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Homework Help: Determining an error due to non point like detector

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Recently I have conducted an experiment using a gamma source (22Na) and a scintillation detector (NaI), at the start of the experiment one of the desired outcomes was to see that the intensity of the source fell off as 1/r^2.

    Seeing as the detector is not point like (I am assuming that the source is) , how would I go about correcting/calculating an error from this?

    2. Relevant equations
    [tex]\Omega = 2\pi(1- \frac{d}{\sqrt{d^2+a^2}})[/tex]

    Where d is the distance from the source and a is the radius of the detector.
    3. The attempt at a solution

    I have worked out the solid angle for each distance from the source: equation given above. How do I turn this into a correction?
    Last edited: Feb 6, 2016
  2. jcsd
  3. Feb 6, 2016 #2
    Your relevant equation is not dimensionally consistent. Is this your solid angle?
  4. Feb 6, 2016 #3
    Sorry, some bad latex skills on my behalf there. Should be ok now.
  5. Feb 6, 2016 #4
    looks good. So if you want to show an inverse square law dependence then there needs to be and explicit 1/d2 in your equation for the count rate. How can you do that?
  6. Feb 6, 2016 #5
    I know that at large distances from the source the a term becomes negligible and the detector becomes point like, hence giving the 1/r^2 dependency; however I'm just not sure how to correct for the finite size of the detector at ranges where I can't neglect the size of the detector.
  7. Feb 6, 2016 #6
    You know in the limit of large d then Ω → K/d2 where K becomes constant and is related to the size of the detector. Now if you factor out 1/d2 from your expression for Ω so you get Ω = K(a,d)/d2. How do you interpret K(a,d)?
  8. Feb 7, 2016 #7
    According to Knoll, when d>>a it tends to Ω → K/d2 as you have said, where [tex]K=\pi a^2[/tex] In the original equation if I factor out 1/d2, I get [tex]\Omega = \frac{1}{d^2}(2\pi d^2-\frac{2\pi d^3}{\sqrt{d^2+a^2}})[/tex] In this case K(a,d) is the term inside the bracket, correct?
    Last edited: Feb 7, 2016
  9. Feb 7, 2016 #8
    Not quite. First note your result has dimensions of m-4. Ω should be dimensionless.

    Check your answer by putting 1/2 factor back into the bracket, you should get your original Ω
  10. Feb 7, 2016 #9
    Looking at my math, you'd wonder why I am even asking this question! Sorry, it's been a long few days! I think that's correct now?
  11. Feb 7, 2016 #10
    That alright I sometimes have similar problems. OK look good.

    This correction however is only necessary for d ≤ 5a for d greater than that the error is ≤ 2%.. I don't know your set up for the experiment so depending on the source crystal distance and the size of the crystal the intrinsic efficiency of the crystal can vary significantly. e.g. for a 1 MeV gamma ray and a 1½ x 1 inch crystal the intrinsic efficiency various from about 32% at 10cm to about 38% at 50 cm. The ISL varies only about 2% over this range. So you may see a deviation from the ISL depending on you equipment, the setting, total counts.
  12. Feb 7, 2016 #11
    I have two cylindrical sodium iodide detectors with a radius of 1.4cm, the experiment was set up so as only gamma radiation from positron annihiliation was recorded and the distances between the two detectors vary from 1-30cm. Hence the source-detector distance was 0.5-15cm.

    I'm not sure how I include this correction in my results however? What I have done so far is plotted the log of the countrate vs the log of the separation of detector and source. In an ideal scenario this would give a slope of -2, however mine is less than this. I know that the size of the detector contributes but I can't see how using the above equation helps.
  13. Feb 7, 2016 #12
    You've taken your count rate and divided each reading by the respective K(a,d)?
  14. Feb 7, 2016 #13
    Ahhh, no I havent, I just plotted the countrate! Doing that should get a value closer to the ISL?
  15. Feb 7, 2016 #14
    It should yes since you are eliminating the part of the geometric factor related to the crystal diameter.
  16. Feb 7, 2016 #15
    Apart from an extreme outlier at d=0.5cm, the data now gives me a value a lot closer to the inverse square law! Thank you!
  17. Feb 7, 2016 #16
    Your a/d is 2.8 and if your source if fairly strong and because you are so close you have to be careful of the count rate, especially when counting all pulses
    ( no window set) which could increase dead time, pulse pile up resulting in two pulse being counted as one, unanticipated scatter problems due to source size or nearby structures.and of course that intrinsic efficiency problem I mentioned.
  18. Feb 7, 2016 #17
    I have noticed that when doing some error propagation, that even though the two parts to K have small relative errors, since the two parts are so similar, when I subtract one from the other, the relative error is huge! Is there any way to correct this?
  19. Feb 7, 2016 #18
    Could you show me what you are doing?
  20. Feb 7, 2016 #19
    I'll give an example:
    For K, given d=14cm and a=1.4cm , I get a value of 6.11
    [tex]2 \pi d^2=1231.5[/tex] with a relative error of 1.4%
    [tex]\frac{2\pi d^3}{\sqrt{a^2+d^2}}=1225.4[/tex] with a relative error of 2.3%

    However when I take one from the other I get K=6.11 with a relative error of 536%

    Is there anyway to prevent this?
  21. Feb 7, 2016 #20
    If you are trying to calculate the error in K then first determine the differential of K wrt d i.e., dKd = (∂K/∂d)dd evaluate for the values of a and d you are using and set dd = to the estimated measurement error for d.

    dKd is the error in K resulting from the error in d.

    Repeat for a to get dKa

    Find he total error for K, dK from

    (dK)2 = (dKd)2 +(dKa)2

    and rel error of K = dK/K
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