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21 Nuclear Chemistry Solutions to Exercises 21.37 (a) Analyze/Plan. + 1 α particle is produced for each that decays. Rate = kN. Calculate the number of particles in the 10.0 mg sample. Calculate and k in min, then rate in dis/min, then the number of disintegrations in 5.0 min. Solve. 10.0 mg Ra 1g 1 mol Ra 6.022 1 10²³ mol Ra Ra atoms = 2.6646 10¹⁹ = 2.66 10¹⁹ atoms 1000 mg 226 g Ra Calculate k in min⁻¹. 1600 yr 365 1 yr X 24 1d hr X 60 1hr min = 8.410 10⁸ min⁻¹ k = 0.693 = 8.410 0.693 10⁸ min = 8.241 10⁻¹⁰ min⁻¹ Rate = kN = (8.241 X 10⁻¹⁰ X 10¹⁹ atoms) = 2.20 10¹⁰ atoms/min (2.20 10¹⁰ atoms/min )(5.0 min) = 1.1 X 10¹¹ atoms decay in 5.0 min 1.1 10¹¹ α particles emitted in 5.0 min (b) Plan. From part (a), the rate is 2.20 X 10¹⁰ disintegrations/min. Change this to dis/s and apply the definition 1 Ci = 3.7 10¹⁰ dis/s. 2.20 1 min 10¹⁰ dis 1 min X 3.7 10¹⁰ 1Ci dis/s X 1000 Ci mCi = 9.891 = 9.9 mCi 21.38 (a) Proceeding as in Solution 21.37, calculate number of atoms and k in 3.75mg Co 1000 1g mg X 1 60 mol g Co Co 6.022 X 1 mol 10²³ Co Co atoms = 3.76375 10¹⁹ = 3.76 10¹⁹ 5.26 yr 1 yr X 24 1d hr X 3600 1hr sec = 1.659 X 10⁸ = 1.66 10⁸ k = 0.693 = 1.659 0.693 10⁸ = 4.178 X 10⁻⁹ = 4.18 X 10⁻⁹ Rate = kN = (4.178 X atoms) = 1.57 10¹¹ atoms/s (1.57 10¹¹ atoms/s )(600 s) = 9.43 10¹³ atoms decay in 600 S 9.43 X 10¹³ particles emitted by a 3.75 mg sample in 600 S (b) 1.57 10¹¹ dis 1Bq = 1.57 10¹¹ Bq S The activity of the sample is 1.57 10¹¹ Bq. 654