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: 10

Answer

a) The Brackett and Paschen series overlap. b) No overlap of Balmer series with Paschen series.

Work Step by Step

(a) We know that for $n_1=3$, $n=\infty$ and $n_1=4$ and $n=\infty$, we have $\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{3^2})-\frac{1}{\infty^2}}=8.204\times 10^{-7}m=820nm$ Similarly $\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{4^2}-\frac{1}{\infty^2})}=14.585\times 10^{-7}m=1458.5nm$ Thus, $820nm\lt \lambda_{Paschen}\lt 1875nm$ $\space 1458.5nm\lt \lambda_{Brackett}\lt 4051nm$ We conclude that the Brackett and Paschen series overlap. (b) We know that for Balmer series and Paschen series $\lambda_3=\frac{1}{(1.097\times 10^7)(\frac{1}{2^2}-\frac{1}{3^2})}=0.6563\mu m$ and $\lambda_4=\frac{1}{(1.097\times 10^7)(\frac{1}{3^2}-\frac{1}{4^2})}=1.875\mu m$ Now, we check the above both scenarios for $n=\infty$ $\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{2^2}-\frac{1}{\infty^2})}=364.6nm$ and $\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{3^2}-\frac{1}{\infty^2})}=0.8204\mu m$ Thus, $364.6nm\lt \lambda_{Balmer}\lt 656.33nm \space 820.4nm\lt \lambda_{Paschen}\lt 1875nm$ We conclude that there is no overlap of Balmer series with Paschen series.
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