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 $(\frac{λ_{A}}{λ_{B}})$ of their wavelengths $λ_{A}$ and $λ_{B}$ at which the peaks occur in their respective radiation curves is

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Solution

From Stephan's law:
$\frac{Δ Q}{Δ t} = σ e A T^{4}$

Given $(\frac{Δ Q}{Δ t})_{A} = 10^{4} (\frac{Δ Q}{Δ t})_{B}$

$σ e A_{A} T_{A}^{4} = 10^{4} σ e A_{B} T_{B}^{4}$

${(\frac{T_{A}}{T_{B}})}^{4} = 10^{4} \frac{A_{B}}{A_{A}}$

$\frac{T_{A}}{T_{B}} = 10 {(\frac{A_{B}}{A_{A}})}^{\frac{1}{4}}$

$\frac{T_{A}}{T_{B}} = 10 {(\frac{π R_{B}^{2}}{π R_{A}^{2}})}^{\frac{1}{4}}$

Given $\frac{R_{A}}{R_{B}} = 400$
$\frac{T_{A}}{T_{B}} = 10 {(\frac{1}{400^{2}})}^{\frac{1}{4}}$

$\frac{T_{A}}{T_{B}} = \frac{10}{20} = \frac{1}{2}$

From wein's displcament law
$λ_{m} T = b$

$λ \propto \frac{1}{T}$

$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}} = 2$
Hence, correct answer is 2.

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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 $(\frac{λ_{A}}{λ_{B}})$ of their wavelengths $λ_{A}$ and $λ_{B}$ at which the peaks occur in their respective radiation curves is