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Topic: Einstein Coefficents  (Read 4412 times)

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MarkDH

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Einstein Coefficents
« on: May 05, 2006, 02:31:11 PM »
Hey,

This is my first post here, but I've searched the forum for related threads and found nothing. I've got a past exam question that I'm stuck with, as the information doesn't seem to be in my lecture notes, and my textbook seems to say I need more information than given.

The question says low pressure Hg vapour is electronically excited to Hg* by a single laser pulse, wavelength 193nm. The half life was determined by monitoring the decay of fluorescence after the pulse, and found to be 80us. When it was repeated with a mixture of Ar and Hg, the half life was found to be 40us. The first question asked is calculate the Einstein Coefficient for absorption for Hg at 193nm.

I'll tell you what I *think* I've worked out so far. Theres only a single laser pulse, so I thought that it would only be spontaneous emission as opposed to stimulated emission. Hence (dN1/dt) B12 = -(dN1/dt) A21. So as you know the half life of excited state, you could work out the rate constant for emission via t1/2 = 0.693/k. This would then mean you could work out (dN1/dt) as it is equal to k[N1]. This is where I get stuck however as the question does not give you a concentration of the Hg vapou, and I'm just assuming theyre all in ground state before the pulser. And thats only to get the rate of change, never mind whats going on in the rest of the absorption coefficient equation.

To be honest I don't know if I'm doing this the completely wrong way using kinetics as I'm trying to work out the method myself using various textbooks. I'm more of an organic chemist (don't hate me please  ;)), so any help with what methodology could be used to work out the Einstein coefficent would be a great help.

Cheers,

Mark

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Re: Einstein Coefficents
« Reply #1 on: May 06, 2006, 11:29:21 AM »
The question says low pressure Hg vapour is electronically excited to Hg* by a single laser pulse, wavelength 193nm.

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MarkDH

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Re: Einstein Coefficents
« Reply #2 on: May 07, 2006, 08:21:20 AM »
Ok, so PV = nRT? I don't have a specific pressure, a volume of the vapour, nor the temperature, so where do I go from here?

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