Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1152: 24

Answer

$1.38Mev$

Work Step by Step

First of all, we find the mass difference as $\Delta m=m_f-m_i$ $\implies \Delta m=210.98726u-210.9887u=-0.00148u$ Now we can find the energy released as: $E=|\Delta m|c^2$ We plug in the known values to obtain: $E=0.00147(931.5Mev/c^2)c^2=1.38Mev$
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