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Re: [Phys-l] thorium-228



On Dec 26, 2009, at 12:29 PM, Brian Whatcott wrote:

ludwik kowalski wrote:
... confirm ...
...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: 3.11*10^-4 Ci. [slightly corrected now]

One curie is that quantity of material of which 3.7E10 atoms
disintegrate/sec

That is true.

given that this benchmark quantity is provided by 1 gm of radium.
Radium mass: 226 (let us say) Half-life 1600 yr
Thorium mass: 228 (let us say) Half-life 1.91 yr
Curie equivalent mass of Thorium: 1 / 842 gm = 1.19E-3 gm

Yes, the ratio of half-lives is close to 842. And yes, the shorter the half-live the highest the activity, for the same number of radioactive atoms (regardless of their mass).


This is not the usual way to calculate the activity; but I see nothing wrong with it, provided ratios of the numbers of atoms are used.


Water mass: 16 + 2 = 18 (let us say)

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.


Reviewing the working of this result, it is clear to me that either the
definition
of the Curie given here is faulty, or Ludwik's definition of the
question as an activity without a mass, is incomplete. Considering that
he has some background in radioactivity, it is more likely I have
misused the definition of activity

The decay rate R (in decays per second) is usually calculated as R=k*N, where k is the decay constant (for a given half-life, T) and N is the number of atoms, for example Th-228. The decay constant is ln(2)/T=0.693/T(s). For Th-228 k happens to be 1.15*10^-8 (s^-1).

1 mole of water (18 grams) contains 6.02*10^23 molecules. This amounts to 3.34*10^22 molecules per gram of water. How many atoms of Th is there in 1 gram of water, if its concentration is 0.03 ppb?

N'=(3.34*10^22)*(0.03*10^-9)=0.10*10^13=1.0*10^12 (atoms of thorium per gram of water)

For one liter, N=1000*N'=1.0*10^15 atoms

Thus the decay rate is R=k*N=(1.15*10^-8) * (1.0*10^15) = 1.15*10^7 decays per second.

The activity in Curies is:

A= R/(3.7*10^10) = (1.15*10^7) / (3.7*10^10 ) = 3.11*10^-4 Ci

P.S. Why did I ask?
In a recently published note (in Physics Letters A) I numerically goofed (orders of magnitude) and I now must ask for the erratum to be printed. I did not want to goof again. That is why I wanted to compare my answer against another calculation.
Someone privately confirmed that my new result is correct, except for a small difference due to rounding.

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.