How Does a GaAs Solar Collector Manage Light Absorption and Re-radiation?

[Math Jeans]

Homework Statement

A flat and thin GaAs solar collector of 100 micrometer thickness is irradiated with 100mW of 650nm light (incident normal to the surface). The absorption coefficient is about 10^4 cm^-1 and the bandgap is 1.4 eV. Assume that the GaAs has ~100% internal quantum efficiency.

How much of the light is reflected?

How much is absorbed and turned into heat?

How much is re-radiated?

How much of the re-radiated light can escape the 17 degree escape angle from total internal reflection?

(Some more questions about it, but I can answer these if I can get the first parts)

Homework Equations

The Attempt at a Solution

I was able to get the first part using the equation for the Fernel coefficient, and got 28.6% of the light.

The main issue that I'm having is with parts 2 and 3:
-I calculated the absorbance of the material using the beer lambert law and got an I/I0 approximately equal to 0 (on the order of 10^-44), so I can assume that all of the light it absorbed.
-Based on how the questions are stated, I'm fairly sure that they're asking for the amount of non-radiative recombination in the second part, and the amount of radiative recombination in the third part.

However, if I'm assuming 100% internal quantum efficiency, doesn't that mean that no recombination should occur because all electrons in the conduction band are going towards current? What am I missing here?

Any help would be appreciated.

Thanks,
Jeans

 

[KataKoniK]

Dear Jeans,

Thank you for your post. It seems like you have a good understanding of the first part of the problem. Let's address your questions about parts 2 and 3.

First, it is correct that the absorbance of the material is essentially 100% due to the high absorption coefficient and thinness of the material. This means that all of the incident light is being converted into heat.

Second, the 100% internal quantum efficiency does not necessarily mean that there is no recombination occurring. Internal quantum efficiency refers to the efficiency of converting absorbed photons into charge carriers, but it does not account for the recombination of those charge carriers. In fact, there will always be some non-radiative recombination occurring in any material, and this will result in the conversion of some of the absorbed light into heat instead of electricity.

As for the third part, the amount of re-radiated light depends on the efficiency of radiative recombination, which is typically lower than internal quantum efficiency. This means that some of the absorbed light will be converted into heat, while the rest will be re-radiated as photons. The exact amount of re-radiated light will depend on the specific properties of the material and the conditions of the experiment.

I hope this helps clarify the problem for you. Let me know if you have any further questions.