# Light absorption in a semiconductor

## Homework Statement:

Silicon Heterojunctions

## Relevant Equations:

Beer's law: I = I0 * exp (-alpha*x)
The doped a-Si: H layers in a HIT solar cell do not contribute to the photocurrent. The light they absorb (according to their absorption curve below) is lost.

For a doped a-Si: H layer at the front side of the cell that is 25nm thick, what percentage of light at 400nm will be lost due to absorption in this layer? For this question, we can ignore the effect of reflection at the front interface. Express your answer as a number between 0 and 100.

My guess:
using beer's law: exp (-alpha*x) = I/I0 = percentage of light lost due to absortion
alpha = 5.8*10^5 cm-1 (because 400 nm = 3.1 eV)
x = 25*10^-7 cm
light lost due to absorption = 23.46 %
But I'm told that this isn't the correct answer, I'm so confused can someone help me please?

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phyzguy
I think you did the arithmetic correctly, but you didn't think through what it means. If I/I0 = 0.23, what percentage of the light is lost?

I think you did the arithmetic correctly, but you didn't think through what it means. If I/I0 = 0.23, what percentage of the light is lost?
For me it means that I = 0.23×I0, that means 23% of the light got absorbed (lost) and 1-0.23 = 77 % didn't get absorbed.

For me it means that I = 0.23×I0, that means 23% of the light got absorbed (lost) and 1-0.23 = 77 % didn't get absorbed.
Nvm after further thinking, it means that 23% got through the semiconductor without getting absorbed that means 77% got absorbed.
But does that mean that 77% of the light is lost? Thats seems to high.

phyzguy
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