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

Answer

C

Explanation

Given maximum wavelength
` lamda _ m = 289.8 nm = 298.8 xx 10 ^( - 9) m `
` = 2898 xx 10 ^( - 10) m `
Stefan's constant ` sigmas = 5.67 xx 10 ^( - 8 ) Wm ^ ( - 2) K ^( - 4 ) `
Wien's constant , b ` = 2898 mu m K = 2898 xx 10 ^( - 6 ) ` mK
According to Wien's displacement law, the maximum wavelength is given by
` lamda _ m = ( b ) / ( T ) rArr T = ( b ) / ( lamda _ m ) " " `...(i)
Substituting given values in Eq. (i), we get
`T = ( 2898 x 10 ^( - 6 ) ) /( 2898 xx 10 ^( - 10) ) = 10 ^ 4 K " " `... (ii)
According to Stefan's law, the energy radiated from a source is given by
` E = sigma Ae T ^ ( 4 ) " " `...(iii)
where, A = area of source
e = emissivity ( value between 0 to 1 )
The intensity of radiations emitted is equal to energy radiated from a given surface area, i.e.,
`I = ( E ) /( A) = sigma e T ^ (4 ) " " ` [from Eq. (iii) ]
As e is very small , so
` I = sigma T^ 4 " " `...(iv)
Substituting the value of T from Eq. (ii) in Eq. (iv), we get
`I = sigma ( 10 ^ 4 ) ^ 4 = 5.67 xx 10 ^( - 8 ) xx 10 ^( 16)`
` [ because sigma = 5.67 xx 10 ^( -8 ) ` (given)]
` = 5.67 xx 10 ^( 8 ) Wm^( - 2) `