Context: Experiment investigating the relationship between radiation and distance using a Geiger counter, stopwatch and metre ruler.

After doing some research (page 3) on how to find the uncertainty of count rate, I came across this as the best estimate: N/T ± √N/T - where N is the number of counts and T is the time. (T in this case was 10 seconds measured on a stopwatch).

What would the uncertainty be given that there is also an uncertainty in T due to human reaction time (my exam board tends to go with about ±0.1s I believe), which leads to an uncertainty within an uncertainty? (N/T±0.1 ±√N/T±0.1)

  • $\begingroup$ To be conservative, maximise the uncertainty, that is, take the smallest $T$ consistent with your data. $\endgroup$ Feb 5, 2018 at 17:26
  • $\begingroup$ The only value of T I had was 10s, I'm just wondering how to deal with the extra uncertainty on T within the ±√N/T±0.1 $\endgroup$
    – Oliver
    Feb 5, 2018 at 17:33
  • $\begingroup$ You don't need to have an uncertainty in the uncertainty. $\endgroup$
    – stafusa
    Feb 5, 2018 at 17:34
  • $\begingroup$ Oh right, am I just misunderstanding how to apply the √N rule? $\endgroup$
    – Oliver
    Feb 5, 2018 at 17:39
  • $\begingroup$ I think that the subject you are interested in is called "propagation of errors". Here's a link with some information ( ipl.physics.harvard.edu/wp-uploads/2013/03/… ). You might also try googling "propagation of errors" for more info. $\endgroup$
    – user93237
    Feb 5, 2018 at 18:28

1 Answer 1


How big is $N$? Since you know $T$ to 1 part in 100, if

$$N >> 100^2 = 10000,$$

then $\delta T$ is the dominant source of error. (And of course, if $N << 10000$, then the statistical error dominates.)

In general, if the sources of error are uncorrelated you can add the them in quadrature:

$$ (\delta\omega)^2 = (\frac{\partial\omega}{\partial N}\delta N)^2 + ( \frac{\partial \omega}{\partial T}\delta T)^2 $$

where the rate is:

$$\omega(N, T) = N/T$$


$$ \delta\omega = \sqrt{\frac{\omega} T[1+\frac{\omega} T(\delta T)^2]}$$

  • $\begingroup$ Actually, could you explain going from (δω)^2 to δω please? $\endgroup$
    – Oliver
    Feb 5, 2018 at 19:54
  • 1
    $\begingroup$ @Oliver If the typo in the exponent threw you off, I fixed it. Otherwise, just take the derivatives and express in terms of $\omega$. $\endgroup$
    – JEB
    Feb 5, 2018 at 23:24

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