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

[tigrus]

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.

Homework Statement

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

Homework 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

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.

 

[lippyka]

Thank you for your replies. I think I understand the concept better now. I was missing the fact that dose rate and intensity can be considered the same here, meaning that attenuation factor is related to dose rate. To solve this I used the equation I=I_0*e^(-yx) where y is the attenuation factor and x is the thickness of the wall. This gave me the thickness of 13.7cm for Co-60 and 10.2cm for Cs-137. The values I got were slightly off from the ones given in the question, but I assume this is due to the approximations I had to do in the attenuation factors. Once again, thank you all for helping me out.