Question

The spectral emissive power $E_{\lambda }$ for a 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 $9A$. Then $\frac{{\lambda }_1}{{\lambda }_2}=$

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

Answer 1

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

The area under the black body curve gives the total power radiated $\left(P\right)$.

From Stefan - Boltzmann law,
$P=e\sigma A{T}^4$ therefore, $T^4\propto \frac{1}{A}$

$P_1\propto T_1^4\propto 9A$
$P_2\propto T_2^4\propto A$
$\Rightarrow \frac{T_1}{T_2}=\frac{\sqrt{3}}{1}$

From Wein's displacement law,

${\lambda }_1{T}_1={\lambda }_2{T}_2$
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{T_2}{T_1}=\frac{1}{\sqrt{3}}$