College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 27 - Quantum Physics - Learning Path Questions and Exercises - Exercises - Page 935: 3

Answer

wavelength = 10622.71 nm , frequency = 2.82$\times10^{13} hertz$

Work Step by Step

Temperature of Blackbody T = 0 $^{0}C$ Temperature of blackbody in absolute kelvin scale T = temperature in $^{0}C$ + 273 Temperature of blackbody in absolute kelvin scale T = 0 $^{0}C$ + 273 Temperature of blackbody in absolute kelvin scale T = 273 K From Wiens displacement law $\lambda_{max} T = 2.90\times10^{-3} m.K$ $\lambda_{max} = \frac{2.90\times10^{-3} m.K}{T}$ $\lambda_{max} = \frac{2.90\times10^{-3} m.K}{273 K}$ $\lambda_{max} = 0.01062271\times10^{-3}$ m $\lambda_{max} = 10622.71\times10^{-9}$ m $\lambda_{max}$=10622.71 nm since frequency $ f= \frac{c}{\lambda}$ where c is speed of light $ c= 3\times10^{8}$m/s so $ f= \frac{3\times10^{8}m/s}{10622.71\times10^{-9} m}$ $ f= 2.824138\times10^{13}/s$ $ f= 2.824138\times10^{13} hertz$
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