Question

The spectral emissive power $E_{\lambda }$ for a black body at temperature $T_1$ is plotted against the wavelength and area under the curve is found to be $A$. At a different temperature $T_2$, the area is found to be $9A$. Than ${\lambda }_1/{\lambda }_2$

• A. $3$
• B. $1/3$
• C. $1/\sqrt{3}$
• D. $\sqrt{3}$

Answer 1

The correct option is D $\sqrt{3}$

Area $=\int ydx=\int \frac{dE}{d\lambda }\times d\lambda =\int dE$
Area $\left(A\right)=E=\sigma T^4=\sigma {\left(\frac{b}{\lambda }\right)}^4$ {Since black body is mentioned in the question}
$\frac{{\text{Area}}_1}{{\text{Area}}_2}={\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^4\Rightarrow \frac{1}{9}={\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^4$
$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{3}$