709. Problem 49.18 (RHK) Given that an ideal radiator has a spectral radiancy at 400 nm that is 3.50 times its spectral radiancy at 200 nm. (a) We have to find the temperature of the radiator. (b) We have to calculate the temperature if the spectral radiancy of the radiator at 200 nm were 3.5 times its spectral radiancy at 400 nm. 
Solution: Click For PDF Version According to Planck’s radiation law the spectral radiancy of a radiator at temperature T and wavelength is given by the following function: where the Boltzmann constant , and the Planck constant
(a) It is given that at temperature T the radiancy at and at are related as follows: . We therefore have the equation The above equation readily simplifies to the form Root is unphysical. We find temperature T corresponding to the root (b) We next solve the second part of the problem. The condition is We once again make the substitution For this case, we solve the following equation for finding the temperature of the radiator: We select the physical root
