# Beta Particle Detection

1. Feb 13, 2009

### FastLineare

We have a surface that is 10 cm x 10 cm. That surface is uniformly covered with a radioactive material. When radioactive material decays, it has a equal probability of emitting radioactive particles, in this case beta particles, in any direction (360 degrees). We have a detector that, mathematically, you can think of as a plane. It is a circle with a diameter of 3.9 cm. The detector is located 0.47625 cm above the center of the 10 cm x 10 cm surface. If we assume that we have an equal probability that radiation particles may emit in any direction, that all surfaces of the 10 x 10 plate have equal levels of radioactivity, and that the emissions are taking place from all surfaces at constant rates, find the probability that emissions will hit the detector. By probability, what I mean is, what percentage of total emissions from the source will hit the detector.

Any thoughts or recommended algorithms would be appreciated. I've tried a few different approaches but am not confident in the theoretical results I've obtained.

Thanks,

FastLineare

2. Feb 13, 2009

### malawi_glenn

10x10cm^2 / area of sphere

3. Feb 13, 2009

### FastLineare

Percentage = 100_cm^2/(11.9_cm^2) = 8.40 = 840%

On the other hand, if I flip the ratio,

Percentage = 11.9_cm^2/100_cm^2 = 11.9%

I thought of trying this yesterday; however, I figured this was too good to be true. Shouldn't the percentage be dependent of the offset distance as well? I also struggled with the idea that all points on the emission surface are emitting particles (in reality they are not but to obtain a worst case senerio this assumption was made). Therefore, this would cover all of Real space except what is taken in by the detector.

Thanks,

FastLineare

4. Feb 13, 2009

### malawi_glenn

oh, dude, I read 3.9 METERS !! ;-)

Consider this picture in the x-y plane (wait til approved by moderator). find which angles theta and phi the detector spans.

You know spherical coordinates yes? http://mathworld.wolfram.com/SphericalCoordinates.html

Thus integral
$$\int _ {\phi = 0}^{2\pi}d\phi \int _ {\theta = 0}^{\pi}sin \theta d\theta = 4\pi$$

You need to know how big integral will be when integrating over the angles considered in the picture. Really easy, answer is 11.8%.

5. Feb 13, 2009

### FastLineare

Ok...I think I see how to handle this.

What I did was computed the dbl integral above symbolically to get,

$$\int _ {\phi = 0}^{2\pi}d\phi \int _ {\theta = 0}^{\pi}sin \theta d\theta = -(phi-2*pi)*(cos(theta)+1) = 4\pi$$

Now I can describe phi and theta in terms of the offset distance between the source and the detector so that I can optimize the efficiency.

Thanks,

FastLineare

6. Feb 13, 2009

### FastLineare

One question I have for you though...How did you come up with the 11.8%? I'm also curious where f(theta) = sin(theta) comes from in this formula?

7. Feb 13, 2009

### malawi_glenn

don't you know how integration of volumes in spherical coordinates work?

I suggest this introduction:

http://en.wikipedia.org/wiki/Spherical_coordinates

(look for surface element, keep r fixed)

I got 11.8% by integrating phi from 0 to 2times the angle spanned by the detector in the picture. And then the same thing for theta integration. Then I took the result and divided with the full solid angle (4pi).

8. Feb 13, 2009

### FastLineare

I really appreciate the advice! The last time I computed double integrals was in Calc 3 a couple years ago. I now see that using integrals in practice is slightly different than when computing them in a class. I understand the basic concepts, but am rusty on the details.

Thanks alot!