Nuclear Chemistry

 

 36) t1/2 = 5.26 yr       

       (a) t = 45.5 s          N0 = 2.44 mg Co-60

 

            t1/2 = 5.26 yr x 365 days/1 yr x 24 hr/1 day x 3600 s/1 hr = 1.66 x 108 s

            k = 0.693/t1/2 = 0.693/(1.66 x 108 s) = 4.17 x 10-9 s-1

 

            ln(Nt/N0) = -kt

            ln(Nt/N0) = -4.17 x 10-9 s-1 x 45.5 s = -1.90 x 10-7

            Nt/N0 = e-1.90 x 10-7

 

            Because of the small value involved, an alternative method is used to preserve significant figures and accuracy.

            Nt/N0 = e-1.90 x 10-7

            Nt/N0 = 1.00 - 1.90 x 10-7

            Nt = N0(1.00 - 1.90 x 10-7) = N0 - 1.90 x 10-7N0

            ΔN = N0 - Nt = N0 - (N0 - 1.90 x 10-7N0) = 1.90 x 10-7N0

            ΔN = 1.90 x 10-7 x 2.44 mg = 4.64 x 10-7 mg

 

             = 4.64 x 10-7 mg Co-60 x 1 g/103 mg x 1 mol Co-60/60.00 g Co-60 x 6.02 x 1023 Co-60 nuclei/1 mol Co-60 x 1 β/1 Co-60 nuclei

            #β = 4.66 x 1012  β

 

      (b) #Bq = 4.66 x 1012  dis/45.5 s x 1 Bq/1 dis/s = 1.02 x 1011 Bq