Two bodies A and B have thermal emissivities of 0.01 and 0.081 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength lambda_(B) corresponding to maximum spectral radiancy in the radiation from B is shifted radiancy in the radiation from Aby 1.00 mu m. If the temperature of A is 5802 K, calculate (a) the temperature of B and (b) wavelength lambda_(B)

Two bodies A and B have thermal emissivities of 0.01 and 0.081 respect

1.	Two bodies A and B have thermal emissivities of 0.01 and 0.081 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength lambda_(B) corresponding to maximum spectral radiancy in the radiation from B is shifted radiancy in the radiation from Aby 1.00 mu m. If the temperature of A is 5802 K, calculate (a) the temperature of B and (b) wavelength lambda_(B)
Answer» Solution :`e_(A) = 0.01, e_(B) = 0.81, T_(A) = 5802 K` `\|lambda_(B) - lambda_(A)\| = 10^(6) m` (a) Total radiant POWER is same `rho = e_(A) sigma A T_(A)^(4) = e_(B) sigma A T_(B)^(4)` `e_(A) T_(A)^(4) = e_(B) T_(B)^(4)` `T_(B) = ((e_(A))/(e_(B)))^((1)/(4)) T_(A) = ((0.01)/(0.81))^((1)/(4)) (5802)` ` = (1)/(3) xx 5802` `= 1934 K` (b) According to Wien's LAW `lambda_(A) T_(A) = lambda_(B) T_(B)` `lambda_(A) xx 1934 = lambda_(B) xx 5802` `lambda_(A) = 3 lambda_(B)` `lambda_(A) - lambda_(B) = 10^(-6)` `3lambda_(B) - lambda_(B) = 10^(-6)` `lambda_(B) = 0.5 xx 10^(-6) m = 0.5 mu m` `lambda_(A) = 3 lambda_(A) = 1.5 mu m`