Probability to create an electron - hole pair inside Si Given a slab of Silicon and a photon flux $\Phi[\frac{\gamma}{cm^2sec}]$ incident on the slab of thickness $h$, the incident power $I$ given by $\Phi$ will be absorbed following the Beer - Lambert law:
$$I(x)=I_0\exp(-kx)$$ where $x$ is the vertical axis from the surface of the semiconductor. What is the probability for a photon $\gamma$ of wavelength $\lambda$ to create an electron - hole pair inside the slab? Thanks.
 A: We will have to assume several things:
1) That your external Quantum Efficiency is unity (assuming you have a good AR coating).
- or conversely we are dealing only with the photons that make it into the substrate.

2) That your Si deep enough that all the photons are absorbed and that yo are concerned only with the photons absorbed  within a certain thickness h.
- other wise you'll have to be concerned about reflections at the other interface too.

3) and we'll assume there is insignificant recombination
Since Si is an indirect bandgap material it's absorption coefficient does depend upon wavelength quite strongly.
In semi-conductor physics the coefficient is called alpha, here is a plot.
From the chart you pick your wavelength, which gives you your alpha. The absorption from the surface to a depth of h is simply: $$ P=P_0(1-e^{-{\alpha(\lambda)}h}) $$.  This is the cumulative absorption in that depth of material (i.e. the integral).
Using the above data and assigning a 450 nm to "blue", 556 nm to "green", 630 nm to "red" and 850 nm to "NIR" we derive the following plot of internal QE.  The total absorption is the area under the curve of each color.  You will notice that blue gets absorbed closer to the surface.  with NIR (Near IR) penetrates much deeper.

