The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm . The radiation intensity for the star is : (Stefan’s constant `5.67 xx 10^(-8)Wm^(-2)K^(-4)`, constant `b = 2898 mu m K`)-

Options

A. `5.67xx10^(8)Wm^(-2)`
B. `5.67xx10^(-12)Wm^(-2)`
C. `10.67xx10^(-7)Wm^(-2)`
D. `10.67xx10^(-14)Wm^(-2)`

Answer

A

Explanation

Here, `lamda_(m)=289.8nm= 289.8xx10^(-9)m`
`sigma=5.67xx10^(-8)Wm^(-2)K^(-4)`
`b=2889mumK=2889xx10^(-6)mK`
If T is temperature of star , then
according to Wien 's law `lamdamT=b`
`T=(b)/(lamdam)=(2889xx10^(-6))/(289.8xx10^(-9))=10^(4)`
From stefan 's law `E=sigmaT^(4)` ltbrge `5.67xx10^(-8)(10^(4))^(4)`
`=5.67xx10^(8)Wm^(-2)`