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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2015 · Paper 1 · Physics · Thermal Properties of Matter

Two spherical stars A and B emit blackbody radiation. The radius of...

Two spherical stars AandBA and B emit blackbody radiation. The radius of AA is 400 times that of BandAB and A emits 104{10}^{4} times the power emitted from BB . The ratio (λAλB)\left(\dfrac{{\lambda}_{A}}{{\lambda}_{B}}\right) of their wavelengths λAandλB{\lambda}_{A} and {\lambda}_{B} at which the peaks occur in their respective radiation curves is

Correct answer

2

Step-by-step solution

P=e\sigmaAT4P = e\sigmaA{T}^{4}
\lambdaT=\lambdaT = constant
PAPB=eσAATA4eσABTB4=AAABTA4TB4=rA2rB2.λB4λA4\dfrac{{P}_{A}}{{P}_{B}} = \dfrac{e\sigma{A}_{A}{T}_{A}^{4}}{e\sigma{A}_{B}{T}_{B}^{4}} = \dfrac{{A}_{A}}{{A}_{B}}\dfrac{{T}_{A}^{4}}{{T}_{B}^{4}} = \dfrac{{r}_{A}^{2}}{{r}_{B}^{2}} . \dfrac{{\lambda}_{B}^{4}}{{\lambda}_{A}^{4}}
104=4002.(λBλA)4{10}^{4} = {400}^{2} . {\left(\dfrac{{\lambda}_{B}}{{\lambda}_{A}}\right)}^{4}
10416×104=(λBλA)4\dfrac{{10}^{4}}{16 \times {10}^{4}} = {\left(\dfrac{{\lambda}_{B}}{{\lambda}_{A}}\right)}^{4}
λAλB=(161)14=2\dfrac{{\lambda}_{A}}{{\lambda}_{B}} = {\left(\dfrac{16}{1}\right)}^{\dfrac{1}{4}} = 2
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