# Half life - Calculate the length of time

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1. Mar 19, 2017

### Faiq

1. The problem statement, all variables and given/known data
A source having a half life of 5.27 years is calibrated and found to have an activity of 3.5*10^5 Bq.The uncertainty in the calibration is +- 2%

Calculate the length of time in days after the calibration has been made for the stated activity to have a maximum possible error of 10%

2. Relevant equations
$$A=A_oe^{-\omega t}$$
$$t_{1/2} = \frac{ln2}{\omega}$$
3. The attempt at a solution
I can't find a possible correlation between the uncertainty and activity.

2. Mar 19, 2017

### Staff: Mentor

Currently the actual activity can be up to 2% above or 2% below the given value.

Can the actual activity get 10% higher than the given value?
Can the actual activity get 10% lower than the given value?
If yes, when does that happen the earliest (what is the worst case)?

3. Mar 19, 2017

### TJGilb

One way to approach this is to plot the activity over time using the measured $3.5*10^5Bq$ value and then also plot the activity over time using plus and minus 2% of this value. Track how these curves diverge from each other over time.

4. Mar 19, 2017

### Staff: Mentor

I'm quite sure the 2% uncertainty refer to the activity, not the half-life. The curves won't diverge.

5. Mar 19, 2017

### haruspex

I don't think @TJGilb thought it referred to the half-life. Rather, it looks like TJ thought the 10% referred to the discrepancy between the actual acitivity at a later time and the predicted activity at that time.

6. Mar 19, 2017

### TJGilb

Yeah, I think I misinterpreted what the problem was asking him to find. It looks like it may be asking him to find when the activity will be potentially 10% less than the original calibration. Of course, you still want to use the lower curve and the higher curve for the worst case scenario.

7. Mar 20, 2017

### Faiq

The main problem I am having is I can't see why the activity and the uncertainty are related by the equation A = U + 2 rather than A = 2U

8. Mar 20, 2017

### haruspex

It's neither.
If the measured activity is A and the uncertainty is ±2% then the actual activity is anything from A(1-.02) to A(1+.02).

9. Mar 20, 2017

### Faiq

I know that
What I am asking is, it is given in the answer that the source has to decay a further 8% to reach the uncertainty of 10%. My question is how is that relation derived?

10. Mar 20, 2017

### Staff: Mentor

The source could have started at 98% the quoted activity, it will exceed the 10% deviation once it reduces its activity to 90% of the quoted activity, which is a $\frac{0.08}{0.98} \approx 0.08$ relative reduction.

11. Mar 20, 2017

### Faiq

So for every percent decay, there is a 2% error?

12. Mar 20, 2017

### Faiq

Oh so wait a minute are they asking length of time after which there would be a decrease of 10% from the original value?
Due to which we will subtract 2% since there is already a deviation of 2% present?

13. Mar 20, 2017

### Staff: Mentor

Not a 10% decrease the original activity (because that is not known exactly), but a decrease to 10% below the measurement value.
It is not exactly 2%, but to a good approximation: Yes.

14. Mar 20, 2017

### Faiq

Yes thank you very much.
So if they want to ask "Calculate the length of time in days after the calibration has been made for the stated activity to have a maximum possible error of x%"
All I have to do is use x-2/100 = e^-xt
Right?

15. Mar 20, 2017

### Staff: Mentor

What is x? No matter what it is it should not appear both in an exponential and outside.

16. Mar 20, 2017

### Faiq

Oh sorry the first x was an arbitrary number, the second is decay constant

17. Mar 20, 2017

### Staff: Mentor

With a lot of interpretation and a new variable name, that might lead to a possible solution.

18. Mar 20, 2017

### Faiq

Okay thank you .

19. Mar 20, 2017

### Staff: Mentor

Which question?

20. Mar 20, 2017

### Faiq

A radioactive source emits alpha particles at a constant rate 3.5x10^6. The particles are collected for a period of 40 days.
BY reference to the half life of the source, suggest why it may be assumed that rate of emission of alpha particles remain constant?