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 $\frac{{\lambda }_1}{{\lambda }_2}$ is :

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

Answer 1

Answer: (D)

$\sqrt{3}$

The spectral emissive power is directly proportional to the fourth power of the absolute temperature.
$E_{\lambda }=k{T}^4$
$E_1=k{T}_1^4=A$
$E_2=k{T}_2^4=9A$
$\frac{E_2}{E_1}=(\frac{T_2}{T_1}{)}^4=9$
$\frac{T_2}{T_1}=\sqrt{3}$
$T_2=\sqrt{3}{T}_1$

According to Wien's displacement law:
$\lambda T=b=constant$
${\lambda }_1{T}_1={\lambda }_2{T}_2$

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