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The energy radiated by a black body at 2...

The energy radiated by a black body at `2300K` is found to have the maximum at `a` wavelength `1260nm`, its emissive power being `8000Wm^(-2)`. When the body is cooled to a temperature `TK`, the emissive power is found to decrease to `500 Wm^(-2)`. Find :
`(i)` the temperature `T`
`(ii)` the wave length at which intensity of emission in maximum at the temperature `T`.

Text Solution

Verified by Experts

`(i)` `lambdam_(1)=1260nm`
`lambdam_(1)=1260xx10^(-9)m`, `T_(1)=2300 K`, `T_(2)=TK`
`(E_(1))/(E_(2))=((T_(1))/(T_(2)))^(4)`
`((E_(1))/(E_(2)))^(1//4)=(T_(1))/(T_(2))`, `T_(2)=T_(1)xx((E_(2))/(E_(1)))^(1//4)`
`T_(2)=T_(1)xx((E_(2))/(E_(1)))^(1//4)`
`T_(2)=2300xx[(500)/(8000)]^(1//4)=2300xx(1)/(2)=1150k`
`T_(2)=1150k` Ans.
`(ii)` by using Wein law
`T_(1)lambda_(m_(1))=T_(2)lambda_(m_(2))`, `lambda_(m_(2))=((T_(1)lambda_(m_(1)))/(T_(2)))`
`lambda_(m_(2))=(2300xx1260xx10^(-9))/(1150)=2250xx10^(-9)M`
`lambda_(m_(2))=2520nm`
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