College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 13 - Problems - Page 499: 63


$T = 1546~K$

Work Step by Step

We can find the temperature: $\frac{3}{2}~k~T = \overline{KE}$ $T = \frac{2~\overline{KE}}{3~k}$ $T = \frac{(2)(3.20\times 10^{-20}~J)}{(3)(1.38\times 10^{-23}~J/K)}$ $T = 1546~K$
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