Introductory Chemistry (5th Edition)

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

Chapter 17 - Radioactivity and Nuclear Chemistry - Exercises - Cumulative Problems - Page 639: 90

Answer

Series of nuclear equation is as follows:

Work Step by Step

Here, at the beginning Al-27 (atomic number) reacts with neutron and form Al-28. (When an element is bombarded with neutron, its neutron number is increased by 1 unit. Hence mass number is increased by 1 unit. That's why Al-27 is converted into Al-28). This process is represented by the first nuclear equation written above. After formation of Al-28, it undergoes an alpha decay and as a result its atomic number will decreased by 2 and mass number is decreased by 4 unit. The new element has the atomic number = 13-2 = 11 and mass number = 28-4 = 24. The new element is Na-24 (atomic number 11). This process is also written in the form of nuclear equation above. (see the 2nd nuclear equation) Lastly, Na-24 undergoes a beta decay and as a result its atomic number increases by 1 unit whereas its mass number remains same. So, the new element has the atomic number = 11+1 = 12. The new element formed is Mg-24 (atomic number = 12). This process is also represented by the last nuclear equation written above.
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