# Determine the amount of americium-241 remaining after 15 years.

1. Jun 27, 2013

### dannie

1. The problem statement, all variables and given/known data
The half life of americium-241 is 457.699 years. A typical smoke detector contains 33.1 kBq of americium-241. Determine the amount of americium-242 remaining after 15 years.

2. Relevant equations
A=Aoe - - 0.693t/T
1/2

3. The attempt at a solution
=33.1xe - (0.693x15/457.699)
A=32.1kBq

2. Jun 27, 2013

### rude man

About right, although you should use more significant digits than ln(0.5) = 0.693. I got 32.4 for my answer.

3. Jun 27, 2013

### HallsofIvy

Staff Emeritus
Rather than use that formula, you could simply use the fact that "The half life of americium-241 is 457.699 years" means that Americium reduces by 1/2 every 457.699 years. 15 years is 15/457.699= 0.033 "half lives" so the sample is reduced by $$(1/2)^{0.033}= 0.98$$ times: 0.98(33.1)= 32 mg.

I am using two significant figures for my answer because "15 years", given to 2 significant figures, is the least accurate figure.

4. Jun 27, 2013

### rude man

"15 years" could have meant 15.000 years,; it probably did.

Not extending log(0.5) to more than 3 sig. digits means an error of 0.4/32.357 = 1.24%.

5. Jun 27, 2013

### Staff: Mentor

You cannot compare the digits like that.
A more extreme example: if the initial radiation is 33.10000 kBq, what is the radiation after 1 second? Sure, the time has just one digit, but the activity will be 33.10000 kBq after 0 seconds and 2 seconds, too. 33kBq would be a wrong answer.

6. Jun 27, 2013

### rude man

Yes indeed.
Assume it was for 30 years instead of 15. The answer to 2 sig. digits would still have been 32 kBq!

Good point, enforcing precision on the 15 yrs was pointless; but log(0.5) should have been extended to more than 3 sig. digits. Agreed?

7. Jun 27, 2013

### haruspex

Good point - this is exponentiation, not multiplication or division. A 6 month error in the 15 years (though I agree with rude man that the question probably does not intend you to assume that) would be ±3.3%, but the resulting error in 2-n is only 0.08%.
OTOH, if the question had asked how much had decayed, there would be a subtraction involved, and the error would now be back to ±3.3% (not a coincidence).

8. Jun 27, 2013

### rude man

A 6 month error in the 15 years would have given an error of 0.077%:

let τ = -457.699/ln(0.5) = 660.32

A = 33.1 exp(-t/τ)
A = 33.1 exp(-15/660.32) = 32.357
A' = 33.1 exp(-15.5/660.32) = 32.332
(A - A')/A = 0.077%.

9. Jun 27, 2013

### haruspex

Wasn't 0.08% near enough?

10. Jun 27, 2013

### rude man

You said 3.3%, maybe I misinterpreted?

11. Jun 27, 2013

### haruspex

I wrote that ±6 months on 15 years was a 3.3% error, producing a 0.08% error in the answer.

12. Jun 28, 2013

### Staff: Mentor

I would use more digits (or 2^...), but that is a small correction. The difference between your answers is just a calculation error in the first post. I get 32.35671 with the exact value and 32.35655 with the ln(2)-approximation.
The relative error on ln(2) can be compared to the relative error of those 15 years - the answer is not very sensitive to it.

13. Jun 28, 2013

### rude man

You're right, the OP's first (& only so far) post did not do the arithmetic correctly & I hadn't checked it.

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