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

Here's the question..

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!

*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 theEstimate 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

----

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!

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