Radiation Exposure - Safe Working Distance

1. Mar 22, 2006

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?

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:

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!

2. Mar 22, 2006

imabug

exposure also follows the inverse square law

3. Mar 23, 2006

Hootenanny

Staff Emeritus
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

4. Mar 23, 2006

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).

5. Mar 23, 2006

quicknote

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