College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 27 - Problems - Page 1043: 45


$a_0 = 5.29\times 10^{-11}~m$

Work Step by Step

We can calculate the value of $a_0$: $a_0 = \frac{\hbar^2}{m_e~k~e^2}$ $a_0 = \frac{(1.0546\times 10^{-34}~J~s)^2}{(9.1\times 10^{-31}~kg)(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)^2}$ $a_0 = 5.29\times 10^{-11}~m$
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