Question

Measuring Radiation

Part A

At a distance of 40 ft , an ionizing radiation source delivers 7.0 rem of radiation. How close could you get to the source and still have no biological effects?

Express your answer to two significant figures and include the appropriate units.

Part B

A nurse administered 2.50 mL of a radioisotope solution that has an activity of 165 μCi/mL . What total dose of the radioisotope did the patient receive?

Express your answer to three significant figures and include the appropriate units.

Answer 1

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 25/7 = 40^2/r1^2

Solve for r1:

7/25 = r1^2/40^2
0.28 = r1^2/1600
r1^2 = 0.28*1600 =448
r1 = 21.167 feet = 21feet (answer)

Checking:
21.167^2 / 40^2 = 448 / 1600 = 0.28 Just what we wanted it to be.

B)

Look at the units.

uCi/mL * mL = uCi (mL cancels out just like 25 miles/gallon * 5 gallons = 125 miles)

So just multiply the volume by the uCi/mL concentration.
165*2.5=412.5uCi=412uCi