InterviewSolution
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InterviewSolution
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Assuming the spectral distribution of thermal radiation energy of obey wien's formula u(omega,T) = A omega^(3)exp (-1 omega//T), where a = 7.64 ps.K, find for a temperature T = 2000K the most probable (a) radiation frequency, (b) radiation wavelength. |
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Answer» Solution :We are GIVEN `u(omega,T) = AOMEGA^(3) exp(-aomega//T)` (a)Then `(du)/(d omega) = ((3)/(omega)-(a)/(T))u = 0` so `omega_(Pr) = (3T)/(a) = (6000)/(7.64) xx 10^(12)s^(-1)` (b) We determine the spectral distribution in wavelength. `-overset~(u)(lambda,T)d lambda = u(omega,T)d omega` But `omega = (2pic)/(lambda)` or `lambda = (2pic)/(omega) = (C')/(omega)` so `d lambda =- (C')/(omega^(2))d omega, d omega=- (C')/(lambda_(2))d lambda` (we have put a minus sign before `d lambda` to subsume just this fact `d lambda` is -ve where `d omega` is `+ve`). `overset~(u) (lambda, T) = (C')/(lambda^(2))u ((C')/(lambda),T) = (C^(4)A)/(lambda^(5))exp(-(aC')/(lambdaT))` This is maximum when `(deloverset~(u))/(del lambda) =6 overset~(u) [(-5)/(lambda)+(aC')/(lambda^(2)T)]` or `lambda_(Pr) = (aC')/(5T) = (2pica)/(5T) = 1.44mu m` |
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