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Science Chemistry Chemistry: An Atoms-Focused Approach

To predict: The modes of decay of the radioactive isotopes given in the problem. Solution: The decay mode for 10 C is either positron emission or electron capture. The decay mode for 19 Ne is either positron emission or electron capture. This nuclide of 50 Ti is stable. Explanation: Concept: The neutron-to-proton ratio of nuclei tells us about the stability of the nuclei. If the ratio is greater than or less than one, the nuclei are unstable. The mode of decay depends on this ratio too. If the ratio is slightly more than one, it will be electron capture process or beta decay; if the ratio is significantly higher than one, then it undergoes alpha decay. Since the element is given, we can find the atomic number from the periodic table and calculate the number of neutrons in them. Formula: Mass number A = n + p Given: Mass number of Carbon = 10 Mass number of Neon = 19 Mass number of Titanium = 50 Calculations: We know that mass number of an isotope is the sum of neutrons and protons present in the nuclei. Therefore, the number of neutrons is n = A - p The atomic number of C is 6 . For C 6 10 , the number of neutrons is n = A - p = 10 - 6 = 4 The neutron-to-proton ratio of C 6 10 is n p = 4 6 = 0.67 Since the ratio is lower than one, it will undergo electron capture or positron emission. The atomic number of N e is 10 . And, for N e 10 19 , the number of neutrons is n = A - p = 19 - 10 = 9 The neutron-to-proton ratio of N e 10 19 is n p = 9 10 = 0.9 Since the ratio is lower than one, it will undergo electron capture or positron emission. The atomic number of T i is 22 . For T i 32 50 , the number of neutrons is n = A - p = 50 - 22 = 28 The neutron-to-proton ratio of T i 32 50 is n p = 28 22 = 1.27 The nuclide is stable in nature. Conclusion: The mode of decay depends on the ratio of the neutrons and protons present in the nuclei. If the ratio is very high, it will undergo alpha decay, and if the ratio is less, it will undergo either beta decay or electron capture.

To predict: The modes of decay of the radioactive isotopes given in the problem. Solution: The decay mode for 10 C is either positron emission or electron capture. The decay mode for 19 Ne is either positron emission or electron capture. This nuclide of 50 Ti is stable. Explanation: Concept: The neutron-to-proton ratio of nuclei tells us about the stability of the nuclei. If the ratio is greater than or less than one, the nuclei are unstable. The mode of decay depends on this ratio too. If the ratio is slightly more than one, it will be electron capture process or beta decay; if the ratio is significantly higher than one, then it undergoes alpha decay. Since the element is given, we can find the atomic number from the periodic table and calculate the number of neutrons in them. Formula: Mass number A = n + p Given: Mass number of Carbon = 10 Mass number of Neon = 19 Mass number of Titanium = 50 Calculations: We know that mass number of an isotope is the sum of neutrons and protons present in the nuclei. Therefore, the number of neutrons is n = A - p The atomic number of C is 6 . For C 6 10 , the number of neutrons is n = A - p = 10 - 6 = 4 The neutron-to-proton ratio of C 6 10 is n p = 4 6 = 0.67 Since the ratio is lower than one, it will undergo electron capture or positron emission. The atomic number of N e is 10 . And, for N e 10 19 , the number of neutrons is n = A - p = 19 - 10 = 9 The neutron-to-proton ratio of N e 10 19 is n p = 9 10 = 0.9 Since the ratio is lower than one, it will undergo electron capture or positron emission. The atomic number of T i is 22 . For T i 32 50 , the number of neutrons is n = A - p = 50 - 22 = 28 The neutron-to-proton ratio of T i 32 50 is n p = 28 22 = 1.27 The nuclide is stable in nature. Conclusion: The mode of decay depends on the ratio of the neutrons and protons present in the nuclei. If the ratio is very high, it will undergo alpha decay, and if the ratio is less, it will undergo either beta decay or electron capture.

Solution Summary: The author explains the neutron-to-proton ratio of nuclei and the mode of decay of the isotopes.

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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