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Science Chemistry Chemistry: An Atoms-Focused Approach

The time taken for 99 % of the C 11 injected into a patient to decay in a positron-emission tomography. Solution: The time taken for 99 % of the C 11 injected into a patient to decay in a positron-emission tomography = 136 m i n Explanation: Concept: We are asked to calculate the time required for 99 % of the C 11 injected in patient to decay. This means if initial amount injected is considered as 100 % then the amount of C 11 that remains after time (t) will be 1 % The half-life of C 11 is given as 20.4 min. Using the decay equation, we can calculate the time required (t). Formula: N t N 0 = 0.5 t t 1 2 Given: i) t 1 2 = 20.4 m i n ii) N 0 = 100 % , iii) N t = 1 % Calculation: N t N 0 = 0.5 t t 1 2 1 100 = 0.5 t 20.4 0.01 = 0.5 t 20.4 Calculating log on both sides, we get: log ⁡ 0.01 = log ⁡ 0.5 t 20.4 log 0.01 = t 20.4 × log ⁡ ( 0.5 ) - 2 = t 20.4 × ( - 0.3010 ) - 2 = t × - 0.01475 t = - 2 - 0.01475 = 135.59 t = 135.59 = 136 m i n . Thus, the time required for 99 % of the C 11 injected into a patient to decay is 136 minutes. Conclusion: The time required for 99 % of the C 11 to decay is calculated using the decay equation and the given half-life.

The time taken for 99 % of the C 11 injected into a patient to decay in a positron-emission tomography. Solution: The time taken for 99 % of the C 11 injected into a patient to decay in a positron-emission tomography = 136 m i n Explanation: Concept: We are asked to calculate the time required for 99 % of the C 11 injected in patient to decay. This means if initial amount injected is considered as 100 % then the amount of C 11 that remains after time (t) will be 1 % The half-life of C 11 is given as 20.4 min. Using the decay equation, we can calculate the time required (t). Formula: N t N 0 = 0.5 t t 1 2 Given: i) t 1 2 = 20.4 m i n ii) N 0 = 100 % , iii) N t = 1 % Calculation: N t N 0 = 0.5 t t 1 2 1 100 = 0.5 t 20.4 0.01 = 0.5 t 20.4 Calculating log on both sides, we get: log ⁡ 0.01 = log ⁡ 0.5 t 20.4 log 0.01 = t 20.4 × log ⁡ ( 0.5 ) - 2 = t 20.4 × ( - 0.3010 ) - 2 = t × - 0.01475 t = - 2 - 0.01475 = 135.59 t = 135.59 = 136 m i n . Thus, the time required for 99 % of the C 11 injected into a patient to decay is 136 minutes. Conclusion: The time required for 99 % of the C 11 to decay is calculated using the decay equation and the given half-life.

Solution Summary: The author calculates the time taken for 99% of the _11C injected into a patient to decay in positron-emission tomography.

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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