Question

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 $\left(\frac{{\lambda }_A}{{\lambda }_B}\right)$ of their wavelengths ${\lambda }_A$ and ${\lambda }_B$ at which the peaks occur in their respective radiation curves is

Answer 1

From stephan's law:
$\frac{\Delta Q}{\Delta t}=\sigma eA{T}^4$
Given, $(\frac{\Delta Q}{\Delta t}{)}_A={10}^4(\frac{\Delta Q}{\Delta t}{)}_B$

$\Rightarrow \sigma e{A}_A{T}_A^4={10}^4\sigma e{A}_B{T}_B^4$
$(\frac{T_A}{T_B}{)}^4={10}^4\frac{A_B}{A_A}$
$\frac{T_A}{T_B}=10(\frac{A_B}{A_A}{)}^{\frac{1}{4}}$

where, $A=\pi r^2$
so, $\frac{T_A}{T_B}=10(\frac{\pi R_B^2}{\pi R_A^2}{)}^{\frac{1}{4}}$
Given, $\frac{R_A}{R_B}=400$
$\frac{T_A}{T_B}=10(\frac{1}{{400}^2}{)}^{\frac{1}{4}}$
$\frac{T_A}{T_B}=\frac{10}{20}=\frac{1}{2}$ .....(1)

From Wein's displcament law:
${\lambda }_{m}T=b$
where, $\lambda \propto \frac{1}{T}$

from eq.(1)
$\frac{{\lambda }_A}{{\lambda }_B}=\frac{T_B}{T_A}=2$
Hence, correct answer is 2.