Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 37 - The Foundation of Modern Physics - Exercises and Problems - Page 1081: 7


$\lambda = 2.42~\mu m$

Work Step by Step

We can find the temperature of the cube: $P = e~\sigma~A~T^4$ $T^4 = \frac{P}{e~\sigma~A}$ $T = \sqrt[4] {\frac{P}{e~\sigma~A}}$ $T = \sqrt[4] {\frac{630~W}{(1)~(5.67\times 10^{-8}~W~m^{-2}~K^{-4})~(6)~(3.0\times 10^{-2}~m)^2}}$ $T = \sqrt[4] {2.057613169\times 10^{12}~K^4}$ $T = 1197.7~K$ We can use Wien's law to find the peak wavelength: $\lambda = \frac{2.9\times 10^6~nm~K}{T}$ $\lambda = \frac{2.9\times 10^6~nm~K}{1197.7~K}$ $\lambda = 2.42\times 10^{3}~nm$ $\lambda = 2.42\times 10^{-6}~m$ $\lambda = 2.42~\mu m$
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