Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 28 - The Electric Potential - Exercises and Problems - Page 836: 52

Answer

$4\times 10^7\;\rm m/s$

Work Step by Step

Let's assume that the system [the proton+ iron nucleus] is isolated. So the energy is conserved. $$E_i=E_f$$ $$K_i+U_i=K_f+U_f$$ We know that the iron nucleus remains constant when the proton is fired and when the proton is on its surface since it reaches it at a speed of 0 m/s. Hence, $$\frac{1}{2}m_pv_i^2+U_i=0+U_f$$ initially, the proton was very far away $r\rightarrow \infty$, and when the proton on the surface of the nucleus $r=r_{nucleus}$. Let's assume that the proton is dimensionless as a point charge. $$\frac{1}{2}m_pv_i^2+0= \dfrac{k(e)(26e)}{r_{nucleus}}$$ $$ v_i^2 = \dfrac{2k (26e^2)}{m_pr_{nucleus}}$$ $$ v_i = \sqrt{\dfrac{2k (26e^2)}{m_pr_{nucleus}}}$$ Plug the known; $$ v_i = \sqrt{\dfrac{2(9\times 10^9) (26)(1.6\times 10^{-19})^2}{(1.67\times 10^{-27}) (4.5\times 10^{-15})}}$$ $$v_i=\color{red}{\bf 4\times 10^7}\;\rm m/s$$
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