[Halogen876]

I've been trying to figure out this question and I've been having some problems. Here is the question:

Sodium atoms absorb strongly at around 589nm. Suppose that a monochromator also passes a neon fill gas line at 588.2nm that has an intensity of 10% of the sodium line when no absorbance occurs. What is the absorbance measured on the AA spectrometer when the true absorbance of the sodium is 0.100? When it is 1.00?

I got the correct answer for the first part (when the true absorbance is 0.100). I simply took 0.100 and multiplied it by 0.9 (which I got from 90% which I got just by doing 100%-10%). The answer I got for this part of the question when I did that math was 0.09 which is the correct answer.

I repeated my process for the second part by doing 1.00x0.9 which is obviously just 0.9 so I thought that would be my answer but the correct answer is 0.74 so either my method for solving the first question was wrong and I just got the correct answer by chance, or there's something different about the second question that I'm missing...either way, any help would be greatly appreciated

[Furanone]

I'm thinking it has something to do with dilute vs non-dilute systems so where 0.1 abs is dilute and that part of %transmittance works out to be linear and proportional with absorbance to make the 90% equal 0.9 absorbance, but at 1.00 abs where it is not dilute then %transmittance will not be linear at this point and therefore not proportional with absorbance. since ABS = 1/Transmittance

[mjc123]

Absorbance is not 1/transmittance. Absorbance is -log₁₀(transmittance); transmittance = 10^-(absorbance).

either my method for solving the first question was wrong and I just got the correct answer by chance

Yup. Pure coincidence.

I simply took 0.100 and multiplied it by 0.9 (which I got from 90% which I got just by doing 100%-10%).

Why? What's the physical basis of it? Do you understand what this means, in terms of what's physically going on in the spectrometer? (I had to think about it for a while, because I don't regularly do AA.)

Suppose that a monochromator also passes a neon fill gas line at 588.2nm that has an intensity of 10% of the sodium line when no absorbance occurs

Here's what I think it means. The spectrometer is passing light from a sodium source, plus the neon line, through the empty sample chamber. With no absorption, the total light intensity reaching the detector is
I₀(apparent) = I₀(Na) + I(Ne) = I₀(Na) + 0.1I₀(Na) = 1.1I₀(Na)
With a sample containing enough Na to give a (true) absorption of 0.1, the fraction of Na light transmitted is 10^-0.1 = 0.794. So
I(apparent) = 0.794I₀(Na) + 0.1I₀(Na) = 0.894I₀(Na)
So the apparent transmission is I(apparent)/I₀(apparent) = 0.894/1.1 = 0.813
and the apparent absorbance is -log(0.813) = 0.090.

Now try it with true absorbance = 1.0 (It works).

[Halogen876]

Thanks a lot mjc123 - that makes perfect sense now