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Topic: Uranium-238 to Lead-206 decay  (Read 552 times)

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Offline uzumi99

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Uranium-238 to Lead-206 decay
« on: February 21, 2022, 11:24:25 AM »

Sorry if this in the wrong forum. I needed a little bit of help with an assignment.

The question is asking me determine the length of a half-life (t 1/2) for the U-238 to Pb-206 parent-daughter pair. Assuming I have 1,000 atoms of U-238 and 0 atoms of Pb-206 at t0, how many atoms will I have after one, two, three, and four half-lives?

I also feel like I might be vastly over complicating this in my head.

Now, my assumption is that after one half-life (4.468 billion-ish years), I'll have 500 U atoms and 500 Pb atoms. The second half-life is another 4.468 billion years later (8.936 total) leaving us with 250 U and 750 Pb, third half-life (13.4 by) will equal 125 U and 875 Pb, and the fourth half life (17.872 billion) will leave us with 62.5 (63?) Uranium atoms and 937 Lead atoms.

Is that correct or am I mistaken?

Thanks in advance

Offline Babcock_Hall

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Re: Uranium-238 to Lead-206 decay
« Reply #1 on: February 21, 2022, 08:16:05 PM »
I did not see any problems with what you wrote.

Offline sjb

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Re: Uranium-238 to Lead-206 decay
« Reply #2 on: February 22, 2022, 01:46:53 AM »
What about the byproducts? (This may not be where you're heading).

Offline Enthalpy

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Re: Uranium-238 to Lead-206 decay
« Reply #3 on: February 23, 2022, 02:42:16 PM »
The possible hairsplitting I see:

Half-life is defined properly only for one decay, not for a chain. But here, one decay is much slower than all others. Also, 206Pb is the only final product. So, besides naming details, I'd agree with the proposed figures.

Decay is random. With moles, the figures are well defined. With only 1000 atoms, they are not, due to the very nature of radioactivity. Hesitating between 62.5 and 63 results from that. The standard deviation on 62.5 should be sqrt(62.5) or nearly 8, and 3 s.d. are not so uncommon, so figures between 38 and 86 can be reasonably expected. A more precise prediction like "63" would probably fail.

Here I computed as if distributions were Gaussian, but even this is wrong with such small figures.

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