College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1042: 39


$\lambda = 1.10\times 10^{-6}~m$

Work Step by Step

We can find the energy of the radiation that is emitted: $E = E_6-E_3$ $E = \frac{E_0}{6^2}-\frac{E_0}{3^2}$ $E = E_0~(\frac{1}{6^2}-\frac{1}{3^2})$ $E = (-13.6~eV)~(-\frac{1}{12})$ $E = 1.13~eV$ We can find the wavelength of the radiation: $E = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.13~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda = 1.10\times 10^{-6}~m$
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