1. Mar 14, 2004

### EasyStyle4747

well, this is a strange question that i have to do a presentation on. I thought it would be easy but i cant find any information after looking through 2 major physics books and searching online. I even asked a tutor about this and still couldnt get any equations i could use. Plz dont bash me if this is the wrong forum. Just direct me to the right one then. This is the problem:
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Radioactive Emission Ranges: For a 35 becquerel (Bq) source of alpha radiation, calculate the rate of particles arriving at a 5.0x10^-4m^2 detector located 20.0 cm, 50.0 cm, 100.0cm, and 130 cm from the source. Repeat the calculations for the same detector at the same positions for a 48 Bq source, a 125 Bq source, and a 1753 Bq source.

Tabulate and graph your results on a detection rate (Bq) versus distance (cm) graph, with source activity(Bq) as an extra parameter.
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Just need an equation or equations to solve this. You don't have to solve it. Im not tryin to cheat, i just need to know where to even start with this thing. I cant find any equation that includes distance along with detection rate and area. Both books dont have anything. Am i suppose to use several equations together or wat? If anyone has any idea how this is done, plz help me.

2. Mar 15, 2004

### ZapperZ

Staff Emeritus
You have, for the 35 Bq source, 35 disintegrations per second. You have to assume that these are emitted isotropically, meaning that they are emitted uniformly in all directions over a solid angle of 4pi.

What you need to do is figure out, at a particular radius (detector location), what is the ratio of the area of your detector to the total spherical area. The ratio will tell you what fraction of the particles will enter your detector per second, giving you the detection rate. Do this for all the radius, and repeat for the different decay rate.

Zz.

3. Mar 15, 2004

### EasyStyle4747

hmm, so is this what u mean: (rate of bqs) * A / 4(PIE)r^2

And why is it the spherical area and not the volume?

4. Mar 15, 2004

### ZapperZ

Staff Emeritus
The question is, why should it be the volume? What logical justification calls for that based on the question given?

How much of the particles that enters the detector depends not on any volume, but rather the area of the detector to catch those particles. The closer the detector is to the source, the large the ratio of the area that it is catching.

I suppose if one is a masochist, one can find the volume of the solid cone with the surface of the detector as the base, and then find the ratio of that with respect to the total spherical volume, but why bother with that much work? But in doing that, you have to assume the shape of the detector's surface area, which you are not given..... Again, way too much unnecessary work.

Zz.

5. Mar 15, 2004

### EasyStyle4747

ok, i think i know what u mean. So i assume theres nothing wrong with the equation i have then right?

6. May 2, 2004

### VBPhysics

there is nothing wrong with that equation. The 35 dps have to be uniformly(evenly) distributed over a sphere whose origin is the 35 dps source. This means that you will detect an amount of dps equal to the percentage of the sphere you are monitoring.