Physics: Principles with Applications (7th Edition)

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

Chapter 29 - Molecules and Solids - General Problems - Page 855: 39

Answer

a. $3.1\times10^4K$ b. $9.3\times10^2K$

Work Step by Step

Use the relationship that is given. $$KE=\frac{3}{2}kT$$ a. $T = \frac{2(KE)}{3k}=\frac{2(4.0 eV)(1.60\times10^{-19}J/eV)}{3(1.38\times10^{-23}J/K)}=3.1\times10^4K$ b. $T = \frac{2(KE)}{3k}=\frac{2(0.12 eV)(1.60\times10^{-19}J/eV)}{3(1.38\times10^{-23}J/K)}=9.3\times10^2K$
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