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Question:

The black body spectrum of an object O_1 is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200nm. Another object O_2 has the maximum radiant intensity at 600nm. The ratio of power emitted per unit area by source O_1 to that of source O_2 is?

1:9

1:81

9:1

81:1

Solution:

Correct option is D. 81:1
The wavelength of peak of blackbody spectrum is related to the temperature by:
λmaxT = k
where k is a constant
Hence λ1T1 = k
λ2T2 = k
Power emitted per unit area ∝ T4
∴I1/I2 = T14/T24 = (λ2/λ1)4 = (600/200)4 = 34 = 81/1 = 81:1