Is this half-life answer kosher?

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Homework Help Overview

The discussion revolves around the calculation of the half-life of uranium (U) based on its decay to radium (Ra) and ultimately to lead. The original poster presents a mathematical approach involving differential equations to derive the half-life of U from the given data about Ra's half-life and its concentration in a sample.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations used to model the decay process, questioning the definitions and relationships between the constants involved. There are attempts to clarify the distinction between lifetime and half-life, as well as the correct formulation of the equations.

Discussion Status

Some participants have provided feedback on the original poster's calculations, pointing out errors and suggesting corrections. There is an ongoing exploration of the mathematical relationships, with no explicit consensus reached on the correct approach yet.

Contextual Notes

Participants note confusion regarding the transcription of equations and the definitions of certain variables. There is acknowledgment of the complexity involved in the calculations and the potential for misinterpretation of the relationships between decay constants and half-lives.

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Question:
U has a very long half life and decays through a series of daughter products ending with a stable isotope of lead. Very old samples of U ore, which have not undergone physical or chemical changes, would be expected to show an equilibrium between the daughter elements provided that their half life was considerably shorter than that of U. Analysis of an ore of U shows that for each 1.00 g of U there is 0.300 μg of Ra . The half life of Ra is 1602 years.
What is the half life of U ?

Proposed solution:

Let x(t) be the number of U, y(t) the number of Ra

\dot{x}=-ax
\dot{y}=ax-by

To find b, let x=0

y=y_0e^{-ln(2)\frac{t}{1602}}
\dot{y}=-by
b=ln(2)\frac{1}{1602}

Now find equilibrium sol'n of the DE:

\frac{dy}{dx}=0
\Rightarrow \frac{by-ax}{ax}=0
\Rightarrow \frac{by}{ax}=1
\Rightarrow a=\frac{by}{x}
Edit after Bruce W pointed out mistake in transcription/logic:

\Rightarrow a=ln(2)\frac{1}{1602}*3*10^{-7}

\dot{x}=-ax
x=x_0 2^{-t/T}
-\frac{ln(2)}{T}x_0 2^{-t/T}=-a x_0 2^{-t/T}
\frac{ln(2)}{T}=a
T=\frac{ln(2)}{a}=(\frac{1}{1602}*3*10^{-7})^{-1}\approx 5.3*10^9
All times in years

Thank you!
 
Last edited:
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Your calculated "a" is the lifetime, not the half-life, as you can see in the equation ##\dot{x}=-ax##.
 
7*10^9*ln(2) \approx 4.85*10^9
Much closer to the real figure. Thank you.
 
Your working confuses me. b is a constant, but you have written
b=ln(2)\frac{t}{1602}
That 't' should not be there. Also, I agree with the following line:
\Rightarrow \frac{by}{ax}=1
But that doesn't rearrange to this:
\Rightarrow a=\frac{x}{by}
Also, 'a' times ln(2) does not give the half-life of U.
 
BruceW said:
Your working confuses me. b is a constant, but you have written
b=ln(2)\frac{t}{1602}
That 't' should not be there. Also, I agree with the following line:
\Rightarrow \frac{by}{ax}=1
But that doesn't rearrange to this:
\Rightarrow a=\frac{x}{by}
Also, 'a' times ln(2) does not give the half-life of U.

Sorry, the t and rearrangement was poorly transposed from paper- the rest was incompetence and rushing.
 
yeah, I've been there man. hehe. I see your edited version. Nice work. I get the same answer.
 

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