Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1112: 6

Answer

7

Work Step by Step

We know that $\frac{1}{\lambda}=\frac{1}{400nm}(R)(\frac{1}{2^2}-\frac{1}{n^2})$ $\frac{1}{n^2}=\frac{1}{2^2}-\frac{1}{(400nm)(R)}$ $\frac{1}{n^2}=\frac{1}{4}-\frac{1}{(400\times 10^{-9})(1.097\times 10^7)}$ $\frac{1}{n^2}=0.0221$ $n=\sqrt{\frac{1}{0.0221}}=7$
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