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Calculation: radioactivity [atoms/ccm]

  1. Nov 30, 2007 #1
    1. The problem statement, all variables and given/known data

    The problem:
    4000,000 gal of water remained in a contaminated building at 3 mile island. The principle sources of radioactivity were:
    137Cs at 156E-6 Ci/cubic cm and 134Cs at 26E-6 Ci/ccm. How many atoms/ccm were in the water?

    2. Relevant equations
    137E-6*3.7E10 Ci= 5.77E6 Bq/ccm
    A ctivity= R*N
    R= Ln2/T1/2

    3. The attempt at a solution
    R= Ln2/30.07yrs*3600s = 6.4E-6/sec
    A = R*N
    N = 5.77E6(decays/sec-ccm)/6.4E-6(1/sec) = 9.02E11 atoms/ccm

    The given answers:
    Cs-137: 7.892E15 atoms/cm3
    Cs-134: 9.045E15 atoms/cm3
  2. jcsd
  3. Nov 30, 2007 #2


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    Staff Emeritus
    Science Advisor

    Be careful with conversions. There are 3600 s/hr, 24 hrs/day and 365.25 days/yr, or about 3.156 E7 s/yr.
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