Fundamentals of Physics Extended (10th Edition)

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

Chapter 21 - Coulomb's Law - Problems - Page 629: 73b

Answer

$v=1.09 \times 10^6 m/s$

Work Step by Step

The radius of second smallest orbit is $r_2 = 2^2a_o = 4(5.292 \times 10^{-11} m) = 2.12 \times 10^{-10} m $ $v= \sqrt{\frac{k|e|^2}{m_er} }$ $v= \sqrt{\frac{(8.99 \times 10^9 N . m^2 / C^2) (1.6 \times 10^{-19}C)^2}{(9.11 \times 10^{-31} kg )(2.12 \times 10^{-10} m)} }$ $v=1.09 \times 10^6 m/s$
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