How does the inverse-square law work?

Question

I'm trying to calculated the intensity of a source that I have done some experiments on using inverse-square law.

$\begin{equation} I(r)=\frac{A \cdot E}{4\cdot \pi \cdot r^2} \end{equation}$

But I'm a bit confused. If A is the activity of the source, it would be 370 kBq (Cs-137), and E would for Cs-137 is 0,662 MeV.

But what I don't understand is, if these to values are constants, how will the intensity change form experiment to experiment, since the value will always be the same(depending on radius r)

I have measured the number of gamma pulses hitting the GM-tube in 60 seconds, and I know that I can use it here:

$\begin{equation}A=\frac{n}{t}=\frac{7610,2}{60}=126,83 Bq\end{equation}$

But this number is way smaller than the constant 370.000 Bq, which the manual says the source is.

Can someone please explain what I'm doing wrong.

Answer 1

The estimation of activity is pretty straightforward and not really at the same time.. The correct equation to use is this: $$\dot{N}=A\cdot n_\gamma\cdot \frac{\varepsilon}{4\pi r^2}\cdot \exp(-\mu_\text{air}r) $$ It gives the count rate in the detector $\dot{N}$ (number of counts per second), from the activity of the source $A$, branching ratio of the radionuclide $n_\gamma$, efficiency of the detector $\varepsilon$, and attenuation coefficient for gamma particles in air $\mu_\text{air}$ for a given distance between the source and the detector $r$.

You have missed the efficiency of the detector mainly, and this is the tricky part. It describes how efficient the detector is at detecting the particles in a given geometry. Basically:

$$\text{efficiency of detector}=\frac{\text{Numb. of photons emitted}}{\text{Numb. of photons detected}}$$

Obtaining the efficiency of the detector is called "calibrating the efficiency of the detector" and only after you have performed this procedure, you can estimate the activity of the source. You can do this up to varying degrees of accuracy with a source of documented activity, depending on the accuracy of the documented activity of course, providing you have such source. Do not forget to take into account the decay during the time between the measurement and the documented date.

The efficiency and air attenuation are energy dependent, so they will be different for radionuclides emitting different energy gamma particles. Air attenuation coefficients for gamma particles can be looked up in NIST tables online.

Air attenuation can basically be neglected if dealing with very short distances $r<1\text{m}$.

For any further information, this topic is well covered in any radiation physics book (e.g. Radiation Detection by Knoll).

Answer 2

Gamma rays are notoriously difficult to block. As a rule of thumb you need 1.3-foot-thick lead or 6.6-foot-thick concrete to catch 99.9999999% of gamma rays. The detecting part of a G–M tube is made out of air which is worse at blocking (and therefore detecting) gamma rays than lead and concrete.

I suspect your G-M tube isn't detecting all of the Gamma rays emitted by your Caesium-137 sample. If that's true then that your G-M tube is only catching some of them and the rest are escaping into your lab (or getting absorbed by the non-detecting metal tube part of your G-M tube).

There are two other possibilities. You can test both of them by measuring your source's activity with someone else's G–M tube.

• The half-life of Caesium-137 is 30.2 years. If your radiation source has been hanging out in your lab since the Cold War that could explain why the activity is lower than it's supposed to be.

• Geiger–Müller tubes have limited lifetimes. The lifetime depends on what kind of G–M tube you're using and how much radiation you expose it to.