Wednesday, 21 November 2018

thermal radiation - Black Body Spectrum Plot


I'm Having trouble replicating the Black Body model for sun shown on this plot


enter image description here


To my understanding I should only use Planck's formula:


$$I(\lambda) = \frac{2\pi hc^2}{\lambda^5}\dfrac{1}{exp\left(\frac{hc}{k_bT\lambda}\right)}\cdot 10^{-9}$$


Being $\lambda$ in [m] and having the last factor, $10^{-9}$, added for the units presented. What i get is


enter image description here



Can anyone see why?



Answer



The equation for the black body spectrum is:


$$ B(\lambda,T) = \frac{2\pi hc^2}{\lambda^5}\frac{1}{exp\left(\frac{hc}{k_bT\lambda}\right)-1} $$


where $B$ is the spectral radiance and has units of watts per unit area of the emitter per unit wavelength. So if we multiply by $10^{-9}$ as you have this gives us the power per square metre of the Sun's surface per nm. When I graph this I get:


Black body curve


which is broadly the same as your curve. As Benjamin says there's an error in your equation though as it happens that error doesn't make too much difference.


To get the total power per nm we have to multiply by the surface area of the Sun, $4\pi r^2$, to get:


$$ P_\text{total} = B\,4\pi r^2 $$


where $r$ is the radius of the Sun. Then divide by the area of the sphere with the radius of the Earth-Sun distance, $4\pi R^2$, to get:



$$ P_\text{Earth} = B\,\frac{r^2}{R^2} $$


The factor $r^2/R^2 \approx 2.17 \times 10^{-5}$, and including this in the calculation gives:


Spectrum at Earth


And this is pretty close to the curve you cite.


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