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Science Chemistry SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

The following nuclear equations describing the preparation of isotopes used for nuclear medicine: L i 6 + n 1 → H 3 + ? O 16 + H 3 → F 18 + ? F e 56 + ? → C o 57 + n 1 S b 121 + H 4 e → ? + 2 n 1 Solution: L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 26 56 + H 1 2 → C o 27 57 + n 0 1 S b 51 121 + H 2 4 e → I 53 123 + 2 n 0 1 Explanation: For the balanced nuclear reaction, the sum of charges, i.e., atomic numbers (subscript) and mass numbers (superscript) must be equal on both sides of the reaction (reactants and products side). If the atomic and mass number of the rest of the particles is known, then we can determine the missing particle in the balanced nuclear reaction. L i 6 + n 1 → ? → H 3 + ? The given reaction is L i 6 + n 1 → ? + H 3 Here, the sum of charge on the left side is 3 since the charge on Li (Lithium) is 3 and that on neutron (n) is 0 , and the total charge on right side is 1 . So, the missing particle or nuclide on the right side must have a charge equal to 2 . The sum of mass numbers on the left side is 7 , and that on right side is 3 . So the missing nuclide must have mass number 4 . The nuclide with atomic number 2 and with mass number 4 is He, and it can represented as H e 2 4 or α particle. Therefore, the complete balanced equation is L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 16 + H 3 → F 18 + ? In this equation, the atomic number of O (Oxygen) is 8 , atomic number of H (hydrogen) is 1 , and that of F (Fluorine) is 9 . The sum of charges on both sides is 9 . The sum of mass numbers on the left side is 19 , and that on the right side is 18 . So, the missing particle must have a charge 0 and mass number 1 . Hence we can use the particle with charge 0 and mass number 1, which is n 0 1 . Therefore, the equation is O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 56 + ? → C o 57 + n 1 In this equation, we need to identify the reactant of the first step, so consider the first step, and the equation is F e 56 + ? → C o 57 + n 1 The atomic number of F e is 26 , and that of Co is 27 . The sum of charges on the left side is 26 , and that on the right side is 27 . Therefore, the missing particle must have 1 charge. The sum of mass numbers on the left side is 56, and that on the right side is 58 , so the missing particle must have mass number as 2 . The element having charge 1 and mass number 2 is Deuterium, and it is represented as H 1 2 . The complete balanced equation is F e 26 56 + H 1 2 → C o 27 57 + n 0 1 d) S b 121 + H 4 e → ? + 2 n 1 The atomic number of S b (antimony) is 51 , and that of He is 2 , so the sum of charges on the left side of the reaction is 53 , and that on the right side is 0 since the charge on neutron (n) is 0 . The sum of mass numbers on the left side is 125 , and that on the right side is 2 . So the missing particle or nuclide must have atomic number 53 and mass number 123 , which is iodine, represented as I 53 123 . Therefore, the balanced nuclear equation is S b 51 121 + H 2 4 e → I 53 123 + 2 n 1 Conclusion: We determined the sum of charges and mass numbers on both sides of the reaction to identify the missing particle or nuclide that balanced the nuclear equations.

The following nuclear equations describing the preparation of isotopes used for nuclear medicine: L i 6 + n 1 → H 3 + ? O 16 + H 3 → F 18 + ? F e 56 + ? → C o 57 + n 1 S b 121 + H 4 e → ? + 2 n 1 Solution: L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 26 56 + H 1 2 → C o 27 57 + n 0 1 S b 51 121 + H 2 4 e → I 53 123 + 2 n 0 1 Explanation: For the balanced nuclear reaction, the sum of charges, i.e., atomic numbers (subscript) and mass numbers (superscript) must be equal on both sides of the reaction (reactants and products side). If the atomic and mass number of the rest of the particles is known, then we can determine the missing particle in the balanced nuclear reaction. L i 6 + n 1 → ? → H 3 + ? The given reaction is L i 6 + n 1 → ? + H 3 Here, the sum of charge on the left side is 3 since the charge on Li (Lithium) is 3 and that on neutron (n) is 0 , and the total charge on right side is 1 . So, the missing particle or nuclide on the right side must have a charge equal to 2 . The sum of mass numbers on the left side is 7 , and that on right side is 3 . So the missing nuclide must have mass number 4 . The nuclide with atomic number 2 and with mass number 4 is He, and it can represented as H e 2 4 or α particle. Therefore, the complete balanced equation is L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 16 + H 3 → F 18 + ? In this equation, the atomic number of O (Oxygen) is 8 , atomic number of H (hydrogen) is 1 , and that of F (Fluorine) is 9 . The sum of charges on both sides is 9 . The sum of mass numbers on the left side is 19 , and that on the right side is 18 . So, the missing particle must have a charge 0 and mass number 1 . Hence we can use the particle with charge 0 and mass number 1, which is n 0 1 . Therefore, the equation is O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 56 + ? → C o 57 + n 1 In this equation, we need to identify the reactant of the first step, so consider the first step, and the equation is F e 56 + ? → C o 57 + n 1 The atomic number of F e is 26 , and that of Co is 27 . The sum of charges on the left side is 26 , and that on the right side is 27 . Therefore, the missing particle must have 1 charge. The sum of mass numbers on the left side is 56, and that on the right side is 58 , so the missing particle must have mass number as 2 . The element having charge 1 and mass number 2 is Deuterium, and it is represented as H 1 2 . The complete balanced equation is F e 26 56 + H 1 2 → C o 27 57 + n 0 1 d) S b 121 + H 4 e → ? + 2 n 1 The atomic number of S b (antimony) is 51 , and that of He is 2 , so the sum of charges on the left side of the reaction is 53 , and that on the right side is 0 since the charge on neutron (n) is 0 . The sum of mass numbers on the left side is 125 , and that on the right side is 2 . So the missing particle or nuclide must have atomic number 53 and mass number 123 , which is iodine, represented as I 53 123 . Therefore, the balanced nuclear equation is S b 51 121 + H 2 4 e → I 53 123 + 2 n 1 Conclusion: We determined the sum of charges and mass numbers on both sides of the reaction to identify the missing particle or nuclide that balanced the nuclear equations.

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

2nd Edition

ISBN: 9780393644777

Author: Gilbert

Publisher: MCGRAW-HILL HIGHER EDUCATION

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SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

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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