Answer
$E_{B} = -28.35 MeV$
Work Step by Step
The binding energy of $\alpha $ particle is
$E_{B} = \Delta E_2 - \Delta E_1 $
$E_{B} = (-24.1 MeV) - 4.25 MeV$
$E_{B} = -28.35 MeV$
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 1305: 48d
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