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Visible light is anything between 1.65eV and 3.10eV, so the answer is somewhere between these values. Naively we could average this to 2.375 eV, but that would not be actual average of full spectrum white light.

As I understanding it, what we perceive as white light, in daylight conditions, is the visible spectrum of roughly 5500K black body radiation (which has more energy on the blue side). I'm not sure how we'd get the average eV of the photons that make it up.

If you have another standard measure of "white" sunlight, I could be happy with that too.

edit: I am not a student. I am not studying physics or mathematics. Sorry I cannot give working beyond attempting to average the highest and lowest eV and saying "obviously that's wrong" (which I did in the question already). If you think the answer to this question can be googled then please provide your google query and a link to something that isn't a physics course.

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    $\begingroup$ There are various ways to define “average”, so you need to be more specific. There is the mean energy, the median energy, the most likely energy, etc. $\endgroup$ – G. Smith Oct 6 at 21:50
  • $\begingroup$ The way I read it, there are hints that suggest what's wanted is the average energy of the BB curve when it is expressed in eV. That is, the weighted average energy. However, @G.Smith is correct, there are other possible definitions. If you can't ask anyone, I'd go with what I propose above. $\endgroup$ – garyp Oct 6 at 22:10
  • $\begingroup$ Weighted average, yes. $\endgroup$ – Qubei Oct 6 at 22:47
  • $\begingroup$ The mean. The most common and typical meaning of "average". Just to clarify as I think "weighted average" could have different interpretations. $\endgroup$ – Qubei Oct 6 at 23:09
  • $\begingroup$ I'm voting to close this question as off-topic because it is a question about human biology. $\endgroup$ – G. Smith Oct 6 at 23:12
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The mean photon energy in a photon gas at temperature $T$ is

$$\bar{E_\gamma}=\frac{3\zeta(4)}{\zeta(3)}k_BT\approx 2.70\,k_BT$$

where $\zeta(x)$ is the Riemann zeta function and $k_B$ is the Boltzmann constant.

The derivation of this from Planck's Law is a standard homework exercise, so I won't give it.

For $T=5500\text{ K}$, $\bar{E_\gamma}=1.28\text{ eV}$.

This average includes photons of all frequencies, including those we can't see. Limiting the calculation to the frequencies that we can see would turn it into a biology problem rather than a physics one. However, the two relevant integrals can be done numerically.

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  • $\begingroup$ Removed an unproductive thread of comments. Comments should be used to request clarification or suggest improvements to this answer. Comments about the question should go on the question itself. $\endgroup$ – Chris Oct 6 at 23:34
  • $\begingroup$ Thank you for the partial answer. To clarify, the visible light spectrum was defined in my question as being 1.65eV and 3.10eV, without requiring any reference to human biology. I phrased the question in this specific way so as to make it digestible by physicists but it appears that is impossible. $\endgroup$ – Qubei Oct 7 at 0:03
  • $\begingroup$ The mean in that range is 2.17 eV. $\endgroup$ – G. Smith Oct 7 at 0:15
  • $\begingroup$ Thank you. Can you please add that to answer so I can accept it? $\endgroup$ – Qubei Oct 7 at 0:21
  • $\begingroup$ @Qubei : it is not the case that photons of energy, say 1.64 eV or 3.11eV are totally invisible, while photons of 1.66eV and 3.09eV are "as visible" as those of 2.4eV. The response curve of the retina is smooth. So your question really is a biological one: one should compute an average energy for 5500° BB radiation weighed by the response curve of each of the three types of cones (and the rods),weigh these three values with the proportion of each type, (daltonians have only two types). Or just take the brute range 1.65eV to 3.10eV but IMHO this is not very meaningful. $\endgroup$ – Alfred Oct 7 at 5:42

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