Question

The spectral emissive power $E$ for a body at temperature $T$ is planted against the wavelength and area under the curve is found to be $A$. At a difference temperature $T_2$ the area is found to be $9A$. Then ${\lambda }_1/{\lambda }_2=$`

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

Answer 1

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

$\frac{E_1}{E_2}=\frac{1}{9}={\left(\frac{T_1}{T_2}\right)}^4$

hence,

$\frac{T_1}{T_2}=\frac{1}{\sqrt{3}}$,

Now,

As we know, From win's formula

$\lambda T=$ constant

Hence

${\lambda }_1/{\lambda }_2=T_2/T_1$

or, $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{3}$