[Halogen876]

Hello,

I've been trying to figure out the following question and haven't been getting very far with it. Here's the question:

Assuming that a fluorescent analyte doesn't absorb any emitted light, that excitation light falls evenly on the entire width of the cell and that the entire cell width can be "seen" by the detector, derive an equation that predicts the deviation from linearity of fluorescence as a function of C and ε, assuming b=1.00cm. (Hint- use the ratio of observed fluorescence to "expected" or "ideal" fluorescence and the series expansion of the exponential retaining both the first and second order terms in C.) At what concentration will there be a 2% deviation (98% of the expected fluorescence is observed) if ε=1.00E4M^-1cm^-1?

I know a couple equations that I think might be useful:
F=F₀(1-e^-2.303εbc)
e^x=1+(x/1!)+(x²/2!)+(x³/3!)+...

but other than that, I'm not sure what I should be looking at or thinking about to start this question...any help would be greatly appreciated!
 

[mjc123]

So combine your equations. Expand the expression for F using x = -2.303εbC.

[Halogen876]

So what I've come up with doing that is:

F=F₀[1-1+2.303εbc-(2.303εbc)²/2!+...]

I'm not sure how far I should write out the expression though...I wrote it out to  the term with the 2! in it since the question mentions the first and second order terms in C, so I was thinking maybe that's what they mean by that, but I'm not sure....

Any more help would be greatly appreciated!

[mjc123]

Well, you've virtually got it. 1-1 cancels out, so
F/F₀ = 2.303εbC - (2.303εbC)²/2.... [Use a capital C because that's what they use in the question]
which is a linear dependence with (for low εC) a small non-linear correction. (For completeness, substitute the given value of b = 1.00 cm.)

[Halogen876]

Ok, that's good to know I'm on the right track.

So then  I can plug in 0.98 for F/F₀, 1.00 for b and 1.00E4 for ε so I get:
 0.98=23030C-2.65E8C²
I can rearrange that so that I can use the quadratic formula to solve for C, but when I do that I run into a problem because I get a negative inside the square root....

When I rearrange the equation, I get 2.65E8C²-23030C+0.98=0. so the inside of my square root in the quadratic formula is (-23030)²-4(2.65E8)0.98 which equals -5.08E8

Obviously there's something I'm missing somewhere since it's not all quite coming together, so any more help would be appreciated!

[Corribus]

Be careful with your signs when you do your expansion.

[Halogen876]

I must be missing something, but I have checked my math several times and have gone over it with a math major and I don't see anywhere where I have the wrong sign...if anyone could point out where I am making an error, I would really appreciate it!

[Corribus]

I admit, the numbers have large exponents and there are a lot of signs to keep track of. It's a messy problem.

Seems to me: your x is negative.  We agree that your expression F/F0 = x + x²/2 + x³/6 + x⁴/24 ...

Now put in -2.303 εC in for x (i'm just ditching the b value since it's 1) and 0.98 in for F/F0

Therefore 0.98 = -2.303 εC + [-2.303 εC ]²/2 + [-2.303 εC ]³/6 + [-2.303 εC ]⁴/24 ...

Simplifying using values for ε given in your opening post: 0.98 = -23030 C + 2.65E8 C² ... (cutting off after two terms).

Therefore your quadratic equation is: 2.65E8 C²-23030 C - 0.98 = 0

This equation has real roots. Using my graphing calculator, the positive root gives C ~ 1.18 x 10^-4 M. Basically you don't seem to be taking in to account the extra minus sign when you subtract everything from 1. This gives you a quadratic equation that never crosses x = 0, hence no roots.

NOTE: I'm not sure why the need for the expansion here. It isn't really expressive of anything physical and cutting it off at term 2 as you've been instructed gives you a pretty inaccurate answer. Much easier to just solve the original exponential function directly by taking the natural logarithm of the expression and solving for C, I think. When I do it that way I am getting an answer of ~ 1.7 x 10^-4 M. I wrestled with why that might be so different from the other answer for a while. Granted, it's been a while since I did a lot of series expansions, but it occurs to me that in McLaurin expansions each successive term is not necessarily smaller than the next because of the way the exponentials line up with the factorials. To see why, look at a basic example like e⁴, the value of which ~54.6. If you do the McLaurin expansion using your formula, the first through eight terms have contributing values of 1, 4.0, 8.0, 10.67, 10.67, 8.53, 5.68, 1.63, 0.72 ... You can see right away that the fourth and fifth terms are actually the biggest. If you add all the infinite terms, you would get the true value of 54.6, but if you cut off after the quadratic (third) term, you'll get a value of 13, very far off the mark.  I tried to solve your problem using higher order terms and I do get closer to the value you get by solving directly, but the math gets really cumbersome and expressions with odd terms only have imaginary roots, so I can't demonstrate to you an asymptotic or oscillatory approach using real numbers, unfortunately.

Anyway, I don't think it's a particularly good problem. But as I said it's been a while since my undergrad calculus classes so maybe I'm missing something instructive about it.

[Halogen876]

Thank you for taking the time to work all that out an explain so much. I agree with you, I don't think it's a very good problem either!

I understand how you got both of those answers (1.18E-4M and 1.7E-4M) but the answer in the textbook is 1.74E-6M so maybe the answer in the book is wrong? If anyone has a way to solve the problem and come out with that answer, I'd really appreciate it if you let me know. Or if you really think the answer in the book is wrong, I'd appreciate hearing that too. Thanks!

[mjc123]

I think we're going wrong with the ratio of "observed" to "expected" fluorescence. What is meant by these terms? If we assume that fluorescence is proportional to the amount of light absorbed, then
F = kI₀(1 - e^-αCb)        (where α = 2.303ε)
F = kI₀(1 - 1 + X - X²/2! + X³/3!...) where X = αCb
What is "expected" fluorescence. I can only assume this means fluorescence (and absorption) that is proportional to X, ignoring the reduction in I due to absorption:
F₀ = kI₀X
Then F/F₀ = 1 - X/2 + X²/6.... = 0.98
This quadratic equation has roots X = 0.0405 and 2.959 (but in the latter case truncation of the expansion is unjustified)
and hence C = 1.76e-6 M. (The "wrong" root is 1.29e-4 M)
In this case there isn't a simpler way of doing it, since F/F₀ = (1-e^-X)/X , which isn't analytically soluble.

[Halogen876]

That makes sense now!

Thank you both so much for all your help with this question - it was very much appreciated