Answer
$$28.3 \mathrm{\ MeV}
$$
Work Step by Step
The total binding energy is
$$
\Delta E_{\mathrm{be}}=\Delta E_{1}+\Delta E_{2}+\Delta E_{3}=19.8 \mathrm{MeV}+6.26 \mathrm{MeV}+2.23 \mathrm{MeV}=28.3 \mathrm{MeV}
$$
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 1303: 22d
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