1.

Using Planck's formula, find the power radiated by a unit area of a black body within a narrow wavelength interval Delta lambda = 1.0 nm close to the maximum of spectral radiation density at a temperature T = 3000K of the body.

Answer»

Solution :We write the required power in terms of the raadiosity by considering `DEL`only the enegry radiated in the given range. Then form the pervious problem
`DELTAP = (c )/(4)overset~~(u)(lambda_(m),T)Delta lambda`
`= (4pi^(2)c^(2)cancelh)/(lambda_(m)^(5)) (Delta lambda)/(e^(2ch//klambda_(m)T)-1)`
But `lambda_(m)T = b`
so `DeltaP = (4pi^(2)c^(2)cancelh)/(lambda_(m)^(5))(Delta lambda)/(e^(2ch//kb)-1)Delta lambda`
Using the DATA
`(2pic cancelh)/(kb) = (2pixx3xx10^(8)xx1.05xx0^(-34))/(1.38xx10^(-23)xx2.9xx10^(-3)) = 4.9643`
`(1)/(e^(2pich//kb)-1) = 7.03xx10^(-3)`
and `DeltaP = 0.312W//cm^(2)`


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