Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 30 - Nuclear Physics and Radioactivity - Problems - Page 881: 12

Answer

29 MeV.

Work Step by Step

Energy is conserved. In the beginning, when the particles are very far apart, there is KE but no PE. Assume that the uranium nucleus is at rest, and all of the initial KE belongs to the alpha particle. At the end, there is PE but no kinetic energy (we ignore the negligible recoil of the uranium nucleus). $$KE_i=PE_f$$ The two particles end up at a distance equal to the sum of their radii; use this to calculate the final electrical potential energy. Use equation 30–1 to find the radii. $$KE_i=PE_f=k\frac{q_{\alpha}q_{U}}{ r_{\alpha}+r_{U}}$$ $$=(8.99\times10^9 Nm^2/C^2)\frac{(2)(92)(1.60\times10^{-19}C)^2}{(1.2\times10^{-15}m)(4^{1/3}+ 232^{1/3})}$$ $$=4.56\times10^{-12}J$$ This is about 29 MeV.
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