# Homework Help: Radioactivity homework

1. Mar 29, 2010

### Magma828

Okay this question is in a section of my text book that has no equations, apart from the slightly mathematical statement:

The rate at which a source emits radioactive particles is called its activity, A. An activity of one particle emitted per second is called a becquerel, Bq.

Here's the question..

A small source of gamma radiation is placed at a distance of 160 mm from a detector of area 18 mm^2. The count recorded on the detector after 30 minutes was 15804.

Estimate the activity of the source.

I worked out the counts detected per second (15804/30/60 = 8.78) but I don't know how to deal with the distance between the source and detector, or the area of the detector.

The final answer is supposed to be 1.57x10^5 Bq

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Okay I've got somewhere[ish] with the solution, but didn't get the correct answer:
The gamma source emits radioactivity in all directions, but is only detected by the 18mm^2 detector.

Calculating the total surface area of the sphere in which the detector could potentially be in is: 4 pi r^2 = 4 x 3.14 x 0.16^2 = 0.32m^2

If this surface area was full of detectors, the counts recorded would be 100% of the counts. The number of potential detectors you could fit into this space would be: total surface area / surface area of detector = 0.32/0.018 = 17.87 detectors.

Therefore, if 15804 counts are detected by one detector, the number of counts detected by 17.87 detectors would be 15804 x 17.87 = 282x10^3 counts in 30 minutes.

I'm sure my solution makes sense, but it's wrong!

Last edited: Mar 29, 2010
2. Mar 29, 2010

### Fightfish

The radiation source emits gamma radiation uniformly (or almost uniformly) in all directions. The detector only managed to "capture" a certain percentage of the emitted gamma photons. So, to obtain the activity of the source, we have to calculate backwards using the fraction of photons emitted that were captured by the detector.

Now, since the radiation source emits gamma radiation uniformly in all directions, the gamma photons are spread out over a spherical surface. It thus follows that the fraction of photons captured by the detector is given by the surface area of the detector divided by the surface area of that spherical surface (the fraction of the spherical surface that is covered by the detector)

3. Mar 29, 2010

### Magma828

Does that mean that my solution is correct then? I think I've done what you said there:

Calculating the total surface area of the sphere in which the detector could potentially be in is: 4 pi r^2 = 4 x 3.14 x 0.16^2 = 0.32m^2

If this surface area was full of detectors, the counts recorded would be 100% of the counts. The number of potential detectors you could fit into this space would be: total surface area / surface area of detector = 0.32/0.018 = 17.87 detectors.

Therefore, if 15804 counts are detected by one detector, the number of counts detected by 17.87 detectors would be 15804 x 17.87 = 282x10^3 counts in 30 minutes.

Maybe I made a stupid mistake somewhere..

OOOOH nevermind, my methodology was right but I should have just used the distances and areas in mm (as it's just ratios) instead of incorrectly converting them to metres.

So yeah I've done it, thanks :D

Last edited: Mar 29, 2010