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Homework Help: Radiation & nuclear physics, linear attenuation Co-60, Cs-137-

  1. Apr 27, 2012 #1
    Hello there.

    I've been struggling with this one for quite a long time now and decided to ask if any of you people could point out what I'm missing here.

    There are actually three problems which are related to each other. I managed to solve the first and the second one, but the third is where I run out of ideas.

    1. The problem statement, all variables and given/known data

    1) The measured dose rate at a distance of one meter from source is 7.5uSv/h. Calculate the dose rate at a distance of a) 1.50m b) 0.10m from the source.

    Answers: a) 3.3uSv/h b) 750 ySv/h

    2) What is the total dose per year received by a person who works in average distance of 2.0m from the radiation source referred in previous exercise, if there are 220 working days per year and the length of the working day is 8 hours.

    Answer: 3.3mSv

    Following is the one I have troubles with
    3) How thick concrete wall is needed to protect the worker so that the annual dose is 0.50mSv for a worker described in the previous exercise, if the source is a) Co-60 b) Cs-137

    Answers: a) 14cm b) 10cm

    2. Relevant equations

    Attenuation of y-radiation: I = I_0 * e^(-yx)

    Relationship between activity and dose rate - r (gamma) constant: A=D'd^2/r

    According to wikipedia r seems to be around 0.57721 uSvm^2/MBq

    3. The attempt at a solution

    As far as I understand 0.50mSv/a year is a dose rate itself, but I converted that to uSv/h

    220 * 8h = 1760h = year of work

    Dose rate D' = 0.50mSv/1760h = 0.284uSv/h

    Q_Co-60 = 2.823067Mev (Taken from a table)

    With that energy the factor of linear attenuation in Pb seems to be around 48.756. This was an approximation. For 2Mev the table says 52,2 and for 3Mev 48,0 so I took a value in between.

    From equation I=I_0*e^(-yx) I could calculate the needed thickness by making it x=ln(I/I_0)/-y but I have not found a way to solve the undampened intensity/dose rate. As far as I understand intensity and dose rates can be considered meaning the same thing here. Am I completely mistaken?

    I could also calculate the activity A with the formula A=D'd^2/r meaning A= (0.284uSv/h * 2^2)/0.57721 uSvm^2/MBq giving acticity of 1.96809 MBq but I have not understood how to utilize this information. How would you proceed?

    Any help is appreciated.

    Edited: Alright, one mistake clearly is that I was using attenuation factor for Pb instead of concrete. For concrete 2MeV the factor is 10.5 and 3MeV it is 8.51. Therefore I assume the factor to use is 8.8682. Still how to solve this remain unclear to me.
    Last edited: Apr 27, 2012
  2. jcsd
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