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Re: [Phys-l] thorium-228



On Dec 27, 2009, at 12:03 PM, Michael Edmiston wrote:


Ludwik, (et.al.)

I haven't been able to respond until today.

There is no answer to the original question (see below) because the question
does not state the *quantity* of thoriated water for which you would like to
know the activity. You merely state the concentration of the thorium in
water. The activity would then depend on whether you have a drop of that
water, or a gram, or a mole, or a liter, or whatever.

In addition, another point of insufficient information is the concentration.
The abbreviation ppb just means "parts per billion." You have to specify
whether ppb means atom/atom, or mass/mass, or mass/volume, etc. In my
experience with environmental work all of these are used, so you really need
to specify. When doing nuclear work I most typically encounter atom/atom
or molecule/molecule, or mole/mole (which are the same thing if you assume
you will always treat things in their normal condition, that is, we treat
water as a molecule and treat thorium as an atom).

I also don't understand how you are getting the decay constant.

Let's start with several necessary items...

half-life = 1.9131 year = 6.0372E7 seconds
decay constant = ln2/HalfLife = 1.1481E-8 sec^-1
1 ppb Th in water means one thorium atom per billion water molecules

Let's find the activity of one mole of water containing 228-Th at 0.03 ppb
atom/molecule.

Number of Th atoms in one mole of water = (0.03 ppb)x(N(A)) = 0.03E-9 x
6.022E23 = 1.8066E13 Th atoms

Activity = (decay constant)x(number atoms) = 1.1481E-8 x 1.8066E13 =
2.0742E5 disintegrations per second for 1 mole of thoriated water.

Activity in Curies = dps/3.7E10 = 2.0742E5/3.7E10 = 5.6059E-6 Ci = 5.6059
uCi (per mole water)

Activity per liter of 0.03-ppb-thoriated water = (5.6059 uCi) x (55.51
mol/L) = 311.18 uCi/L = 3.1118E-4 Ci/L

If you do the calculation with the understanding that ppb is mass/ mass, then
1 ppb means one-nanogram of thorium in 1 gram of water. In that case we can
approximate a liter of water as 1000 grams, so 0.03 ppb is 3E-8 grams of
thorium which is 1.316E-10 moles, which is 7.923E13 atoms of Th. Multiply
this by the decay constant and we get 9.0961E5 dps = 2.458E-5 Ci/L

So neither of my calculations agree with yours.

Thank you for very important qualification, Mike. In the erratum I will clearly state that 0.03 ppb stands for "atoms of Th=228 per molecule of water."

Your calculation 3.11E-4 Ci/L does agree with mine (but not the one that was printed, and which I want to correct in the erratum). Now I know that I am not going to goof again.

Ludwik


Ludwik's new book (AUTOBIOGRAPHY) see:

http://pages.csam.montclair.edu/~kowalski/mybook2.html


Share this link with those who might be interested. Thanks in advance.