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Science Chemistry Chemistry: An Atoms-Focused Approach

The half-life of a radionuclide if only 6.25 % of it remains after 8 hours and 20 minutes. Solution: t 1 2 = 125 m i n Explanation: Concept: Assuming that the initial amount of the radionuclide was 100%, we can say that the amount of it remaining after 8 hours and 20 minutes is 6.25 % . Thus, we know the value of N 0 a n d N t . Using the decay equation, we can calculate the half-life of this radionuclide. Formula: N t N 0 = 0.5 t t 1 2 Given: i) t = 8 h o u r s a n d 20 m i n ii) N 0 = 100 % , iii) N t = 6.25 % Calculation: The time is given as 8 hours and 20 minutes. Converting the 8 hours to minutes, we get: 8 h r × 60 m i n 1 h r = 480 m i n Total time = 480 min ⁡ + 20 min ⁡ = 500 m i n Substituting the values in the equation: N t N 0 = 0.5 t t 1 2 6.25 100 = 0.5 500 t 1 2 0.0625 = 0.5 500 t 1 2 Calculating log on both sides, we get: log ⁡ 0.0625 = log ⁡ 0.5 500 t 1 2 log 0.0625 = 500 t 1 2 × log ⁡ ( 0.5 ) - 1.2041 = 500 t 1 2 × ( - 0.3010 ) t 1 2 = 500 × - 0.3010 - 1.2041 t 1 2 = 124.98 m i n Thus, the half-life of the radionuclide is 125 minutes. Conclusion: Using the decay equation and the time given, we can calculate the half-life of the given radionuclide.

The half-life of a radionuclide if only 6.25 % of it remains after 8 hours and 20 minutes. Solution: t 1 2 = 125 m i n Explanation: Concept: Assuming that the initial amount of the radionuclide was 100%, we can say that the amount of it remaining after 8 hours and 20 minutes is 6.25 % . Thus, we know the value of N 0 a n d N t . Using the decay equation, we can calculate the half-life of this radionuclide. Formula: N t N 0 = 0.5 t t 1 2 Given: i) t = 8 h o u r s a n d 20 m i n ii) N 0 = 100 % , iii) N t = 6.25 % Calculation: The time is given as 8 hours and 20 minutes. Converting the 8 hours to minutes, we get: 8 h r × 60 m i n 1 h r = 480 m i n Total time = 480 min ⁡ + 20 min ⁡ = 500 m i n Substituting the values in the equation: N t N 0 = 0.5 t t 1 2 6.25 100 = 0.5 500 t 1 2 0.0625 = 0.5 500 t 1 2 Calculating log on both sides, we get: log ⁡ 0.0625 = log ⁡ 0.5 500 t 1 2 log 0.0625 = 500 t 1 2 × log ⁡ ( 0.5 ) - 1.2041 = 500 t 1 2 × ( - 0.3010 ) t 1 2 = 500 × - 0.3010 - 1.2041 t 1 2 = 124.98 m i n Thus, the half-life of the radionuclide is 125 minutes. Conclusion: Using the decay equation and the time given, we can calculate the half-life of the given radionuclide.

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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Chemistry: An Atoms-Focused Approach

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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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A sample of rock was found to contain 8.23 mg of rubidium-87 and 0.47 mg of strontium-87.. (a) Calculate the age of the rock if the half-life of the decay of rubidium by emission is 4.71010 y. (b) If some S3887r was initially present in the rock, would the rock be younger, older, or the same age as the age calculated in (a)? Explain your answer.
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