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Emission of Gamma and Beta rays : Experimental Problem 
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#1
Jun809, 02:55 PM

P: 17

Hi there!
The experiment: I'm counting Gamma and beta rays emitted from gamma and beta cylindrical sources, for counting I'm using a simple GM counter, which has nearly the same cylindrical shape (i mean diameter). As we all know this are electromagnetic emissions, so they distance must obbey an inverse square law. The problem: When I make the analysis of the data i don't get a 2 in the power of the distance, but a 1.48 exactly all the times. I'm guessing the problem is that the inverse square law applies for puntual emission, not for a "cylindrical" emission, as the one i'm using. ¿How can I solve the problem? I'm trying to get some geometrical factor that might let me get the inverse square law, but i'm not sure. If anyone knows, i'd love to listen. P.D.: Greetings from Colombia 


#2
Jun809, 06:38 PM

Sci Advisor
P: 6,068

You need to more specific about the geometry involved. What is the shape of the source? How far apart are the source and detector? The inverse square law will hold if the separation distance is large compared to the size of the source.
Further question: the attenuation (particularly of beta) due to the atmosphere, assuming the space between source and dectector is simply air. 


#3
Jun809, 06:46 PM

P: 17

For the source imagine a coin, only one side of the "coin" radiates. The detector has an opening of aproximately the same radius as the source. The distance varies from 1cm to 15cm more or less. I think this distances can't be taken as large compared to the source, since the diameter of it it's around 4 cm. 


#4
Jun809, 08:31 PM

P: 986

Emission of Gamma and Beta rays : Experimental Problem
You don't get inverse square law exactly, because your source and your detector are not pointlike and their dimensions are comparable with the distance between them. At close distances, you get fewer hits than predicted by inversesquare law.
If you plot log of hit count against log of distance, with perfect pointlike source and detector you'd get a straight line with a constant 2 slope (n = c*x^2, therefore ln n = ln c  2 ln x). With reallife equipment you get a curved line whose slope approaches 2 for high distances. For an isotropic, homogeneous flat disk source and a perfectly efficient flat disk detector, coaxial and parallel to each other, the dependence should look like this [tex]N \propto \int_0^r \int_0^r \int_0^{2 pi} \int_0^{2 pi} r_1 r_2 dr_1 dr_2 d\phi_1 d\phi_2 h / (h^2 + r_1^2 + r_2^2  2 r_1 r_2 cos (\phi_1  \phi_2))^{1.5}[/tex] where h is the distance between disks. Does not look like it evaluates analytically (not to me anyways  you can eliminate one of [tex]\phi[/tex] and that's about it), but numerical calculation confirms your slope ~1.5 when h ~ r. 


#5
Jun809, 09:30 PM

P: 17

Thank you, great answer. I haven't calculate the integral, but i believe you. Thanks again.



#6
Jun1109, 05:38 AM

P: 121

Hamster, could you elaborate a bit on how you got to that integral? The denominator looks a bit like the cosine law.



#7
Jun1109, 05:08 PM

P: 986

[tex]r_2 dr_2 d\phi_2 cos(\theta) / l^2[/tex] where l is the distance between (r1, phi1) on the emitter and (r2, phi2) on the detector, and [tex]\theta[/tex] is the angle between l and a normal to the surface. [tex]cos(\theta) = h/l[/tex] [tex]l^2 = h^2 + (r_1^2 + r_2^2  2 r_1 r_2 cos(\phi_1  \phi_2))[/tex] 


#8
Jun1209, 05:36 AM

P: 121

Thanks, guy. Nice piece of analytical work.



#9
Jun1309, 08:10 AM

P: 17

Yeah, that was great. Just one thing, I think it's not N what is proportional to the integral, but log(N). Please correct me if I'm wrong.



#10
Jun1509, 04:49 AM

P: 121

What makes you think that?



#11
Jun1509, 08:04 AM

P: 17




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