JEE Advance - Physics (2015 - Paper 1 Offline - No. 2)
Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits
104 times the power emitted from B. The ratio $$\left( {{{{\lambda _A}} \over {{\lambda _B}}}} \right)$$ of their wavelengths $${{\lambda _A}}$$ and $${{\lambda _B}}$$ at which the peaks
occur in their respective radiation curves is
Answer
2
Explanation
Power, $$P = (\sigma {T^4}A) = \sigma {T^4}(4\pi {R^2})$$
or, $$P \propto {T^4}{R^2}$$ ..... (i)
According to Wien's law,
$$\lambda \propto {1 \over T}$$
($$\lambda$$ is the wavelength at which peak occurs)
$$\therefore$$ Eq. (i) will become,
$$P \propto {{{R^2}} \over {{\lambda ^4}}}$$
or, $$\lambda \propto {\left[ {{{{R^2}} \over P}} \right]^{1/4}}$$
$$ \Rightarrow {{{\lambda _A}} \over {{\lambda _B}}} = {\left[ {{{{R_A}} \over {{R_B}}}} \right]^{1/2}}{\left[ {{{{P_B}} \over {{P_A}}}} \right]^{1/4}}$$
$$ = {[400]^{1/2}}{\left[ {{1 \over {{{10}^4}}}} \right]^{1/4}} = 2$$
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