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. `5.67 xx 10^(8) W m^(-2)`
B. `5.67 xx 10^(12) W m^(-2)`
C. `10.67 xx 10^(7) W m^(-2)`
D. `10.67 xx 10^(14) W m^(-2)`

Answer

A

Explanation

According to Wien's displacment law,
`lambda_(max) T = b`
`:. T = (b)/(lambda_(max)) = (2898 mu mK)/(289.8 nm)`
`= (2898 xx 10^(-6) mK)/(289.8 xx 10^(-9) m) = 10^(4) K`
According to Stefan Boltzmann law,
Radiation intensity, `E = sigma T^(4)`
`E = (5.67 xx 10^(-8) Wm^(-2) K^(-4))(10^(4) K)^(4)`
`= 5.67 xx 10^(8) W m^(-2)`