Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 20 - Exercises - Page 975: 89

Answer

$\,^{14}_{7}N$

Work Step by Step

The nuclear equation can be written as $^{238}_{92}U+\,^{A}_{Z}X\rightarrow ^{247}_{99}Es+ 5\,^{1}_{0}n$ where $X$ is the symbol of the bombarding particle, $A$ is its mass number and $Z$ is the charge. Since sum of the masses and the charges should be equal on both the sides, we have $238+A=247+5(1)$ and $92+Z=99+5(0)$ $\implies A=14$ and $Z=7$ The particle with $Z=7$ is $N$. Therefore, the bombarding particle is $\,^{14}_{7}N$.
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