# Uncertainty in activity

Hello, one of my friends came over to get some help on a homework problem, and it went past over my head im pretty good with stats, just dont know how to approach this to help him out.

So a radionuclides' activity (Cesium 137) is measured in the year of 1981, (no date just year), and its 60mCi, and its present activity is measured on 02/05/2016.

so i calculated the present act to be 26.84mCi.

however, how do i find totaI uncertainty in the activity.

I was thinking may be it was (InitialA^2 + FinalA^2)^2? However, there does seem to be error in the decay time due to not having a date opposed to just a year.

any help?

BvU
Homework Helper
Hello Smigglet,

If the original activity is given as 60 mCi it will be hard to give an answer in 4 digits. 1% is a reasonable estimate of the error in original activity.
And the uncertainty in time is 6 months.
Apply the error propagation rule $$\left (\Delta f(x,y) \right )^2 = \left ( \partial f\over \partial x\right )^2 \left (\Delta x\right )^2 + \left ( \partial f\over \partial y\right )^2 \left (\Delta y\right )^2$$

And PF greetings to your friend

Thank you for the reply :)

How would i go about applying the error propogation rule this scenario, i normally have detector data, or number of counts and time that use i the propagation rule. Not too sure how i would apply it here, and excuse me for dumb my question, i cannot see how the uncertainty in time is 6 months if anything id think it would be the difference from date to date, assuming the measurement was taken 02/05/1981, so i would think the uncertainty would be 1 month since we dont know if the measurement was on january.

I think im confusing myself here lol

BvU
Homework Helper
Like most years, 1981 runs from January 1 to December 31; so if you use ##t_0## June 1 you're at most 6 months off.
The text as you render it (no date just year) gives me no reason to expect that ##t_0## is May 2, 1981.
It's a conservative estimate for the error in the ##t_0## (a better estimate might be the sigma for a uniform distribution -- 3.5 months)

If someone gives you 60 for the activity, you may assume it's not 61 and not 59, so ##60.0 \pm 0.5## is a reasonable interpretation.

You are confusing yourself if you interpret the 60 as a number of counts (in which case ##\sqrt {60}## would be the estimated standard deviation)

Anyway, you have two contributions of about 1 % so reporting ##\ 27\pm 0.5## mCi or just plain ##27## mCi would be reasonable IMHO.

Like the pirates say: it's more what you'd call "guidelines" than actual rules.

Not too sure how i would apply it here
If ##\ \displaystyle {\ f = A_0\, 2^{t-t_0\over t_{1/2}}} \ ## then you can differentiate and calculate. I just did the differentiation numerically by calculating the power of 2 for ##t_0 = ## 1-1-81, 1-6-81, 1-12-81 and adding 1% (in ##A_0## ) and 1% (in the power of 2) in quadrature: about 1.4% of 27, so 0.5

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