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Determine extinction coefficients in glass for Fe2+/Fe3+

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ACvub
#1
Aug28-13, 03:14 AM
P: 9
Hello everybody,

I want to determine the extinction coefficients of Fe2+ and Fe3+ in glass.
There are literature data (e.g. Weyl's book "coloured glass"), so I know what kind of curves I should expect. As I am studying a slightly different soda-lime-silicate system, I want to recalculate the curves.

I have glasses with different Fe2+/Fe3+ ratio and total iron concentration [Fe].
I have measured the absorbance of two glasses and I have solved a simple two equations system starting from Lambert-Beer law: A = Ʃ εCd
where A is absorption, ε is the extinction coefficient, C is the concentration and d is the thickness of my glass.

Unfortunately, I get negative values in one of the two extinction coefficient curves. Obviously, this doesn't make sense.

Anyone sees what is wrong in my reasoning? I can't figure it out.

Thank you very much in advance,
Mark

PS For a close look at the system I have made, see the attachment (.pdf)
Attached Files
File Type: pdf System_Solution.pdf (81.5 KB, 17 views)
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chemisttree
#2
Aug28-13, 11:24 PM
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Try solving for Fe+3 in terms of Fe+2 and total Fe.
ACvub
#3
Aug29-13, 01:50 AM
P: 9
Thank you chemisttree for replying.
However, I don't get what you mean. How can I solve in terms of total Fe?
In the system I have the concentrations of the two absorbing species:
Fe3+ (CFe3+) and Fe2+ (CFe2+)

Where should total Fe appear in the equations?

Borek
#4
Aug29-13, 02:44 AM
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Determine extinction coefficients in glass for Fe2+/Fe3+

You said you know total Fe for the glass, so you can write an equation for the mass balance:

Fe2+ + Fe3+ = Ftotal
ACvub
#5
Aug29-13, 05:41 AM
P: 9
You are right! I understand.
But even if I substitute CFe3+ with (CFetotal - CFe2+), I don't see how it would solve the problem. At the end it is always the same value.
ACvub
#6
Aug29-13, 06:58 AM
P: 9
the unknown variable are the ε for the two ionic species.
so adding the mass balance equation does not add any value.
chemisttree
#7
Aug29-13, 03:46 PM
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Sooo, you're not even going to try it my way?
ACvub
#8
Aug30-13, 02:12 AM
P: 9
I am sorry chemisttree, maybe I hadn't explain well myself. I didn't want to be disrespectful.
I have tried to use your advice. But I don't see how.
Substituting ferric concentration with the subtraction of ferrous from total iron concentration, the final solution remain unchanged.

I attached the spectra of the two glasses I am using. Also, I have plot quickly (read as "I haven't add units and axis names") the absorption coefficients I obtained for Fe2+ and Fe3+. You see that εFe3+ is negative, which doesn't have any physical meaning.

Below you find the concentration data for both GlassA and GlassB.

GlassA:
CFe2+ = 0.056 wt%
CFe3+ = 0.125 wt%
CFetotal = 0.181 wt%

GlassB:
CFe2+ = 0.127 wt%
CFe3+ = 0.285 wt%
CFetotal = 0.412 wt%

If I substitute in GlassA CFe3+ = CFetotal - CFe2+, I get the same value: 0.181 - 0.056 = 0.125 wt%.

I would appreciate if you could keep helping me.
Thank you.
Attached Thumbnails
GlassAbsorbances.PNG   GlassExtCoeff.PNG  
ACvub
#9
Sep4-13, 01:45 AM
P: 9
Nobody has an idea on what's wrong?
chemisttree
#10
Sep4-13, 12:15 PM
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For the attached spectra, what value of extinction coefficient did you use for Fe+2 and Fe+3? Did you hold the extinction coefficient for Fe+2 constant? Or did it vary with wavelength?
ACvub
#11
Sep4-13, 12:34 PM
P: 9
they both vary with wavelength. In the 2-equations system I have attached to my first thread, the unknowns are the two extinction coefficients.
chemisttree
#12
Sep4-13, 12:45 PM
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So you have the spectra in the form of wavelength and absorbance? Can't help you without raw data.
ACvub
#13
Sep5-13, 12:40 AM
P: 9
Yes, my data are absorbance as a function of wavelength. NOw I don't have access to the hard drive where data data are. I will post the raw data later on today so you can have a look.
thank you again!
ACvub
#14
Sep6-13, 03:26 AM
P: 9
sorry for the delay. here is the file with the absorbance for both glasses.
Attached Files
File Type: txt AbsExtCoeff.txt (37.2 KB, 4 views)


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