Answer
$_{83}^{235}Bi$
Work Step by Step
Bi-235 reacts with a neutron to form Tc-137, Zr-97 and 2 neutrons.
$_{83}^{235}Bi + _{0}^{1}n → 2 \space _{0}^{1}n + _{43}^{137}Tc + _{40}^{97}Zr$
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 - General Questions - Page 817d: 70a
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