How many grams of CO60 result in 1 Millicuire of activity Ho

How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Years

Solution

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663e+8 s

activity dN/dt = 1 mCi = 3.7 X 10⁷ decay/s

activity ,   dN/dt = N*lambda

                dN/dt = N*[ln2/T_1/2]

                    N = (dN/dt )(T_1/2) / ln2

                       = ( 3.7 X 10⁷ )(1.663e+8 ) / ln2

                       = 8.877e+15

Number of moles:

      n = N/N_A = 8.877e+15 / 6.022X10²³ = 1.474e-8 mol

mass of the CO-60 is,

    m = n*M = [1.474e-8 mol]*[59.93 grams /mol] = 8.83e-7 grams

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     time t = -[T_1/2 / ln2]*ln[N/N₀]

              = - [5.3 years / ln2]*ln[1x10^-6/1x10^-3]

              = 52.8 years