How much Plutonium 239 will remain after 20,000 years?

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Homework Help Overview

The discussion revolves around the decay of Plutonium 239 (P-239) over a period of 20,000 years, focusing on the calculation of the remaining quantity from an initial sample size of 10 grams, given a decay rate of 0.00284 percent per year.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of the decay formula A=A0e^-rt, with initial attempts yielding results that raise doubts about their validity. Questions arise regarding the interpretation of the decay rate as a percentage and its proper conversion for calculations.

Discussion Status

There is an ongoing exploration of the correct interpretation of the decay rate, with some participants suggesting alternative expressions for the percentage. A revised calculation that yields a more plausible result has been proposed, although explicit consensus on correctness has not been reached.

Contextual Notes

Participants are grappling with the conversion of the decay rate from a percentage to a decimal form, which is critical for accurate calculations. The original poster expresses uncertainty about the initial results, indicating a need for clarification on the mathematical approach.

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Homework Statement



Plutonium 239 (P-239) decays at a rate of 0.00284 percent per year. If the initial sample size
of P-239 is 10 g, how much will remain as P-239 after 20,000 years?

Homework Equations



A=A0e^-rt

The Attempt at a Solution


A=10e^-0.00284*20000
lnA=ln10+lne^-56.4
lnA=1-56.4
A=8.7*10^-25
This answer doesn't seem reasonable and I don't have confidence that I did everything correctly.
 
Last edited:
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I tried the question again.
If A=A0*e^-rt
I want to find A
If I directly substitute and solve, I'll get..
A=10e^(-.00284*20000)
A=2.14*10^-24grams
 
I think the problem is that you are putting 0.00284 in your equation. The 0.00284 is a percentage. So how else could you express that?
 
Its a percentage so .00284 becomes .0000284?
A=10*e^(-.0000284*20000)
A=5.66grams
That sounds more realistic. Is this correct?
 
Seems reasonable to me.
 
Thank you.
 

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