Re: Question on radiolysis of H20

[rkathren@tricity.wsu.edu (Ron L. Kathren)]

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
>
>
>