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)

$\frac{E_{1}}{E_{2}}=\frac{1}{9}=\left(\frac{T_{1}}{T_{2}}\right)^{4}$
$\lambda _{1}T_{1}=\lambda _{2}T_{2}\Rightarrow \frac{T_{1}}{T_{2}}=\frac{\lambda _{2}}{\lambda _{1}}$
$\frac{1}{9}=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{4} \Rightarrow \frac{\lambda_{2}}{\lambda_{1}}=\frac{1}{\sqrt{3}}$
$\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{3}$