Question

The spectral emissive power ${\text{E}}_{\lambda }$ for a body at temperature $T_1$ is plotted against the wavelength (see figure) and area under the curve is found to be $A$ . At a different temperature $T_2$ the area is found to be $9A$ . Then ${\lambda }_1/{\lambda }_2=$

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

Answer 1

Answer: (D)

$\frac{E_1}{E_2}=\frac{1}{9}={\left(\frac{T_1}{T_2}\right)}^4$
${\lambda }_1{T}_1={\lambda }_2{T}_2\Rightarrow \frac{T_1}{T_2}=\frac{{\lambda }_2}{{\lambda }_1}$
$\frac{1}{9}={\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^4\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{1}{\sqrt{3}}$
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{3}$