Q. The adjoining diagram shows the spectral energy density distribution $E_{\lambda}$ of a black body at two different temperatures. If the areas under the curves are in the ratio 16 : 1, the value of temperature $T$ is
A) $32,000\ K$
B) $16,000\ K$
C) $8,000\ K$
D) $4,000\ K$
Explanation
(d) $\frac{A_{T}}{A_{2000}} = \frac{16}{1}$ [Given]
Area under $e_{\lambda} - \lambda$ curve represents the emissive power of body and emissive power $\propto T^{4}$
[Hence area under $e_{\lambda} - \lambda$ curve) $\propto T^{4}$
$\Rightarrow \frac{A_{T}}{A_{2000}} = \left( \frac{T}{2000} \right)^{4} \Rightarrow \frac{16}{1} = \left( \frac{T}{2000} \right)^{4} \Rightarrow T = 4000K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.