- #1
henpen
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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
[tex]\dot{x}=-ax[/tex]
[tex]\dot{y}=ax-by[/tex]
To find b, let x=0
[tex]y=y_0e^{-ln(2)\frac{t}{1602}}[/tex]
[tex]\dot{y}=-by[/tex]
[tex]b=ln(2)\frac{1}{1602}[/tex]
Now find equilibrium sol'n of the DE:
[tex]\frac{dy}{dx}=0[/tex]
[tex]\Rightarrow \frac{by-ax}{ax}=0[/tex]
[tex]\Rightarrow \frac{by}{ax}=1[/tex]
[tex]\Rightarrow a=\frac{by}{x}[/tex]
Edit after Bruce W pointed out mistake in transcription/logic:
[tex]\Rightarrow a=ln(2)\frac{1}{1602}*3*10^{-7}[/tex]
[tex]\dot{x}=-ax[/tex]
[tex]x=x_0 2^{-t/T}[/tex]
[tex]-\frac{ln(2)}{T}x_0 2^{-t/T}=-a x_0 2^{-t/T}[/tex]
[tex]\frac{ln(2)}{T}=a[/tex]
[tex]T=\frac{ln(2)}{a}=(\frac{1}{1602}*3*10^{-7})^{-1}\approx 5.3*10^9[/tex]
All times in years
Thank you!
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
[tex]\dot{x}=-ax[/tex]
[tex]\dot{y}=ax-by[/tex]
To find b, let x=0
[tex]y=y_0e^{-ln(2)\frac{t}{1602}}[/tex]
[tex]\dot{y}=-by[/tex]
[tex]b=ln(2)\frac{1}{1602}[/tex]
Now find equilibrium sol'n of the DE:
[tex]\frac{dy}{dx}=0[/tex]
[tex]\Rightarrow \frac{by-ax}{ax}=0[/tex]
[tex]\Rightarrow \frac{by}{ax}=1[/tex]
[tex]\Rightarrow a=\frac{by}{x}[/tex]
Edit after Bruce W pointed out mistake in transcription/logic:
[tex]\Rightarrow a=ln(2)\frac{1}{1602}*3*10^{-7}[/tex]
[tex]\dot{x}=-ax[/tex]
[tex]x=x_0 2^{-t/T}[/tex]
[tex]-\frac{ln(2)}{T}x_0 2^{-t/T}=-a x_0 2^{-t/T}[/tex]
[tex]\frac{ln(2)}{T}=a[/tex]
[tex]T=\frac{ln(2)}{a}=(\frac{1}{1602}*3*10^{-7})^{-1}\approx 5.3*10^9[/tex]
All times in years
Thank you!
Last edited: