## Occupational Safety and Health Administration OSHA

### Example Problems

1. The half-life of Po-210 is 138.4 days. An anti-static plate is obtained with an activity of 300 µCi. What is the activity 3 years later?

3 yr   =   3(365)   =   1095 days

At   =   A0e • A0e-2t   =    0.693 / τ1 / 2

At   =   (300)e  -0.693•1095 / 138.4 =  (300)e-5.48

=   (300)(0.00416)   =   1.25µCi

An alternative method of calculating the answer is to calculate the number of half lives that have passed and used An = A0 (1/2)n.

n = 1095days / 138.4 days/ half - life =  791 half lives

An   =   300(1/2)7.91 = 300 (0.00416)   =   1.25 µCi

2. A source has an intensity of 500 mr/hr at 6 feet. What is the intensity at 50 feet?

I1 / I2 = [d1 / d2]2  I1 = 500    d1 = 6    d2 = 50

I2 = [I1 / d2 / d1]2 = 500 / (50/6)2 = 500 / (833)2 = 500 / 69.4 = 7.2 mR/hr

3. The half-value layer for a shielding material is 5 mm. If the intensity of the radiation beam striking the shield is 100 R/hr and the shield is 2 inches thick, what is the intensity is mR/hr on the other side of the shield?

I = I0e  -0.693•x / HVL = (100)e  -0.693(2")(25.4 mm/inch) / 5mm

=   (100)e -7.04 =  100 (0.000876)  =  0.088 R/hr

= 88 mR/hr

4. Calculate the absorbed dose rate produced in bone (f = 0.922) by 1 MeV gamma radiation which produced an exposure rate of 0.5 mR/hr.

D = 0.869 ƒ X (R) rads

= (0.869)(0.922)(0.5 x 10-3 R/hr)

5. Calculate the absorbed dose rate in soft tissue (f = 0.965) for the condition in problem 4.

D = (0.869)(0.965)(0.5 X 10-3 R/hr)