Two spherical starts A and B emit black body radiation. The radius of A is 400 times that of B and A emits `10^(4)` times the power emitted from B. The ratio `(lambda_(A)//lambda_(B))` of their wavelengths `lambda_(A)` and `lambda_(B)` at which the peaks oc cur in their respective radiation curves is :

Answer

2

Explanation

Here, `R_(A) =400 R_(B), E_(A)=10^(4) E_(B)`
as `E =sigma T^(4)A = sigma T^(4) 4 pi R^(2) , i.e., E prop T^(4) R^(2)`
`:. (E_(A))/(E_(B))=(T_(A)^(4)R_(A)^(2))/(T_(B)^(4)R_(B)^(2))` or `10^(4) = ((T_(A))/(T_(B)))^(4) xx(400)^(2)`
or `(T_(A))/(T_(B))=[(10^(4))/((400)^(2))]^(1/4) = (1)/(2)`
From Wien's displacement law
`(lambda_(A))/(lambda_(B)) = (T_(B))/(T_(A)) = (2)/(1)`