# How should I treat background radiation in an experiment?

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 $$\thicksim -2$$.

• 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. – user137289 Mar 3 '20 at 10:12
• @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? – SRSR333 Mar 3 '20 at 13:49
• 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. – 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 $$\sigma_2 = \sqrt{Var[X_1 + X_2] - Var[X_1]} = \sqrt{E[X_1 + X_2] - E[X_1]}$$