Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 18 - Nuclear Chemistry - Problem Solving Practice 18.3 - Page 792: c


$^{20}_{9}F\rightarrow \,\,^{0}_{-1}e+\,\,^{20}_{10}Ne$

Work Step by Step

There are excess neutrons in $\,^{20}_{9}F$, so beta emission is probable. In the case of $\beta$ emission, mass number does not change, but the atomic number increases by one unit. Therefore, the resulting nuclide is one with the mass number 20 and atomic number=$9+1=10$. $Ne$(neon) is the element with atomic number 10. Thus, we can write the nuclear equation for decay as $^{20}_{9}F\rightarrow \,\,^{0}_{-1}e+\,\,^{20}_{10}Ne$
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