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Science Chemistry Chemistry: An Atoms-Focused Approach

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.

Chemistry: An Atoms-Focused Approach

Chemistry: An Atoms-Focused Approach

14th Edition

ISBN: 9780393912340

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

Publisher: W. W. Norton & Company

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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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Chemistry: An Atoms-Focused Approach

Ch. 21 - Prob. 21.1VP Ch. 21 - Prob. 21.2VP 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 Ch. 21 - Prob. 21.11VP

Ch. 21 - Prob. 21.12VP Ch. 21 - Prob. 21.13QA Ch. 21 - Prob. 21.14QA Ch. 21 - Prob. 21.15QA Ch. 21 - Prob. 21.16QA Ch. 21 - Prob. 21.17QA Ch. 21 - Prob. 21.18QA Ch. 21 - Prob. 21.19QA Ch. 21 - Prob. 21.20QA Ch. 21 - Prob. 21.22QA Ch. 21 - Prob. 21.23QA Ch. 21 - Prob. 21.24QA Ch. 21 - Prob. 21.25QA Ch. 21 - Prob. 21.26QA Ch. 21 - Prob. 21.27QA Ch. 21 - Prob. 21.28QA Ch. 21 - Prob. 21.29QA Ch. 21 - Prob. 21.30QA Ch. 21 - Prob. 21.31QA Ch. 21 - Prob. 21.32QA Ch. 21 - Prob. 21.33QA Ch. 21 - Prob. 21.34QA Ch. 21 - Prob. 21.35QA Ch. 21 - Prob. 21.36QA Ch. 21 - Prob. 21.37QA Ch. 21 - Prob. 21.38QA Ch. 21 - Prob. 21.39QA Ch. 21 - Prob. 21.40QA Ch. 21 - Prob. 21.41QA Ch. 21 - Prob. 21.42QA Ch. 21 - Prob. 21.43QA Ch. 21 - Prob. 21.44QA Ch. 21 - Prob. 21.45QA Ch. 21 - Prob. 21.46QA Ch. 21 - Prob. 21.47QA Ch. 21 - Prob. 21.48QA Ch. 21 - Prob. 21.49QA Ch. 21 - Prob. 21.50QA Ch. 21 - Prob. 21.51QA Ch. 21 - Prob. 21.52QA Ch. 21 - Prob. 21.53QA Ch. 21 - Prob. 21.54QA Ch. 21 - Prob. 21.55QA Ch. 21 - Prob. 21.56QA Ch. 21 - Prob. 21.57QA Ch. 21 - Prob. 21.58QA Ch. 21 - Prob. 21.59QA Ch. 21 - Prob. 21.60QA Ch. 21 - Prob. 21.61QA Ch. 21 - Prob. 21.62QA Ch. 21 - Prob. 21.63QA Ch. 21 - Prob. 21.64QA Ch. 21 - Prob. 21.65QA Ch. 21 - Prob. 21.66QA Ch. 21 - Prob. 21.67QA Ch. 21 - Prob. 21.68QA Ch. 21 - Prob. 21.69QA Ch. 21 - Prob. 21.70QA Ch. 21 - Prob. 21.71QA Ch. 21 - Prob. 21.72QA Ch. 21 - Prob. 21.73QA Ch. 21 - Prob. 21.74QA Ch. 21 - Prob. 21.75QA Ch. 21 - Prob. 21.76QA Ch. 21 - Prob. 21.77QA Ch. 21 - Prob. 21.78QA Ch. 21 - Prob. 21.79QA Ch. 21 - Prob. 21.80QA Ch. 21 - Prob. 21.81QA Ch. 21 - Prob. 21.82QA Ch. 21 - Prob. 21.83QA Ch. 21 - Prob. 21.84QA Ch. 21 - Prob. 21.85QA Ch. 21 - Prob. 21.86QA Ch. 21 - Prob. 21.87QA Ch. 21 - Prob. 21.88QA Ch. 21 - Prob. 21.89QA Ch. 21 - Prob. 21.90QA Ch. 21 - Prob. 21.91QA Ch. 21 - Prob. 21.92QA Ch. 21 - Prob. 21.93QA Ch. 21 - Prob. 21.94QA Ch. 21 - Prob. 21.95QA Ch. 21 - Prob. 21.96QA Ch. 21 - Prob. 21.97QA Ch. 21 - Prob. 21.98QA Ch. 21 - Prob. 21.99QA

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