College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1043: 46


$E_1 = -13.6~eV$

Work Step by Step

We can calculate the value of $E_1$: $E_1 = \frac{-m_e~k^2~e^4}{2\hbar^2}$ $E_1 = \frac{-(9.1\times 10^{-31}~kg)(9.0\times 10^9~N~m^2/C^2)^2(1.6\times 10^{-19}~C)^4}{(2)(1.0546\times 10^{-34}~J~s)^2}$ $E_1 = (-2.1717\times 10^{-18}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $E_1 = -13.6~eV$
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