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 200 nm. Another object $O_2$ has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source 01 to that of source 02 is

• A. 1 : 81
• B. 1 : 9
• C. 9 : 1
• D. 81 : 1

Answer 1

The correct option is D 81 : 1

The wavelength of peak of blackbody spectrum is related to the temperature by:

${\lambda }_{max}T=k$

where $k$ is a constant.

Hence ${\lambda }_1{T}_1=k$

and ${\lambda }_2{T}_2=k$

Power emitted per unit area $\propto T^4$

Therefore $\frac{I_1}{I_2}=\frac{T_1^4}{T_2^4}$ $=\frac{{\lambda }_2^4}{{\lambda }_1^4}$ $=81$