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Paperthickness measured by radiation

  1. Nov 13, 2006 #1
    Hello,

    This is my first post here, and I must say I'm in trouble caused by the last problem in my homework assignment. I live in Denmark so excuse me if some of the terms are not correct English.

    What I do know: Source: Ti-204, Decay: Beta-, Activity: 3,1MBq,
    Half-life period: 3,8 years

    The relationship between paperthickness (x) and intensity (measured decay/sec) is declining linearly (not sure what it's called). I've measured a, the coefficient, to: -40 counts/mm.

    The problem is this: At a measurement 25 months after the source started decaying, the detector measures 478 decays/sec.
    - What is the thickness of the paper?

    I am troubled because I'm not sure of the relationship between the measured decay (Intensity) and the other parametres. I know what the activity should be, but not the intensity.

    Any help would be appreciated.

    /Thomas M.
     
  2. jcsd
  3. Nov 13, 2006 #2

    OlderDan

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    I think you have left out some information. There must have been a detector count at the begining of the 25 month. The usual relationship between radiation intensity and barrier thickness assumes that the fractional change in count is proportional to the thickness. If a certain thickness reduces the intensity by 10%, a second layer would reduce the lowered intensity by another factor of 10% for a net reduction of 19%. Intensity vs thickness is analogous to intensity vs time.
     
  4. Nov 13, 2006 #3

    I've got a graph: 805 counted decays at the beginning of the 25 months. This graph is declining linearly with a = -40 counts/mm.
    e.g. 765 counts at 1 mm. paper, 725 at 2 mm. and so forth. What do I do with that info.? I think it's my understanding of measured intensity (counts/time), or lack thereof, that is the main problem. What say you?

    /Thomas M.
     
  5. Nov 13, 2006 #4

    OlderDan

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    I would say that your graph is probably not really linear. Surely the data points you fit to find the slope of the line had some scatter, and you drew the best straight line you could through those points. What do you think you would have found if the source intensity had been 205 counts instead of 805 counts? Do you think you would have seen 165 at 1mm, 125 at 2mm and so forth? What is far more likely is that you would have seen the same fractional reduction in intensity, or 195 at 1mm, and 185 at 2mm etc.

    If you still have your original data, try fitting it to an exponential decay curve instead of a line. The form of the equation will be

    I = I_o*exp(-kx) where I_o is the source intensity and x is paper thickness.

    If you don't have all the data, you can still use the couple of points you just quoted to find the k in this equation. Once you have that, calculate the intensity of your source after 25 months and calculate the ratio of measured intensity to source intensity. That ratio should be the same for a given thickness of paper no matter how strong the source. If you find the value of k from your original data, you can find x from the intensity ratio measured at any time.
     
  6. Nov 13, 2006 #5
    I've made a big mistake: I wasn't aware that the graph was printed on "exp-paper", that's why it made a straight line. :yuck:
    I should have known this, since I made an experiment 2 weeks ago, that resulted in an exponential declining function.

    I know that k = ln(2)/T½
    I know both ln(2) obviously, and as stated before T½ = 3,8 years. Thank you for the help!

    Best regards,
    /Thomas.
     
  7. Nov 13, 2006 #6

    OlderDan

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    I know what you mean, but in this case k is related to thickness rather than time. There is a "half-thickness" if you will that reduces the intensity to half its original value, analogous to the "half-life" that reduces the source intensity by half as time passes. You might label the half-thickness as X½.
     
  8. Nov 13, 2006 #7
    Yep, I am aware of that. The problem recided in the proper use of intensity - I wasn't sure if the 805 counts/sec. should be understood as intensity. You've been a great help,

    Thanks alot.
     
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