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. $32000\text{K}$
• B. $16000\text{K}$
• C. $8000\text{K}$
• D. $4000\text{K}$

Answer 1

The correct option is D $4000\text{K}$

$\frac{A_T}{A_{2000}}=\frac{16}{1}$ [Given]
area under $E_{\lambda }-\lambda$ curve represents the emissive power of body
Emissive power $\propto T^4$
(Hence, area under $E_{\lambda }-\lambda$ curve) $\propto T^4$
$\Rightarrow \frac{A_T}{A_{2000}}={\left(\frac{T}{2000}\right)}^4$
$\Rightarrow \frac{16}{1}={\left(\frac{T}{2000}\right)}^4$
$\Rightarrow T=4000\text{K}$