Two spherical bodies A (radius 6cm) and B (radius 18cm) are at temperature `T_(1)` and `T_(2)` respectively The maximum intensity in the emission spectrum of A is at `500 nm` and in that of `B` is at `1500nm` considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of `B` . ?

Options

A. 9
B. 12
C. 3
D. 6

Answer

A

Explanation

Accordding to Wien.s displacement law, `lamda_(m)T=` constant
`:. (lamda_(m))_(A) T_(A) = (lamda_(m))_(B) (T_(B))`
or `(T_(A))/(T_(B)) = ((lamda_(m))_(B))/((lamda_(m))_(A)) = (1500nm)/(500nm) or (T_(A))/(T_(B)) =3` …(i)
According to Stefan Boltzmann law, rate of energy radiated by a black body
`E= sigma AT^(4) = sigma4pi R^(2) T^(4)` [Here, `A= 4pi R^(2)`]
`:. (E_(A))/(E_(B)) = ((R_(A))/(R_(B)))^(2) ((T_(A))/(T_(B)))^(4) = ((6cm)/(18cm))^(2) (3)^(4) = 9` (Using (i))