Radiation Exposure - Safe Working Distance

[quicknote]

The Canadian Nuclear Safety Commission sets upper limits on the amounts of radiation allowable in the workplace. The maximum permissible dose for radiation workers is 5 rem (50 mSv) per year. Cobalt-60 is a radioisotope whose gamma rays are sometimes used in radiation treatment.
What is the safe working distance from an unshielded 60Co source that produces 0.800 rad per hour at a distance of 1m?

Quote your answer in metres.

Assume a 40 hour week, with a three-week holiday period per year. You will need to think about how the intensity falls off with distance; you may assume that the 60Co is a point source, and that the radiation is emitted isotropically.

I'm totally stumped with this question. I have no idea how to approach it.
I know that I have to get the units to match:
1 rad = 0.01 Gy

The equivalent dose of 50mSv per year would be 50mGy per year (assuming that the weight factor equals 1).

A friend of mine told me that the intensity is proportional to the 1/ (distance squared)... but how do I get the intensity?

Any help would be greatly appreciated!

 

[imabug]

exposure also follows the inverse square law

 

[Hootenanny]

Intensity is defined as the power radiated over a given area and as imabug says follows the inverse square law. Mathematically intesity can be defined as;

$$\left| I \right| = \frac{I_0}{r^2} = \frac{P}{4\pi r^2}$$

Where I is instensity at a point r displaced from the source. $I_0$ is the intensity at the source, P is the power of the source. This formula assumes that no energy is lost to the medium through which the radiation is travelling.

Hope this helps
-Hoot

 

[daveb]

Perhaps the easiest way to approach it is to calculate the total exposure over the course of a year (in mrem). This gives exposure at 1 meter distance. Since exposure falls off at 1/r^2, just calculate the distance that would be required to reduce the exposure to the 5000 mrem limit. You can ignore the effects of decay since Co-60 has half life of 5.27 years (OK, you could be more precise by correcting for decay, but it would only decay a little over 12% by the time the year ended).

 

[quicknote]

Thanks for the help!
I understand it a bit better now.