Two stars A and B of same size, have thermal emissivities of 0.2 and 0.64 respectively. Both stars emit total radiant power at same rate. If the temperature of A is 5000 K and the wavelength `lamda_(B)` corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in radiations from A by 2.0 `mu`m, then find the temperature of star B and wavelength `lamda_(B)`.

Explanation

According to Stefan's law the power radiated by a body is given by
`P=epsi sigmaAT^(4)`
Now, according to given problem, both stars emit total radiant power at same rate
`thereforeP_(A)=P_(B)`
Also both have same size `thereforeA_(A)=A_(B)`
`therefore epsi_(A)T_(A)^(4)=epsi_(B)T_(B)^(4)`
or `0.2(5000)^(4)=(0.64)T_(B)^(4)`
`T_(B)^(4)=(0.2)/(0.64)(5000)^(4)`
`T_(B)=5000[(0.2)/(0.64)]^(1//4)=5000[(1)/(3.2)]^(1//4)`
`=2500[5]^(1//4)=3740K`
Now accoring to Wien's displacement law
`lamda_(A)T_(A)=lamda_(B)T_(B)`
or `lamda_(B)=(5000)/(3740)xxlamda_(A)`
or `lamda_(B)=1.337lamda_(A)`
also `lamda_(B)-lamda_(A)=2mum`
`thereforelamda_(B)-(lamda_(B))/(1.337)=2mum`
`thereforelamda_(B)=((1.337)/(0.337))2mum`
`lamda_(B)=7.93mum`