A) At a distance of 38 ft , an ionizing radiation source delivers 4.5 rem of radiation. How close could you get to the source and still have no biological effects?
B) A nurse administered 4.00 mL of a radioisotope solution that has an activity of 195 μCi/mL . What total dose of the radioisotope did the patient receive?
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This is a similar problem posted in Yahoo Answers 2 years ago just adjust the values and the uCi in place of mRem
Best Answer: A) At a distance of 41 feet, the radiation source delivers 4.5 rem of radiation.
The dose delivered in the period of time in question is inversely proportional to 1/r^2. So if you cut the distance in half, you increase the dose by 2 squared. If you cut it down to a third, the dose is increased by a factor of 3 squared.
We want to keep the dose at or below 25 rem. This means we can increase the dose by a factor of up to 25/4.5 = 5.55 = 50/9
D = k/r^2
If we call the different doses D1 and D2 at distances r1 and r2, then the relation between the doses and the distances is:
D1/D2 = r2^2/r1^2
So 25/4.5 = 41^2/r1^2
Solve for r1:
4.5/25 = r1^2/41^2
0.18 = r1^2/1681
r1^2 = 0.18*1681 = 302.58
r1 = 17.39 feet
17.39^2 / 41^2 = 302.41 / 1681 = 0.18 Just what we wanted it to be.
B) A solution's activity will not be measured in mL. It may be measured in Ci/mL or rem/mL, but it has to be in the form of units of radioactivity divided by volume so that when you multiply by the volume used you're left with units of radioactivity.
If you have a solution that delivers 165 mrem/mL and inject 5 mL into a patient, the patient receives a dose of
165 mrem/mL * 5 mL = 825 mrem * mL/mL = 825 mrem.
Karl · 2 years ago