Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 28 - Quantum Mechanics of Atoms - General Problems - Page 828: 65

Answer

Copper.

Work Step by Step

Use the equation developed in Example 28-7 (page 818) to calculate the wavelength of the $K_{\alpha}$ line. $$\frac{1}{\lambda}=(1.097\times10^7m^{-1})(Z-1)^2(\frac{1}{(n’)^2}-\frac{1}{n^2})$$ Solve for Z. $$Z=1+\sqrt{\frac{1}{\lambda}(1.097\times10^7m^{-1})^{-1}(\frac{1}{(n’)^2}-\frac{1}{n^2})^{-1}}$$ The $K_{\alpha}$ line is for electrons moving from n = 2 down to n’ = 1. $$Z=1+\sqrt{\frac{1}{0.154\times10^{-9}m}(1.097\times10^7m^{-1})^{-1}(\frac{1}{1^2}-\frac{1}{2^2})^{-1}}$$ $$Z=1+28.1\approx 29$$ The unknown material has Z=29, and corresponds to copper.
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