Nuclear Chemistry

 36) t_1/2 = 5.26 yr       

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

            t_1/2 = 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/t_1/2 = 0.693/(1.66 x 10⁸ s) = 4.17 x 10^-9 s^-1

            ln(N_t/N₀) = -kt

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

            N_t/N₀ = e^{-1.90 x 10-7}

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

            N_t/N₀ = e^{-1.90 x 10-7}

            N_t/N₀ = 1.00 - 1.90 x 10^-7

            N_t = N₀(1.00 - 1.90 x 10^-7) = N₀ - 1.90 x 10^-7N₀

            ΔN = N₀ - N_t = 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

            #β = 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