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Science Chemistry CHEM:ATOM FOC 2E CL (TEXT)

The half-life of a radionuclide if 87.5 % of it decays in 6.6 d a y s . Solution: The half-life of the radionuclide is 2.2 d a y s . Explanation: Concept: All radioactive decay processes follow first order kinetics. Each radioactive nuclide has a characteristic half life ( t 1 2 ) , the time interval during which the quantity of radioactive particles decreases to half of its initial quantity. Formula: N t N 0 = 0.5 t t 1 2 Given: 87.5 % of the radionuclide decays in 6.6 days. Calculations: Let’s consider that the initial radioactivity of a radionuclide is N 0 = 100 . As 87.5 % of the original quantity decays in 6.6 days, we have to subtract the 87.5 from the initial radioactivity 100 to get the final radioactivity ( N t ) of a radionuclide. So final radioactivity of a radionuclide is N t = 100 - 87.5 = 12.5 after a time interval of 6.6 d a y s . The integrated rate law for the radioactive decays is N t N 0 = 0.5 t t 1 / 2 Taking the natural log of both sides yields l n N t N 0 = - 0.693 × t t 1 / 2 Substituting the given values N 0 = 100 , N t = 12.5 , time interval t = 6.6 d a y s yields the equation l n 12.5 100 = - 0.693 × 6.6 d a y s t 1 / 2 - 2.0794 = - 4.5738 t 1 / 2 t 1 / 2 = - 4.5738 - 2.0794 t 1 / 2 = 2.199 d a y s = 2.2 d a y s Conclusion: The half-life of the nuclide is calculated from the given percentage, time, and integrated rate law

The half-life of a radionuclide if 87.5 % of it decays in 6.6 d a y s . Solution: The half-life of the radionuclide is 2.2 d a y s . Explanation: Concept: All radioactive decay processes follow first order kinetics. Each radioactive nuclide has a characteristic half life ( t 1 2 ) , the time interval during which the quantity of radioactive particles decreases to half of its initial quantity. Formula: N t N 0 = 0.5 t t 1 2 Given: 87.5 % of the radionuclide decays in 6.6 days. Calculations: Let’s consider that the initial radioactivity of a radionuclide is N 0 = 100 . As 87.5 % of the original quantity decays in 6.6 days, we have to subtract the 87.5 from the initial radioactivity 100 to get the final radioactivity ( N t ) of a radionuclide. So final radioactivity of a radionuclide is N t = 100 - 87.5 = 12.5 after a time interval of 6.6 d a y s . The integrated rate law for the radioactive decays is N t N 0 = 0.5 t t 1 / 2 Taking the natural log of both sides yields l n N t N 0 = - 0.693 × t t 1 / 2 Substituting the given values N 0 = 100 , N t = 12.5 , time interval t = 6.6 d a y s yields the equation l n 12.5 100 = - 0.693 × 6.6 d a y s t 1 / 2 - 2.0794 = - 4.5738 t 1 / 2 t 1 / 2 = - 4.5738 - 2.0794 t 1 / 2 = 2.199 d a y s = 2.2 d a y s Conclusion: The half-life of the nuclide is calculated from the given percentage, time, and integrated rate law

Solution Summary: The author calculates the half-life of a 199Au if 16.5% of it decays in 168hours.

CHEM:ATOM FOC 2E CL (TEXT)

CHEM:ATOM FOC 2E CL (TEXT)

2nd Edition

ISBN: 9780393284218

Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss

Publisher: WW Norton & Co

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Writing and Balancing Nuclear Equations

A nuclear equation is a symbolic representation of a nuclear reaction using certain symbols. It is studied in nuclear chemistry and nuclear physics.

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CHEM:ATOM FOC 2E CL (TEXT)

Ch. 21 - Prob. 21.1VP Ch. 21 - Prob. 21.2VP Ch. 21 - Prob. 21.3VP Ch. 21 - Prob. 21.4VP Ch. 21 - Prob. 21.5VP Ch. 21 - Prob. 21.6VP Ch. 21 - Prob. 21.7VP Ch. 21 - Prob. 21.8VP Ch. 21 - Prob. 21.9VP Ch. 21 - Prob. 21.10VP

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