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Re: Time needed for Rn-222 daughters to reach equilibrium withtheirparent?
Franz and Steve,
We usually let radon progeny grow in for 3 hours to reach an equilibrium
factor of 0.99. Theoretically, ten half lives would bring the daughter to
99.9% of the parent activity (1-2^-10=0.999).
In the case of radon progeny, we have to consider that Bi-214 follows Pb-214
and they have similar 1/2 lives, so it takes a bit longer to grow in than
just a single daughter.
We have put a program on the web that, among other things, calculates the
equilibrium factor as a function of time. It is at:
http://members.shaw.ca/eic/Tools/RoomRadonTime.htm
Keep in mind that the calculation is done numerically in 15 second time
increments, so there is an inherent error and equilibrium factors >1 are
reported after long decay times. Hope this helps,
Kai
----- Original Message -----
From: "Franz Schoenhofer" <franz.schoenhofer@CHELLO.AT>
To: "Frey, Steven R." <sfreyohp@SLAC.STANFORD.EDU>;
<radsafe@list.vanderbilt.edu>
Sent: Wednesday, January 14, 2004 2:32 PM
Subject: AW: Time needed for Rn-222 daughters to reach equilibrium with
theirparent?
>
>
> -----Ursprüngliche Nachricht-----
> Von: owner-radsafe@list.vanderbilt.edu
> [mailto:owner-radsafe@list.vanderbilt.edu]Im Auftrag von Frey, Steven R.
> Gesendet: Mittwoch, 14. Jänner 2004 20:23
> An: RADSAFE (radsafe@list.vanderbilt.edu)
> Betreff: Time needed for Rn-222 daughters to reach equilibrium with
> theirparent?
>
>
> Dear Radsafers -
>
> How long does it take Radon-222 daughter products to reach equilibrium
with
> their parent,
> assuming one starts with only Radon-222 gas (no daughters yet having begun
> ingrowth)?
>
> -------------------------------------------
>
> Steve,
>
> Provided that the parent nuclide is longer-lived than the daughters as a
> rule of thumb more than 99% equilibrium will be reached after ten
> half-lifes. I suppose you are looking at the s h o r t -lived daughters
of
> Rn-222. The short-lived daughter with the longest half-life is Pb-214 with
> 26.8 minutes half-life, so the time required will be approximately 270
> minutes. The system has to be closed and undisturbed.
>
> Best regards,
>
> Franz
>
>
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