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# Flash photolysis - Beer Lambert law Watch

I see that I'm going to have to re-arrange the question to get ln(I/I0)/(-sigma x l) = [OH] but I dont have I or I0. The only thing I can think to do is use the fact that 10% of the initial light is absorbed so would this mean that I/I0 = 1/0.9 ?

Also I dont understand why I am told the wavelength, it doesnt seem necessary!

Thanks
Attached Images

2. anyone?
3. (Original post by DonnieBrasco)

I see that I'm going to have to re-arrange the question to get ln(I/I0)/(-sigma x l) = [OH] but I dont have I or I0. The only thing I can think to do is use the fact that 10% of the initial light is absorbed so would this mean that I/I0 = 1/0.9 ?

Also I dont understand why I am told the wavelength, it doesnt seem necessary!

Thanks
Solve it as a function of I 0... You don't have enough information, I.e. I0 to be able to solve it.
Btw I0 is the initial intensity... Doesn't come out well on the app

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4. How exactly would you solve it as a function of I0? Can you assume that I is 0.9 since 10% of the light is absorbed? Then shouldn't I0 be 1.0?
5. (Original post by JMaydom)
Solve it as a function of I 0... You don't have enough information, I.e. I0 to be able to solve it.
Btw I0 is the initial intensity... Doesn't come out well on the app

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This is what I thought, I dont really see how the question is possible to calculate. I'll just have to ask my lecturer,

Thanks though
6. (Original post by ldfire10035)
How exactly would you solve it as a function of I0? Can you assume that I is 0.9 since 10% of the light is absorbed? Then shouldn't I0 be 1.0?

(Original post by DonnieBrasco)
This is what I thought, I dont really see how the question is possible to calculate. I'll just have to ask my lecturer,

Thanks though
Ok, here goes..... After having actually done it, here is how you solve it.

Now I have attached a picture of solving it terms of I0, but this may be sort of doing the wrong question. I interpreted it as OH being generated by the absorption of UV, rather than the light probing the concentration. I have solved it in terms of OH being generated by absorption of UV radiation. You don't even really need the Beer-Lambert Law!!!!

If it is the OH doing the absorption and we have assumed that the concentration is constant and not affected by the UV, then just divide through by I0, so that I/I0 = 0.1. (done the first bit of that for you also, sure you can generate a concentration from this formula. You will need to pick a volume and work it out from there)

Ahhh, I go to attach the file and it doesn't work!!!! Stupidly large files, grrrr.
Attached Images
7. Beer Lambert problem.pdf (81.8 KB, 62 views)
8. Still a bit confused, in your working you have I0 - I giving 0.9I0 when the quantum yield is 1 and because 10% of the incident light is absorbed by the OH. I assume that, because you took it from the perspective of OH production, that a quantum yield of 1 means that 1 OH is produced per 1 photon absorbed.

Also, since I0 is the initial incident intensity and I is the intensity of being transmitted through the sample, shouldn't it be the other way around?

I0 - I = # photons absorbed = # OH radicals produced.

So: I0 - I = 0.9 IO, would imply that 90% of the initial photon intensity was absorbed by the sample.

If I/I0 = 0.1 as a transmittance, wouldn't that imply that 90% of the light was absorbed by the OH sample?
9. (Original post by ldfire10035)
Still a bit confused, in your working you have I0 - I giving 0.9I0 when the quantum yield is 1 and because 10% of the incident light is absorbed by the OH. I assume that, because you took it from the perspective of OH production, that a quantum yield of 1 means that 1 OH is produced per 1 photon absorbed.

Also, since I0 is the initial incident intensity and I is the intensity of being transmitted through the sample, shouldn't it be the other way around?

I0 - I = # photons absorbed = # OH radicals produced.

So: I0 - I = 0.9 IO, would imply that 90% of the initial photon intensity was absorbed by the sample.

If I/I0 = 0.1 as a transmittance, wouldn't that imply that 90% of the light was absorbed by the OH sample?
Sorry, got that bit wrong. Should probably pay more attention to what I'm doing! Yeah should be 0.1 I0 but pretty sure the rest is correct

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10. Yup, everything seems to be good from what I saw haha. Thanks for the explanation
11. (Original post by JMaydom)
Ok, here goes..... After having actually done it, here is how you solve it.
Thank you for the calculation but I'm afraid it is slightly above what I'm doing, I have no idea what quantum yield is!

Can you not do it this way? Although the answer I get is ludicrous!

12. (Original post by DonnieBrasco)
Thank you for the calculation but I'm afraid it is slightly above what I'm doing, I have no idea what quantum yield is!

Can you not do it this way? Although the answer I get is ludicrous!

Quantum yield is a very simple idea. For the example I used all I'm saying is that for each photon absorbed, one radical is produced

As to you calculation... The most difficult bit was always the units in the last bit. Not sure you have that correct. Probably due to you defining the concentration in dm and the absorption coefficient being in cm

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13. (Original post by JMaydom)
Quantum yield is a very simple idea. For the example I used all I'm saying is that for each photon absorbed, one radical is produced

As to you calculation... The most difficult bit was always the units in the last bit. Not sure you have that correct. Probably due to you defining the concentration in M per dm and the absorption coefficient being in cm

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14. OK thanks, but provided I can convert my units correctly, does the calculation look plausible?
15. (Original post by DonnieBrasco)
OK thanks, but provided I can convert my units correctly, does the calculation look plausible?
You got the ratio of I/I0 right but it looks to me that you got it the wrong way
So should be correct from then onwards, apart from units of course!

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