The spectral emissive power `E_(lambda)` for a body at temperature `T_(1)` is poltted againist the wavelenght and area under the curve is found to be `A.At` a different temperature `T_(2)` the area is found to be A then `lambda_(1)//lambda_(2)=`
.

Options

A. 3
B. `1//3`
C. `1//sqrt3`
D. `sqrt3`

Answer

D

Explanation

Area of graph
`=E`
`E alpha T^(4)` and `lambda alpha (1)/(T) rArr E alpha (1)/(lambda^(4))`
Area `alpha (1)/(lambda^(4))` .