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-2t   =   A0e 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 = 500 (50/6)2 = 500 (833)2 = 500 69.4 = 7.2 mR/hr [ ]2

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)

= 0.4 x 10-3 rads/hr

= 0.4 m rad/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)

= 0.419 m rad/hr