Answer
$ _{94}^{239}Pu + 2 \space _{0}^{1}n → _{95}^{241}Am + _{-1}^{0}e$
Work Step by Step
Pu-239 reacts with 2 neutrons to form Am-241 with the release of a beta particle.
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 18 - Nuclear Chemistry - Questions for Review and Thought - Topical Questions - Page 817c: 45
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