# Homework Help: Is this half-life answer kosher?

1. Apr 4, 2013

### henpen

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: Apr 4, 2013
2. Apr 4, 2013

### Staff: Mentor

Your calculated "a" is the lifetime, not the half-life, as you can see in the equation $\dot{x}=-ax$.

3. Apr 4, 2013

### henpen

$$7*10^9*ln(2) \approx 4.85*10^9$$
Much closer to the real figure. Thank you.

4. Apr 4, 2013

### BruceW

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.

5. Apr 4, 2013

### henpen

Sorry, the t and rearrangement was poorly transposed from paper- the rest was incompetence and rushing.

6. Apr 4, 2013

### BruceW

yeah, I've been there man. hehe. I see your edited version. Nice work. I get the same answer.