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: 5

Answer

temperature of object= 693.67 $^{0}C$

Work Step by Step

Measured frequency of red hot object $ f= 1.0 \times10^{14} Hz$ wavelength corresponding to above frequency $\lambda=\frac{c}{f}$ speed of light $c=3\times10^{8} m/s$ so $\lambda=\frac{3\times10^{8} m/s}{1.0 \times10^{14} Hz}$ $\lambda=3\times10^{-6} m$ $\lambda=3000\times10^{-9} m$ $\lambda=3000 nm$ assuming this wavelength as wavelength corresponding to maximum emission $\lambda_{max}= 3\times10^{-6} m$ From Wiens displacement law $\lambda_{max} T= 2.90\times10^{-3}m.K$ $T= \frac{ 2.90\times10^{-3}m.K}{\lambda_{max}}$ $T= \frac{ 2.90\times10^{-3}m.K}{ 3\times10^{-6} m}$ $T=0.966666\times10^{3}K$ $T= 966.67 K$ absolute kelvin temperature = temperature in $^{0}C$ + 273 so temperature in $^{0}C$ = absolute kelvin temperature -273 so temperature in $^{0}C$ =966.67 - 273 so temperature in $^{0}C$ =693.67 $^{0}C$
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