Question

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

Answer 1

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

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

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity ,  $dN/dt = N* \lambda$

              $\frac{dN}{dt} = N* \frac{ln2}{T_1/2}$

                   $N = \frac{(dN/dt )(T1/2)}{ln2}$

                      = ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

                      = 8.877*10^{16}

Number of moles:

     n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

   m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

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    time t = -[T1/2 / ln2]*ln[N/N0]

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

             = 52.8 year