I am performing an experiment to verify the inverse-square law for the intensity of a beam of $\gamma$ rays emitted by a sample of $^{60}\text{Co}$. The set-up is as follows:

  1. A Geiger-Müller tube connected to an electronic counter is set up.
  2. A background reading is first taken, by performing 10 measurements of $100\text{ s}$ each with no source nearby.
  3. The average count is taken: in my experiment, I have a value of $c_{BG} = 28.5$, with a standard deviation, $\sigma_{BG}=3.837$, represented as $28.5 \pm 3.837$.
  4. Now, the radioactive source is placed at a fixed position, and the GM tube is placed at a distance $d\text{ cm}$ away from the source.
  5. 10 readings lasting $100\text{ s}$ each are taken, and the average count for 100 seconds, $\langle c_d \rangle$ is calculated at this distance $d$.
  6. Steps 4 and 5 are repeated for other values of $d$.

Now, for each distance $d_i$, I have an average count over 100 seconds, $\langle c_i\rangle$, and an associated standard deviation of the 10 readings, $\sigma_i$.

I then have a set of $\left\{c_i \pm \sigma_i\right\}$.

My question is therefore this: can I simply subtract $c_{BG} \pm \sigma_{BG} $ from each $c_i \pm \sigma_i$, and then propagate the errors to find the average count rate, $\left\langle N\left(c_{{d}_{i}}\right)\right\rangle$ at each $d_i$? If not, how else should I go about dealing with the background radiation?

My end goal is to plot the logarithm of the average count rate, $\ln(\left\langle N\left(c_{{d}_{i}}\right)\right\rangle)$, on $\ln d$ and determine the gradient—ideally $\thicksim -2$.

  • $\begingroup$ When you measured the background in a 1000 seconds, the easy way to estimate the error is the square root of the number of counts during that time: 285 ± 17. $\endgroup$ – user137289 Mar 3 '20 at 10:12
  • $\begingroup$ @Pieter, thanks for the comment. Care to share why the square root of the total number of counts is a good estimate of the error? $\endgroup$ – SRSR333 Mar 3 '20 at 13:49
  • $\begingroup$ That is a basic result from the statistics of counting independent events. I am not a statistics person, this is not something that I could give a proof of. $\endgroup$ – user137289 Mar 3 '20 at 15:40

If you think of the problem in terms of random variables, it becomes easy. You have two random variables: The first one $X_1$ for the background noise, and the second one $X_2$ for the radioactive decay. Both follow a Poisson distribution. \begin{align} X_1 &\sim Pois(\lambda_1) \\ X_2 &\sim Pois(\lambda_2) \end{align} For the Poisson distribution we know that the decay rate is equal to the expectation value as well as to the variance, $\lambda = E[X_1] = Var[X_1]$. Thus, the expectation value is a "good" estimator for the rate parameter. Hence, we use $\lambda_1 \approx \bar{N}_{BG}$.

Now, we can show that the sum of two independent Poissonian random variables is again Poissonian, $$ X_1 + X_2 \sim Pois(\lambda_1 + \lambda_2) $$ (Note that here we assume that the decay product is not radioactive) Thus, the estimate of $\lambda_2$ is given $$ E[X_1 + X_2] = E[X_1] + E[X_2] \Rightarrow \lambda_2 = E[X_2] = \ldots$$ For the standard deviation of $\lambda_2$ we use $Var[X_1 + X_2] = Var[X_1] + Var[X_2]$. Solving for $Var[X_2]$ and taking the square root, we obtain $$ \sigma_2 = \sqrt{Var[X_1 + X_2] - Var[X_1]} = \sqrt{E[X_1 + X_2] - E[X_1]} $$

  • 1
    $\begingroup$ Cobalt-60 decays to Nickel-60, which is stable, so your assumption is correct. $\endgroup$ – PM 2Ring Mar 3 '20 at 22:53

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