Answer
b) -2.42$\times10^{-19} J$
Work Step by Step
Using equation $E_{n} = -R_{H}(\frac{1}{n^{2}})$ ,
We get $E_{3} = -2.18\times10^{-18}(\frac{1}{9}) = -2.42\times10^{-19} J$.
Chemistry (4th Edition)
by Burdge, Julia
Published by
McGraw-Hill Publishing Company
ISBN 10:
0078021529
ISBN 13:
978-0-07802-152-7
Chapter 6 - Section 6.3 - Bohr’s Theory of the Hydrogen Atom - Checkpoint - Page 250: 6.3.1
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