University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 43 - Nuclear Physics - Problems - Exercises - Page 1477: 43.52

Answer

4.69 MeV.

Work Step by Step

Use the atomic masses given to calculate the mass defect of $0.005121u$, the total masses on the left minus the total masses on the right. The positive sign means there was more mass before. This equals an energy release of 4.77 MeV of kinetic energy. Following Example 43.5, conservation of linear momentum tells us how the decay partners share the KE. The $\alpha$ particle has $\frac{226}{230}$ of the kinetic energy, or 4.69 MeV.
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