Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 38 - Quantization - Exercises and Problems - Page 1115: 28


The radius of the atom is $~~1.90~nm$

Work Step by Step

We can find the state $n$: $E_n = \frac{13.6~eV}{n^2}$ $n^2 = \frac{13.6~eV}{E_n}$ $n = \sqrt{\frac{13.6~eV}{E_n}}$ $n = \sqrt{\frac{13.6~eV}{0.378~eV}}$ $n = 6$ We can find the radius: $r_n = n^2~a_B$ $r_6 = (6)^2~(0.0529\times 10^{-9}~m)$ $r_6 = 1.90\times 10^{-9}~m$ $r_6 = 1.90~nm$ The radius of the atom is $~~1.90~nm$
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