Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 17 - Radioactivity and Nuclear Chemistry - Exercises - Problems - Page 638: 66

Answer

After filling the blank, the partial decay series is as follows:

Work Step by Step

See the 1st equation. The parent nuclide is Ra-225 (atomic number 88) and the daughter nuclide is Ac-225 (atomic number 89). So, it is clear that atomic number is increased by (89-88) = 1 unit but the mass number remains same. So, here one beta particle is emitted from the nucleus of Ra-225. The beta particle is written in the form of symbol on the right side of the equation. In 2nd equation, the parent nuclide is Ac-225 (Atomic number 89) and on the right side of the equation we see the symbol of 'He' atom which means one alpha particle is emitted from Ac-225. We have to find the daughter nuclide or element. After emission of alpha from a nucleus, atomic number of the nuclide is decreased by 2 units and mass number of the nuclide is decreased by 4 units. So, atomic number of the daughter nuclide is (89-2) = 87 and mass number is (225-4) = 221. Therefore, the new element will be Fr (Atomic number 87) which can be written in the form of symbol on the right side of the equation. In 3rd equation, we can see that the daughter nuclide is At-217 (Atomic number 85) and an alpha is emitted from parent nuclide. So, the parent nuclide will have atomic number greater than 2 and mass number greater than 4 units from the daughter nuclide. So, the atomic number of parent nuclide is 85+2 = 87 and mass number of parent nuclide is 217+4 = 221. So, the parent nuclide is Fr (Atomic number 87) which can be written in the form of symbol on the left side of the equation. In 4th equation, parent nuclide is given and on the right side of the equation we see that an alpha particle is emitted in the form of He atom. So, the daughter nuclide will have atomic number less than 2 and mass number less than 4 units from parent nuclide. So, the atomic number of daughter nuclide will be 85-2 = 83 and mass number of that nuclide will be 217-4= 213. So, the daughter nuclide is Bi (Atomic number 83) which can be written in the form of symbol on the right side of the equation.
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