How would I calculate the energy (in kilojoules) for 1 mole of a photon having a wavelength of 623nm? If I first calculated the energy of the photon in Joules and got 3.20x10^-37J.

I am wondering if I would need to find out the wavelength of 1 mole of these photons first and then carry out the same steps as I did when looking for the energy of the single photon. So:
if I converted nm to m first, then: 6.23x10^11m X 6.022x10^23 = _____m

and plug that into my E= h * c/wavelength = _____ and convert that to kilojoules?

I am wondering if I would need to find out the wavelength of 1 mole of these photons first and then carry out the same steps as I did when looking for the energy of the single photon. So:
if I converted nm to m first, then: 6.23x10^11m X 6.022x10^23 = _____m

and plug that into my E= h * c/wavelength = _____ and convert that to kilojoules?

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Al E. Share
Feb 6, 2018

Answer:

Your methodology, although not executed, seems sound. However, #623nm ne 6.23*10^11m#, wrong unit conversion. Moreover, you're forgetting we're interested in one mole of photons, not a single photon.

Explanation:

Consider,

#lambda * nu = c#
#=> nu = c/lambda#

Moreover, consider the frequency of one photon at that wavelength,

#nu = ((3.0*10^8m)/s)/(623nm * (m)/(10^9nm)) approx 4.8*10^14s^-1#

Now, recall,

#E = h * nu# per photon.

The energy for one photon of that wavelength is then,

#E = (6.626*10^-34m^2kg)/s * (4.8*10^14s^-1) approx 3.2*10^-19J#

Hence,

# 1"mol" * (6.02*10^23"photons")/"mol" * (3.2*10^-19J)/("photon") approx 1.9*10^5J = 190kJ#

of energy are in that many photons of that wavelength.

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