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Question: The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm. The radiation intens...

The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm. The radiation intensity of the star is (Stefan’s constant −5.67×10−8Wm−2K−4- 5.67 \times 10^{- 8}Wm^{- 2}K^{- 4}Wien’s constant b =2898 μmK\mu mK)

A

5.67×108Wm−25.67 \times 10^{8}Wm^{- 2}

B

5.67×1012Wm−25.67 \times 10^{12}Wm^{- 2}

C

10.67×107Wm−210.67 \times 10^{7}Wm^{- 2}

D

10.67×1014Wm−210.67 \times 10^{14}Wm^{- 2}

Answer

5.67×108Wm−25.67 \times 10^{8}Wm^{- 2}

Explanation

Solution

According to Wien’s displacement law,

λmax⁡\lambda_{\max}

∴T=bλmax⁡2898μmK289.8nm\therefore T = \frac{b}{\lambda_{\max}\frac{2898\mu mK}{289.8nm}}

=2898×10−6mK289.8×10−9m=104K= \frac{2898 \times 10^{- 6}mK}{289.8 \times 10^{- 9}m} = 10^{4}K

According to Stefan Boltzmann law,

Radiation intensity,R=σT4R = \sigma T^{4}

R=(5.67×10−8Wm−2K−4)(104K)4=5.67×108Wm−2R = (5.67 \times 10^{- 8}Wm^{- 2}K^{- 4})(10^{4}K)^{4} = 5.67 \times 10^{8}Wm^{- 2}