An indirectly heated filament is radiating maximum energy of wavelength `2.16xx10^-5cm`. Find the net amount of heat energy lost per second per unit area, the temperature of the surrounding air is `13^@C`. Given `b=0.288cm-K.sigma=5.77xx10^-5erg//s-cm^(2)-K^(4)`).

Explanation

Here, `lambda_(m) = 2.16 xx 10^(-5)cm`
`E = ? , T_(0) = 13^(@)C = 13+273 = 286K`
`b= 0.288 cm K`,
From Wien's displacemet law,
`T = (b)/(lambda_m) = (0.288)/(2.16 xx 10^(-5)) = 13333.3K`
As `E = sigma (T_(4)-T_(0)^(4))`
`:. E = 5.77 xx 10^(-5) [(13333.3)^(4)-(286)^(4)]erg s^(-1) cm^(-2)`
`=18.24 xx 10^(8) Js^(-1)m^(-2)`.