Re: [Phys-l] thorium-228

[Brian Whatcott <betwys1@sbcglobal.net>]

Ah, good!
reviewing my earlier effort (extracted below)...

=========================================

[bw]
Let us suppose the specified dilution represents the proportions
by mass.   For unit mass of water, the proportion of Thorium
would be 3E-11       The rate at which unit mass of Thorium decays
is 1600/1.91 as fast as Radium    so Thorium is 842 times as active
as that same mass of Radium
The suggested 3.57E-4 Ci of Thorium activity is provided by
3.57E-4 / 842  gm  X 228/226 Thorium = 4.28E-7 gm

At a dilution of  1 : 3E-11, this would involve a mass of  14.3 kg of 
mixture.

=========================================
       ...I see I calculated that  3.57E-4 Ci of activity in
the specified *mass* concentration would need 14.3 kg of the solution.
This amounts to 3.57E-4 Ci/14.3  per kg, or roughly
2.49E-5 Ci per liter...so I *still* didn't get it quite right
compared with your 2.46E-5 Ci/L.

Thanks

Brian

Michael Edmiston wrote:
> /snip/
>
> 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
>
> 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>
>
> [lk]
> 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.