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

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.

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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