Question

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. $32,000K$
• B. $16,000K$
• C. $8,000K$
• D. $4,000K$

Answer 1

The correct option is D $4,000K$

$\frac{A_T}{A_{2000}}=\frac{16}{1}\left(given\right)$

Area under $e_{\lambda }-\lambda$ curve represents the e missive power of body and e missive power $\propto T^4$

(Hence area under $e_{\lambda }-\lambda$ curve) $\propto T^4$

$\begin{array}{c}\frac{A_T}{A_{2000}}={\left(\frac{T}{2000}\right)}^4=\frac{16}{1}\\ {\left(\frac{T}{2000}\right)}^4=T\\ \therefore T=4000k\end{array}$