So I have an exercise that says that the initial rate of $14\mathrm{C}$ was $13.5$ (per second) and nowadays it $10.8$ (per second). The half-life time is $5730$ years with an uncertainty of $30$ years... Then it's asked to calculate the decay constant ($-1.21 \times 10^-4 \,\mathrm{(year)^{-1}}$) and the time that has passed since the carbon was set free ($1844$ years).

Now it's asked to calculate the uncertainty in the calculation of the time.

I'm having trouble deriving the expressions that allow me to calculate the uncertainty... Can someone help me?



For calculate the uncertainty I need do derivate the formula $t = -\frac{A}{k}$

BUT I also need to multiply this by the uncertainty of $k$

So I need to calculate the uncertainty of $k$

How can I do it?


2 Answers 2


This type of problem is called 'propagation of uncertainty', or 'propagation of error'. You know the uncertainty in one quantity which you are using to calculate another, so you just need to find how to 'propagate' the initial uncertainty into the result. There are numerous good tutorials online (like this one, or this-one).

Let's say that x is the half-life, and f(x) is the duration of time (in this case 1844 years). It ends up being a relationship between derivatives, because you want to know how much the change in your initial quantity (dx) creates a change in your final quantity (df).

What is the exact relationship you are using? How can you apply the 'propagation of error' equations to your particular relationship?

  • $\begingroup$ Hi! I'm trying to derivate the expression t= (-ln(R/Ro)/k) (k is the decay constant)... $\endgroup$ Oct 17, 2015 at 16:55
  • $\begingroup$ @GrangerObliviate So the form of your equation is t(k) = -A/k = -Ak^{-1} for some constant value A. You can use the power-rule to find the derivative $\endgroup$ Oct 17, 2015 at 16:59
  • $\begingroup$ That's what I did... So that part of the uncertainty is well calculated... Maybe I got wrong the "error of k" (I need to multiply for that, right?). What I did was derivate the expression (ln2/half-life time)... Maybe I did it wrong $\endgroup$ Oct 17, 2015 at 17:06
  • $\begingroup$ Sorry, I don't understand your question/confusion. Perhaps you could edit your initial question showing exactly what you've done - and what you're stilling having trouble with. $\endgroup$ Oct 17, 2015 at 17:11
  • $\begingroup$ I edited, tell me if it's clearer now :) $\endgroup$ Oct 17, 2015 at 17:14

Your uncertainty can be determined simply by looking at your data. Both of your detection rates are given as XY.Z seconds. Take your first example, 13.5 per second. This could be anywhere in the range of 13.450000...01 per second to 13.550 per second, assuming you use normal rounding standards. So your uncertainty in each measurement is 0.1 seconds.


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