Is this half-life answer kosher?

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The discussion centers on calculating the half-life of uranium (U) based on its decay to radium (Ra) and ultimately lead. The proposed solution involves differential equations to model the decay process, but there are errors in the transcription and logic regarding constants and relationships between variables. After corrections, it is noted that the calculated lifetime is approximately 5.3 billion years, which is closer to the expected value. Participants emphasize the importance of accurately defining constants and relationships in the equations. The conversation highlights the collaborative effort to clarify the calculations and reach a consensus on the correct half-life of U.
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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!
 
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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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