Two substances A and B have relation `T_(A)=4T_(B)` for their absolute zero temperature. The difference in their wavelength is `3.0mum` when maximum radiation energy emit from them. Wavelength of B for maximum radiation energy is . . . . . ..`mum`.

Options

A. 2
B. `2.5`
C. `4.0`
D. `4.5`

Answer

C

Explanation

`T_(A)=4T_(B)`
`:.T_(A)gtT_(B)`
and `(lamda_(m))_(A)T_(A)=(lamda_(m))_(B)T_(B)` from `(lamda_(m))_(B)gt(lamda_(m))_(A)`
also `(lamda_(m))_(B)-(lamda_(m))_(A)=3xxx10^(-6)m`
Now `((lamda_(m))_(B))/((lamda_(m))_(A))=(T_(A))/(T_(B))`
`((lamda_(m))_(B)-(lamda_(m))_(A))/((lamda_(m))_(A))=(T_(A)-T_(B))/(T_(B))`
`=(4T_(B)-T_(B))/(T_(B))`
`(3xx10^(-6))/((lamda_(m))_(A))=3`
`:.(lamda_(m))_(A)=(3xx10^(-6))/(3)`
`=1xx10^(-6)m`
Now `(lamda_(m))_(B)=(lamda_(m))_(A)+3xx10^(-6)m`
`:.(lamda_(m))_(B)=(lamda_(m))_(A)+3xx10^(-6)m`
`=1xx10^(-6)m+3xx10^(-6)m`
`=4xx10^(-6)m`
`=4mum`