Subjective Type

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 $$\displaystyle \left ( \frac{\lambda_A}{\lambda_B} \right )$$ of their wavelengths $$\lambda_A$$ and $$\lambda_B$$ at which the peaks occur in their respective radiation curves is

Solution

From stephan's law:
$$\dfrac{\Delta Q}{\Delta t}= \sigma e A T^4$$

Given $$(\dfrac{\Delta Q}{\Delta t})_A=10^4(\dfrac{\Delta Q}{\Delta t})_B$$
$$ \sigma e A_A T_A^4=10^4\sigma e A_B T_B^4$$
$$(\dfrac{T_A}{T_B})^4=10^4\dfrac{A_B}{A_A}$$
$${\dfrac{T_A}{T_B}}=10(\dfrac{A_B}{A_A})^\dfrac{1}{4}$$
$${\dfrac{T_A}{T_B}}=10(\dfrac{\pi R_B^2}{\pi R_A^2})^\dfrac{1}{4}$$
Given $$\dfrac{R_A}{R_B}=400$$
$$\dfrac{T_A}{T_B}=10(\dfrac{1}{400^2})^\dfrac{1}{4}$$
$$\dfrac{T_A}{T_B}=\dfrac{10}{20}=\dfrac{1}{2}$$

From wein's displcament law
$$\lambda_mT=b$$
$$\lambda \alpha \dfrac{1}{T}$$
$$\dfrac{\lambda_A}{\lambda_B}=\dfrac{T_B}{T_A}=2$$
Hence, correct answer is 2.

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