Half life - Calculate the length of time

[Faiq]

Homework Statement

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%

Homework Equations

A=A_oe^{-\omega t}
t_{1/2} = \frac{ln2}{\omega}

The Attempt at a Solution

I can't find a possible correlation between the uncertainty and activity.

 

[mfb]

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)?

 

[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.

 

[mfb]

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.

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

 

[haruspex]

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

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.

 

[TJGilb]

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.

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.

 

[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

 

[haruspex]

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

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).

 

[Faiq]

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).

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?

 

[mfb]

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?

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.

 

[Faiq]

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

 

[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?

 

[mfb]

Oh so wait a minute are they asking length of time after which there would be a decrease of 10% from the original value?

Not a 10% decrease the original activity (because that is not known exactly), but a decrease to 10% below the measurement value.

Due to which we will subtract 2% since there is already a deviation of 2% present?

It is not exactly 2%, but to a good approximation: Yes.

 

[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?

 

[mfb]

All I have to do is use x-2/100 = e^-xt

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

 

[Faiq]

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

 

[mfb]

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

 

[Faiq]

Okay thank you .
Can you answer one more question about half life

 

[mfb]

Which question?

 

[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?

 

[mfb]

It depends on the source. Questions unrelated to this homework problem should go in a new thread.

 

[Faiq]

The half life can be determined by
A=ln2/t * 6.03*10^23
Where A is given in the question

Should I start a new thread?

 

[mfb]

Yes please start a new thread for a different homework problem.

 

[John Park]

a constant rate 3.5x10^6

Reference https://www.physicsforums.com/threads/half-life-calculate-the-length-of-time.908200/

What are the units? Are you told anything about the half-life of the source?

 

[Faiq]

Units are s^-1