# Photon energy and power

My question has to do with the solar spectrum. I understand sunlight is composed of light of different wavelegnths, and these wavelegnths have different energies associated with them. I also understand we can integrate across all wavelengths to get the total irradiance (power) in (watts/m2). But what happens if we just use one wavelength of light, say light at 600nm. How would I be able to get the "power" at that wavelength using data from AM1.5Global spectrum?

• What ideas do you have? If you can get the total power by integrating across all the wavelengths, how would you get the power contributed by a small range of wavelengths? – Bill N Apr 16 '18 at 15:48
• It doesn't really make sense to ask about the power at a single wavelength. The spectrum at one wavelength is infinitesimally narrow, and thus has zero power. Look at the units on the $y$-axis of the spectrum. What you really want is the power in a narrow band around 600 nm. The size of the narrow band depends on the application and equipment available. – garyp Apr 16 '18 at 17:00

The total power per unit area from a blackbody radiator can be obtained by integrating the Planck radiation formula over all wavelengths. The radiated power per unit area as a function of wavelength is $$\frac{dP}{d\lambda}\frac{1}{dA}=\frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)}$$ so the integrated power is $$\frac{P}{A} = 2\pi h c^2\int_0^\infty\frac{d\lambda}{\lambda^5(e^{hc/kT}-1)}$$
This makes sense (ish) I'm not gonna claim I fully understand the Planck radiation formula, but I can make assumptions on how to use it. Assuming that that first equation somehow takes a wavelength ($\lambda$) and turns it into a radiated power per unit area ($\frac{dP}{dA}$) for that wavelength of light. The second equation, therefore, basically takes the sum over all wavelengths of light. Looks good so far.
You want a specific set of wavelengths, though. Say, from 380nm to 750nm (the visible spectrum.) Since you're using actual data, an integral like $$2\pi h c^2\int_{\lambda_1}^{\lambda_2}\frac{d\lambda}{\lambda^5(e^{hc/kT}-1)}$$ won't be computable. What you need is discreet integration, so to speak. If you assumed that the set $\{\lambda_n|0\leq n\leq m\}$ represents all $m$ wavelengths in your data, then you could rewrite the integral as a big sigma, or $$\sum_{n=0}^{m}\frac{ 2\pi h c^2}{(\lambda_n)^5(e^{hc/kT}-1)}$$