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  1. The half-life of Po-210 is 138.4 days. An anti-static plate is obtained with an activity of 300 ĩCi. 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