Swany
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Thorium Expirements
Recently I was offered some thorium nitrate, and since there was no trouble acquiring it (shipping) I accepted it. So, now I have a respectable amount
of thorium nitrate and no definite idea of what I want to do with it. It is old, I would guess old enough to be in whatever radioactive equilibrium it
exists in, I recall reading something about that. At an estimated 3300 Bq/g, it is definitely radioactive...
So... any thoughts on what I should do with it besides be very careful? At first glance making the metal seems to be a pain.
I suppose if anyone were interested in a trade, I may be interested.
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pantone159
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I once thought about trying to isolate the Ra-228 daughter, which is itself a source of Ac-228, which would make a 'purer' form of that elusive
element for my collection. If you measure the radioactivity in a way that does not pick up alphas (usually the case, all you would get is the 228-Ra
-> 228-Ac (5.76 y), 228-Ac -> 228-Th (1.913 y), and a couple of betas in a more complicated branched sequence in the last day of its life.
My idea was to precipitate the Ra as the sulfide, using a Ba carrier, then save that in a separate vial. You would catch some Ra-224 as well, but
this would all decay to stable Pb-208 in a few weeks, after that, I think that almost all the beta activity would be 228-Ra and 228-Ac, with the Ac
probably giving a stronger signal on a geiger counter, as it has a shorter half life (6.15 h vs. 5.76 y), and the beta particle is probably similarly
more energetic.
It would be nice to have a gamma ray spectrometer available, to monitor where the various elements really went to, as a check that this worked.
Edit: The key decay involved in whether your sample is in radioactive equilibrium is the 5.76 y decay of Ra-228. (Besides that, the chain has a 1.913
y 228-Th -> 224-Ra decay, and all the others are less than a week.) Which means, if your sample is several times 5.76 y old (say, 20-30 years), it
will be in equilibrium. That is what you want if you want to isolate the daughters.
[Edited on 16-8-2008 by pantone159]
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JohnWW
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Th-232, which you presumably bought as recently-prepared Th(NO3)4, has an half-life of the order of 10^11 years, many times the lifetime of the solar
system and indeed of the universe. So, how long do you expect to wait in order for an isolatable amount of Ra-228 (half-life only 1,620 years, meaning
that an equilibrium low concentration of it will occur) to accumulate by alpha-emission? You will be VERY old by then!
[Edited on 17-8-08 by JohnWW]
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pantone159
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Ra-228 only has a half life of 5.76 y as I stated. (Ra-226 is the one with the c. 1600 y half life.)
To get to 'radioactive equilibrium', you only need, IIRC, several times the half life of the longest lived daughter, the half life of the original
parent does not matter.
This isn't a true equilibrium - the amount of Th-232 is slowly decreasing, the only true state of equilibrium is when it has all turned to lead.
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pantone159
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Radioactive equilibrium calculations
Consider a simplified system, where radioactive species A decays into B, with a very long half life, and B decays into C, with a much shorter
half-life, and C is stable.
The equations describing the amount of A,B,C are:
dA/dt = -kA * A(t)
dB/dt = +kA * A(t) - kB * B(t)
dC/dt = +kB * B(t)
where kA, kB are the decay constants. (kB >> kA), and the initial conditions are A(0)=A0, B(0)=C(0)=0.
The only *true* equilibrium is when dA/dt=dB/dt=dC/dt all equal 0, but this only happens after everything has gone to C. When one speaks of
'radioactive equilibrium', you mean dB/dt = 0 only.
Now lets solve these equations...
A(t) is easy, A(t) = A0 * exp(-kA * t).
So now we have:
dB/dt = kA * (A0 * exp (-kA*t)) - kB * B(t)
Since A decays much more slowly than B, A(t) is effectively a constant, so this equation can be simplified as:
dB/dt = kA * A0 - kB * B(t)
To solve this, change variables, choosing Beta(t) = B - (kA/kB)*A0. Substitutute this in, to replace B with Beta, to get:
dBeta/dt = kA * A0 - kB * (Beta(t) + (kA/kB)*A0)
after cancelling terms, you get:
dBeta/dt = -kB * Beta(t)
This is trivially solved to get:
Beta(t) = Beta(0) * exp (-kB * t)
Rewriting this in terms of B again instead of Beta,
B(t) - (kA/kB)*A0 = (B(0) - (kA/kB)*A0) * exp (-kB * t)
and B(0) = 0, and with some rearrangement:
B(t) = (kA/kB)*A0 * (1 - exp (-kB*t))
The resulting amount of B grows towards the equilibrium amount of B_eq = (kA/kB)*A0, and this happens with a time constant of kB, so the decay rate of
B, not A, is what affects this.
I don't feel like working this out for a general decay sequence right now, but I think the basic result is the same.
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Trifluoroacetic
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Have you thought about separating the isotope based off of they're charge to mass ratio by vaporizing the thorium sample and then accelerating a beam
of ions through a magnetic feild?
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