Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 24 - Electric Potential - Problems - Page 716: 100

Answer

The least center-to-center distance the alpha particle will be from the target nucleus is $~~8.8\times 10^{-14}~m$

Work Step by Step

We can assume that the initial electric potential energy of the system is zero. The alpha particle can reach a distance such that the electric potential energy of the system is equal to the initial kinetic energy of the system. We can find the distance $r$: $K = U$ $K = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{r}$ $r = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{K}$ $r = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~\frac{(92)(1.6\times 10^{-19}~C)(2)(1.6\times 10^{-19}~C)}{0.48\times 10^{-12}~J}$ $r = 8.8\times 10^{-14}~m$ The least center-to-center distance the alpha particle will be from the target nucleus is $~~8.8\times 10^{-14}~m$.
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