GATE-2018 (24) :Wyllie – Time average equation

CSIR-NET 2013

67. If the wavelength of peak emission of black body A is twice that of black body B, then the radiation intensity of A is:

A) 4 times that of B              

B) twice that of B

C) 1/16th of that of B

D) 16 times that of B

solutions

Wien’s displacement law:- states that the black 

body radiation curve for different temperatures peaks

at a wave length inversely proportional to the 

temperature.

λ_max = b / T -----------(1)

Where ,T – absolute temperature in kelvin

b – constant of proportionality called wien’s 

displacement constant( 2.897 * 10^-3 m K)

stefan – Boltzmann law :- 
 

The radiant flux of a black body (f_b) at a kinetic temperature of T_kin is

F_b = σ T ⁴_kin ------------(2)

σ – stefan – Boltzmann constant

( 5.6703 * 10^-8 W / m² K⁴ )

F_b = black body radiant flux

from equation (1) and (2)

F_b = σ T⁴_kin

T⁴_kin = F_b / σ

        = (F_b )^1/4 / σ ---------(3)

equation (3) substitute in equation (1)

λ = b / (( F_b)^1/4 / σ )
 

   = b*σ / (F_b)^1/4

there fore ,

The relation between radient flux and wave length

F_b = 1 / λ⁴ ( all are constants )

Given values ,

     λ_A = 2λ_{B ,}                   λ_A = λ_B

→    F_b(A) = 1 / ( 2λ_B )⁴ ----------(4)

        F_b(B) = 1 / ( λ_B)⁴ ------------(5)

equation (4) / (5)

F_b(A) / F_b(B) = 1 / 16λ_B⁴ / 1 / λ_B⁴ = 1 / 16

There fore ,

F_b(A) = 1 / 16 *F_b(B)

There fore , the radient flux of black body at A is 1 / 16 of B