Two black bodies A and B at temperatures 5802 K and 1934 K emits total radiations at the same rate. The wavelength `lamda_(B)` corresponding to maximum spectral radiancy from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by `1.00mum`. Then

Options

A. `lamda_(A)=3/2mum`
B. `lamda_(B)=0.5mum`
C. `lamda_(B)=3/2mum`
D. `lamda_(B)=3mum`

Answer

C

Explanation

According to Wien's displacement law,
`lamda_(m)`T = constant
`thereforelamda_(A)T_(A)=lamda_(B)T_(B)`
`thereforelamda_(A)=lamda_(B)*[T_(B)/T_(A)]=lamda_(B)xx5802/1934=lamda_(B)/3`
But `lamda_(B)-lamda_(A)=1mumorlamda_(B)-lamda_(B)/3=2/3lamda_(B)=1mum`
`thereforelamda_(B)=3/2mumandlamda_(A)=0.5mum`