Question on radiolysis of H20

[clement@vaxxine.com (Clement)]

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