Answer
$\Delta_{120} = -0.097803~u$
Work Step by Step
We can find the mass excess of $^{120} Sn$:
$\Delta_{120} = M - A$
$\Delta_{120} = 119.902 197~u - 120~u$
$\Delta_{120} = -0.097803~u$
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 1302: 10e
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