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Nuclear Physics - Uranium decay chain and Bateman equation

  • Thread starter Phruizler
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  • #1
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Homework Statement



Calculate the activity of ##^{222}Rn## in an ore sample containing 5g of natural uranium.

Homework Equations



##^{238}U## decay chain (to Radon): ##^{238}U\rightarrow^{234}Th\rightarrow^{234}Pa\rightarrow^{234}U\rightarrow^{230}Th\rightarrow^{226}Ra\rightarrow^{222}Ra##

Bateman equation (for activity of a daughter isotope after a series of decays):

##A_n = N_o\sum\limits_{i=1}^{n} c_ie^{-\lambda_it}## where ##c_i=\frac{\lambda_1\lambda_2 ... \lambda_n}{(\lambda_1 - \lambda_i)(\lambda_2 - \lambda_i) ... (\lambda_n - \lambda_i)} \qquad \left(i \neq n\right)##

The Attempt at a Solution



Well, I used the known decay constants for each isotope in the decay chain and plugged them into the Bateman equation. Unfortunately, ##c_i## turns out to be negative, giving me a negative number of atoms, which is obviously incorrect. The portion where I simply convert 5g of uranium into total number of atoms is fine. Anyone know where I'm going wrong? I'm assuming ##t=0## since time is seemingly irrelevant to the problem as stated.

EDIT: I guess I should add that I am getting my decay constants from Wolfram Alpha, but I don't see why it would be giving me anything but the correct values, as they match with other available literature.
 
Last edited:

Answers and Replies

  • #2
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If you start with pure uranium at t=0 then your Radon decays at t=0 will be zero.
U-238 should be by far the most long-living isotope in your sample. If the sample is old enough (no time is given, so I guess you have to assume that), all other isotopes should be very close to their equilibrium concentrations.
 

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