Answer
$Q_1 = 31.8MeV$
Work Step by Step
For the first reaction,
$^{223}Ra \longrightarrow ^{209}Pb + ^{14}C $
$Q_1 = (m_{Ra} - m_{Pb} - m_c) c^2$
$Q_1 = (223.01850u - 208.98107u - 14.00324u) (931.5 MeV/u) $
$Q_1 = 31.8MeV$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 42 - Nuclear Physics - Problems - Page 1305: 52a
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