Two bodies A and B ahave thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power of the same rate. The wavelength `lambda_B` corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by `1.00mum.` If the temperature of A is 5820K:

Options

A. the temperature of B is 1934K
B. `lambda_B=1.5mum`
C. the temperature of B is 11604K
D. the temperature of B is 2901K

Answer

A, B

Explanation

(a,b)
Energy emitted per second by body `A=epsilon_AsigmaT_A^4A`
Energy emitted per second by body `B=epsilon_BsigmaT_B^4A`
Given that power radiated are equal `epsilon_AsigmaT_A^4A=epsilon_BsigmaT_B^4A`
`rArrT_B=((epsilon_A)/(epsilon_B))^(1//4)xxT_A=1934K`
According to Wein's displacement law `lambda_mprop1/T`
Since temperature of A is more therefore `(lambda_m)_A` is less
`:. (lambda_m)_B-(lambda_m)_A=1xx10^-6m (given).....(i)`
Also according to Wein's displacement law
`(lambda_m)_AT_A=(lambda_m)_BT_B`
`rArr((lambda_m)_A)/((lambda_m)_B)=(T_B)/(T_A)=5802/1934.....(ii)`
On solving (i) and (ii), we get
`lambda_b=1.5xx10^-6m.`