Two bodies Aand B of equal surface area have thermal emissivities of 0.01 and 0.81 respectively . The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies A and B at wavelength `lambda _(A) and lambda _(B)`respectively . Difference in these two wavelengths is `1 mu m`. If the temperature of the body A is 5802 K, then value of `lambda_(B)` is

Options

A. `(1)/(2) mu m`
B. `1 mu m`
C. `2 mum`
D. `(3)/(2) mu m`

Answer

D

Explanation

`lambda_(B)`= ?
According to Stefan's law, `E = e A sigma T^(4)`
Here , `E_(A) =E_(B)`
`e_(A) xx T_(A)^(4) = e_(B) xx T_(B)^(4)`
(because A and B has equal surface area ` A_(A) =A_(B), sigma_(A) - sigma _(B))`
`(0.01) xx (5802)^(4) = 0.81 xx T_(B)^(4) `
`T_(B) = 1934 K`
From Wein's law ,`lambda T` = constant
`lambda _(A) T_(A) = lambda _(B) T_(B)`
`(lambda _(A))/(lambda_(B)) = (T_(B))/(T_(A))`
`1-(lambda_(A))/(lambda_(B)) = implies 1- (T_(B))/(T_(A))`
`implies (lambda _(B)- lambda_(A))/(lambda_(B)) = (T_(A) -T_(B))/(T_(A))`
`implies (1)/(lambda _(B)) = (5802 - 1934)/(5802)`
` therefore lambda_(B) = (3)/(2) mum`