A black body emits radiation at the rate P when its temp. is T. At this temperature the wavelength at which the radiation has maximum intensity is λ. If at another temperature T' the power radiated is P' and wavelength at maximum intensity is λ/2 then?

The options given are-
1) P'T'=32PT
2) P'T'=16PT
3)P'T'=8PT
P'T'=4PT

1 Answer
A08
Feb 21, 2018

This is what I get

Explanation:

From Wien's Displacement Law we know that for a radiating body the product of maximum wavelength radiated #lambda# and its temperature #T# in kelvin is a constant.

#lambdaxxT="Wien's Constant"\ b=2.898xx10^-3\ mK# ....(1)

Also from Stefan-Boltzmann law for radiation from a black body we have Power radiated #P# is

#P=epsilonsigmaAT^4# ......(2)
where #epsilon# is emmissivity of surface which is #=1# for a black body, #sigma# is Stefan's constant and #A# is surface area of the radiating object.

At another temperature #T^'# we have the expression for the black body

#P^'=epsilonsigmaA(T^')^4# .......(4)

Multiplying both sides with #T^'#, we get

#P^'T^'=epsilonsigmaA(T^')^5# ......(5)

From (1) we have

#lambda^'T^'=b#
#=>T^'=b/lambda^'# ......(6)

Inserting given value of wavelength at maximum intensity #lambda^'=lambda/2# in (6) we get

#T^'=b/(lambda/2)#
#=>T^'=2b/(lambda)#

Inserting this in RHS of (5) we get

#P^'T^'=epsilonsigmaA(2b/(lambda))^5#

Rewrite RHS as and then using (1)

#P^'T^'=32[epsilonsigmaA(b/(lambda))^4]xx(b/lambda)#
#P^'T^'=32[epsilonsigmaAT^4]xxT#

Now using (2) we get

#P^'T^'=32PT#

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