# 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?

Feb 6, 2018

Your methodology, although not executed, seems sound. However, $623 n m \ne 6.23 \cdot {10}^{11} m$, wrong unit conversion. Moreover, you're forgetting we're interested in one mole of photons, not a single photon.

#### Explanation:

Consider,

$\lambda \cdot \nu = c$
$\implies \nu = \frac{c}{\lambda}$

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

$\nu = \frac{\frac{3.0 \cdot {10}^{8} m}{s}}{623 n m \cdot \frac{m}{{10}^{9} n m}} \approx 4.8 \cdot {10}^{14} {s}^{-} 1$

Now, recall,

$E = h \cdot \nu$ per photon.

The energy for one photon of that wavelength is then,

$E = \frac{6.626 \cdot {10}^{-} 34 {m}^{2} k g}{s} \cdot \left(4.8 \cdot {10}^{14} {s}^{-} 1\right) \approx 3.2 \cdot {10}^{-} 19 J$

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.