# AM1.5 Spectral Irradiance unit conversion

I have the AM 1.5 spectrum http://rredc.nrel.gov/solar/spectra/am1.5/

Which gives spectral irradience in units of $\frac{W}{m^2 nm}$ vs wavelength in $nm$. For my purposes I need this spectrum in terms of $\frac{W}{m^2 eV}$ vs $eV$.

Converting the x-axis is relatively straightforward: $$\varepsilon \ [eV]=\frac{hc}{\lambda \ [nm] \ q}\times 10^9$$ Where q is the electron charge and h,c are in their SI units. However I can't seem to figure out the correct y-axis conversion. I would expect it to be similar in magnitude to the black body spectrum obtained via plancks law:

$$I(\varepsilon) = \frac{2 \pi}{h^3 c^2} \frac{\varepsilon^3}{\text{exp} \left( \frac{\varepsilon}{kT_{sun}}\right)-1} \times q$$

How can I convert this spectrum between $\frac{W}{m^2 nm}$ and $\frac{W}{m^2 eV}$ ?

Let's denote the flux (power per unit area) per unit wavelength as $F_\lambda$, and the flux per unit energy as $F_\varepsilon$. Then we simply have $$F_\varepsilon = F_\lambda \left\lvert \frac{\mathrm{d}\lambda}{\mathrm{d}\varepsilon} \right\rvert.$$ All you need to do is express wavelength in terms of energy for light. You were getting at this in your x-axis conversion, except your conversion there is wrong confusing. Electron-volts are units of energy, plain and simple; just use the relation $\lambda\varepsilon = hc$, where conveniently $hc = 1.240\times10^3\ \mathrm{eV\cdot nm}$.
• Thank you, that seems to work. The reason for the electron charge appearing is from the conversion between eV and J [$1 eV = q J$]. The conversion can be seen to be equivalent $hc \times 10^9 / q = 1240 eV. nm$, if h,c are in SI units – uqtredd1 Oct 10 '15 at 4:59