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Re: Question on radiolysis of H20



Isn't this what professors are for?  You're paying for their help, and one
on one would seem to me to be better than the cold RADSAFE approach.  I sure
would hope that my students would come to me!

The Little Professor aka Ron Kathren


>OK, right up front let me say that I'm taking a university course in
>Radiobiology, and the following question is part of an assignment.  I've
>tried looking through a few texts on the subject, but I'm still having
>trouble with what looks like a very simple problem.  I figured that RADSAFE
>isn't much different from any other resource, or from asking a professor for
>help - just more convenient - so here goes...
>
>Here's the problem (in several parts):
>
>(a) Calculate the total energy (in eV) deposited in 500 ml of water
>following a dose of 20 krads of x-rays.
>
>Ok, this should be trivial.  Here's my solution:
>1 rad = 100 erg/g
>therefore 2e4 rad = 2e6 erg/g
>multiply by 500 g (1 g/ml for water) to yeild 1e9 erg
>but 1 erg = 6.24e11 eV,
>so total energy deposited is 6.24e20 eV
>
>
>(b)  Given that the G value for the production of H2O2 is 0.55 under these
>particular conditions, calculate the number of H202 molecules produced.
>
>Here's my solution:
>The G value is the number of molecules produced per 100 eV deposited.
>Therefore, the number of H2O2 molecules produced should be
>6.24e20 eV x 0.55 molecules / (100 eV) = 3.43e18 molecules
>
>
>(c) Calculate the concentration of H2O2 formed.
>
>Well, the number of moles of H2O2 formed is
>3.43e18 / (6.023e12 mol^-1) = 5.70e5 mol
>This is in 500 ml (0.5 liter), so the concentration is
>5.70e5 mol / 0.5 liter = 1.14e6 mol/l = 1.14e6 M
>
>
>Ridiculous!  At a molecular weight of about 38 g/mol, 1.14e6 mol of H2O2
>molecules (the number per litre of water) would have a mass of over 40 tonnes!
>
>Where have I gone wrong?!?!
>
>Thanks in advance for your help,
>
>
>Chris
>
>clement@vaxxine.com
>
>
>