Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits ${10}^4$ times the power emitted from B. The ratio $\left({\lambda }_A/{\lambda }_B\right)$ of their wavelengths ${\lambda }_A$ and ${\lambda }_B$ at which the peaks occur in their respective radiation curves is

According to Wien's displacement law, ${\lambda }_A=T_A={\lambda }_B=T_B$

where $T_A$ and $T_B$ are the temperatures of stars A and B. Thus

$\frac{{\lambda }_A}{{\lambda }_B}=\frac{T_B}{T_A}$

According to Stefan's law, the energy radiated per second from the surface of A and B are

$E_A=\sigma T_A^4{A}_A^2 and E_B=\sigma T_B^4{A}_B^2$

where $A_A=$ and $A_B=$ are their respect Therefore
$\frac{E_A}{E_B}={\left(\frac{T_A}{T_B}\right)}^4\times {\left(\frac{R_A}{R_B}\right)}^2$
Given and Substituting these values, we have values, we have ${10}^4={\left(\frac{T_A}{T_B}\right)}^4\times (400{)}^2\Rightarrow \mathrm{ }\frac{T_A}{T_B}=\frac{1}{2}$

Using (ii) in (i), we get $\frac{{\lambda }_A}{{\lambda }_B}=2$