First slide

Radiation

Question

The emission spectrum of a black body at two different temperatures are shown by curyes P and Q (as shown in figure). The ratio of the areas
under the two curves P and Q will be

Moderate

A

1 : 16
B

4 : 9
C

81 : 256
D

16 : 81

Solution

Absolute temperatures of the black body corresponding to curve P and Q are in the inverse ratio of $λ_{m}$ (Wein's displacement law).
$i . e ., \frac{T_{P}}{T_{Q}} = \frac{1987}{2980}$
Area under curves represent the total power radiated by a body and is proportional to the fourth power of absolute temperature (Stefan's law)
$∴ \frac{A_{p}}{A_{Q}} = {(\frac{T_{P}}{T_{Q}})}^{4} = \frac{16}{81}$

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