequency value gained to solve for the energy? $\endgroup$
    – Huda Alnusairi
    Mar 17, 2019 at 12:32
  • $\begingroup$ Yes, you do, but mind the units. If you take $c = \pu{3.00e8 m s-1}$, then make sure you plug in the wavelength in meters. $\endgroup$
    – andselisk ♦
    Mar 17, 2019 at 12:36

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