Re: [Phys-l] thorium-228

[Michael Edmiston <edmiston@bluffton.edu>]

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.


Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton University
1 University Drive
Bluffton, OH 45817
419.358.3270
edmiston@bluffton.edu


--------------------------------------------------
From: "ludwik kowalski" <kowalskil@mail.montclair.edu>
Sent: Friday, December 25, 2009 6:01 PM
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Subject: [Phys-l] thorium-228

> Not trusting myself, I would like someone to confirm that the answer
> below is correct.
>
> Question: Th-228 (half-life 1.913 years) is dissolved in water, at the
> concentration of 0.03 ppb. What is the activity of that isotope?
> (ignore activity of daughters).
>
> Answer: 3.57*10^-4 Ci.
>
> Thanks in advance,
>
> Best wishes to all,
>
> 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.