The Planck blackbody function, radiation per unit area per unit wavelength ## M(\lambda, T)=\frac{2 \pi hc^2}{(\lambda^5)(e^{\frac{hc}{\lambda k T}}-1)} =n(\lambda) \, (\frac{c}{4})( \frac{hc}{\lambda} ) ## where ## n(\lambda) ## is the photon density per unit volume per unit wavelength. To get the photon density per unit volume ## n =\int\limits_{0}^{+\infty} n(\lambda) \, d \lambda ## . This is the easiest way I know of computing it. ## \\ ## Basically I worked backwards from a step in the derivation of ## M(\lambda, T) ##, where the photon density per unit volume per unit wavelength ## n(\lambda) ## has already been computed, and the next (final) step in getting ## M(\lambda, T) ## is to use the effusion rate formula ## R=\frac{n \bar{v}}{4} ## with ## \bar{v}=c ##, and ## n ## is made to be ## n(\lambda) ## because we are working with a spectral density here. This gets multiplied by ## E_p=\frac{hc}{\lambda} ## to give ## M(\lambda, T)=n(\lambda) (\frac{c}{4})(\frac{hc}{\lambda}) ##. The ## M(\lambda, T) ## in the form above (first line) is a very well-known result. This is easier than repeating the steps of the Planck function derivation where ## n(\lambda) ## gets computed after about 9 or 10 steps. ## \\ ## For a rough estimate, we can use ## \int\limits_{0}^{+\infty} M(\lambda, T) \, d \lambda=\sigma T^4 ##,(which is exact), where ## \sigma=5.67 \cdot 10^{-8} ## watts/(m^2 K^4), along with Wien's law, ## \lambda_p T=2.898 \cdot 10^{-3} ## m K, and a good estimate for ## \bar{\lambda} \approx 2 \lambda_p ##. This would get us an estimate for ## n ## without doing the integral of ## n=\int\limits_{0}^{+\infty} n(\lambda) \, d \lambda \approx \frac{\big{(}\int\limits_{0}^{+\infty} M(\lambda,T) \, d \lambda \big{)}\,(4) (2 \lambda_p)}{hc^2}=\frac{(\sigma T^4)(4)(2)(\frac{2.898 \cdot 10^{-3}}{T})}{hc^2} ##. My arithmetic (which I need to check) gets that ## n \approx 6 \cdot 10^{14} ## photons/m^3 . ## \\ ## A numerical (speadsheet) integration of ## n=\int\limits_{0}^{+\infty} n(\lambda) \, d \lambda ## would be more accurate, but I expect it would get a similar result.