Two bodies A and B have thermal emissivities of 0.01 and 0.81 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 from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.0 `mum`. If the temperature of A is 5802 K, calculate (a) the temperature of B, (b) wavelength `lambda_B`.

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

Energy emitted per second by body `A=epsilon_(A)sigmaT_(A)^(4)A` where `A` is the surface area.
Enenrgy wmitted per second by body `A=epsilon_(B)sigmaT_(B)^(4)A`
Diven that power radiated is equal
`epsilon_(A)sigmaT_(A)^(4)A,epsilon_(A)T_(A)^(4)=epsilon_(B)T_(B)^(4)`
`implies T_(B)=((epsilon_(A))/(epsilon_(B)))^(1//4)T_(A)=1934K`
According to Wien's displacement law `(lambda_(m))prop(1)/(T)`
Since temperature of `A` is more, therefore `(lambda_(m))_(A)` is less
`:. (lambda_(m))_(B)-(lambda_(m))_(A)=(lambda_(m))_(B)T_(B)`
`implies (lambda_(m))_(A)/((lambda_(m))_(B))=(T_(B))/(T_(A))=(1934)/(5802)=(1)/(2)` (ii)
On solving Eqs. (i) and (ii),
we get `lambda_(B)=1.5xx10^(-6)m` .