Is this half-life answer kosher?

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In summary, U has a half-life of approximately 5.3*10^9 years, based on the equilibrium between its daughter elements and the amount of Ra present in an ore sample. This is calculated using the equation \dot{x}=-ax, where x represents the number of U atoms and a is equal to ln(2) divided by the lifetime of Ra, which is 1602 years.
  • #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!
 
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  • #2
Your calculated "a" is the lifetime, not the half-life, as you can see in the equation ##\dot{x}=-ax##.
 
  • #3
[tex]7*10^9*ln(2) \approx 4.85*10^9[/tex]
Much closer to the real figure. Thank you.
 
  • #4
Your working confuses me. b is a constant, but you have written
[tex]b=ln(2)\frac{t}{1602}[/tex]
That 't' should not be there. Also, I agree with the following line:
[tex]\Rightarrow \frac{by}{ax}=1[/tex]
But that doesn't rearrange to this:
[tex]\Rightarrow a=\frac{x}{by}[/tex]
Also, 'a' times ln(2) does not give the half-life of U.
 
  • #5
BruceW said:
Your working confuses me. b is a constant, but you have written
[tex]b=ln(2)\frac{t}{1602}[/tex]
That 't' should not be there. Also, I agree with the following line:
[tex]\Rightarrow \frac{by}{ax}=1[/tex]
But that doesn't rearrange to this:
[tex]\Rightarrow a=\frac{x}{by}[/tex]
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.
 
  • #6
yeah, I've been there man. hehe. I see your edited version. Nice work. I get the same answer.
 

Related to Is this half-life answer kosher?

1. What is half-life?

Half-life is the amount of time it takes for half of a given substance to decay or break down into a different substance.

2. Is half-life answer kosher?

This question is not specific enough to provide a definitive answer. "Kosher" typically refers to food that is prepared according to Jewish dietary laws. If this question is referring to a scientific calculation or answer, it would be more appropriate to ask for clarification on the specific context.

3. How is half-life used in science?

Half-life is used to measure the rate of decay of radioactive substances, to determine the age of archaeological artifacts, and in various other scientific fields such as pharmacology and environmental science.

4. What factors affect the half-life of a substance?

The half-life of a substance can be affected by environmental factors, such as temperature and pressure, as well as the characteristics of the substance itself, such as its chemical composition and stability.

5. How is half-life calculated?

The calculation for half-life depends on the type of decay or process being measured. Generally, it involves using the amount of the substance at a given time and the rate of decay to determine the amount remaining after a certain amount of time has passed.

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