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Homework Help: Radiation Dose Calculation

  1. Dec 18, 2017 #1
    1. The problem statement, all variables and given/known data

    I have attached a picture of the question.

    A 2 Ci 137Cs source is equal to 2 x (3.7x1010 Bq) = 7.4x1010 s-1

    8 hours a day, 5 days a week and for 50 weeks is a total duration in seconds of 7.2x106 s

    An 8 cm lead shield = 0.08 m

    IRR99 regulations set an effective dose limit as 20 millisieverts per year. Therefore, 10% of of their effective dose limit is 2 millisieverts = 0.02 m2s-2

    2. Relevant equations


    3. The attempt at a solution

    [7.4x1010 s-1 x 0.08 m] / 7.2x106 s = 0.02 m2s-2 - d

    d = 2.4x10-5 m

    I jiggle the equation about in different ways but keep getting small values for d.


    Attached Files:

  2. jcsd
  3. Dec 18, 2017 #2


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    How does your calculation take into account that there is a lead door, as opposed to some other kind of door, between the source and the worker?
    How does your calculation take into account that the farther away from the source, the less radiation the worker receives?
  4. Dec 18, 2017 #3
    There are a lot of confusion in your calculation :
    Do you know the inverse square rule ? The exponential attenuation ? The intensity of emission (85% for Cs-137) ...
  5. Dec 18, 2017 #4
    You need to find the the specific gamma rate constant/dose rate constant which relates the dose rate (Sv/hr) at a specified distance to the activity(Bq) of the source which a published for radionuclides and sources made from them.
  6. Dec 19, 2017 #5
    I haven't a clue to be honest guys as this isn't in my lecture notes.

    I've got a value of 3.7x107 Bq/hr (source strength?)

    The inverse square law is I = S/4 pie x r2

    I don't know how to work out a value for I (intensity at surface of sphere).
    Last edited: Dec 19, 2017
  7. Dec 19, 2017 #6
    You can have any source with a specific decay rate but the radiation dose rate at a specified distance depends on the isotope. Each isotope has its own specific gamma ray constant and each particular source made from this isotope may have that significantly modified by the encapsulation material due to attenuation of that material.
  8. Dec 19, 2017 #7
    What are the equations I need?

    I have a dose constant of 0.38148 Rem/hr for 137Cs at a distance of 1m from a 1 point curie source.
  9. Dec 19, 2017 #8
    A small, help,
    For Cs137 you have the specific dose rate 84 microSv/h/GBq at one meter
    For 2 Ci it gives 6216 microSv/h
    Your worker works 2000 h and your objective is 2mSv per year so your objective dose rate is 1 microSv/h
    The attenuation coefficient is for lead and 660 keV 1.4 /cm
    With that you must conclude
    I have 30 cm
  10. Dec 19, 2017 #9
    Please explain the 85%.
  11. Dec 20, 2017 #10
  12. Dec 20, 2017 #11
    How do you get a value for the linear attenuation coefficient for lead, anyway?

    I'm close to giving up because I don't know what equations I need / what order to do this question in. Do I use I = I0e-ux then the inverse square law? I have no idea how to approach this question.
  13. Dec 20, 2017 #12


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    Yes, first you use the exponential to find how much radiation makes it through the door, then you use the inverse square law to find how much distance you need to put between the worker and the door to get the desired overall attenuation factor. Look at post #8 for the linear attenuation coefficient for lead.
  14. Dec 20, 2017 #13
    I = I0e(-ux)

    I = (7.4x10-10)e(-1.4x0.08)
    I = 6.61x10-10 s-1
    I = 1.83x10-13 /hr

    I = S/4 pie r2
    r = square root [(10-6)/(4pie)(1.83x10-13)]
    r = 659

    lmao I don't know what im doing. I'm going to quit the course.

    Thanks for the help.
  15. Dec 20, 2017 #14
    The use of the .85 factor is appropriate when you are detecting the 660 keV photons and are trying to determine the activity of the source. It is not appropriate when we are interested in the exposure rate of a given source activity.
  16. Dec 20, 2017 #15
    Of course it is !
    The dose for Cs137 is due to the 662 keV
    You know that
    H=1.5E-10×A×i×E (in mSv/h)
    Ain Bq , E in MeV and i=.85
    See, for example chapter 4 of http://www.springer.com/us/book/9783319486581
  17. Dec 20, 2017 #16

    Only if you are calculating the exposure rate constant from the spectrum of photons emitted. The exposure rate constants quoted in the literature have taken that fact into account or they are measured in which case it is irrelevant.
  18. Dec 20, 2017 #17
    Of course, but it is interessting to know how to calculate this constant dose rate from a source flux, no ?
    It is useful to know with a activity to calculate flux, fluence at a distance d, and dose equivalent with the conversion factor of the ICRP. But it is only my opinion
  19. Dec 20, 2017 #18


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    The attenuation coefficient is in 1.4 cm-1 so the door thickness should be expressed in centimeters not meters. This should change your answer significantly.
  20. Dec 21, 2017 #19
    Using the thickness in cm does change the answer but it's still wrong by a lot. Thanks for the help.

    I need someone to type out the solution as it's the only way I can understand this question. I don't know what I'm doing and have wasted too much time on this one question.
  21. Dec 21, 2017 #20
    Using the following data

    dose at occupied area = 2 mSv
    attenuation coefficient for lead and Cs-137 gamma rays = 1.24 cm ( gives a Half Value Layer of 0.56 cm the published value)
    shielding 8 cm lead.giving an attenuation of 0.492 x10-4
    Dose rate constant = 0.003815 Sv/Ci/hr at 1 meter
    time of occupation = 2000 hrs
    Source activity = 2 Ci

    I get 61.3 cm
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