What fraction of a beam is absorbed in a layer containing 1 kg/m^2 of gas?

Question

Radiation with frequency ν and zero zenith angle passes through a gas with absorption coefficient k_v = 0.01 m²/kg.

(a) What fraction of the beam is absorbed in a layer containing 1 kg/m² of gas?

(b) What mass of gas would the layer have if it absorbed half the incident radiation?

I think that the optical depth = τ = 1, but I'm not sure what to do next! Any help would be greatly appreciated.

Answer 1

Hi Jenna M.,

First thing to do is, calculate the Absorbance of the layer. Just a consideration of the units will tell you that you just multiply, and A = 0.01 absorbance units. Now, I'll assume that you don't have an intuitive grasp what that means at all. Absorbance = 1.0 means you have only 10% of your initial energy passed, absorbance = 2.0 means you have only 1% of your initial energy passed, and so forth. In other words, there's an exponential decay of energy, with increasing absorbance.

So now that you know Energy = 10^(-something), and you have 2 "reference points", can you work out the exact form of that "something" term? It has to do with Absorbance!

Once you've done that, calculating 10^(-0.1) for energy remaining, and then "flipping" that into energy absorbed, you've done part A.

For part B, you just need to plug in to your bolded equation above (with Energy = 0.5), and work your way back through the corresponding Absorbance to a density term. Hint: What's the inverse mathematical operation to the "10^something" mathematical operation?

-- Cheers, -- Mr. d.