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Two spherical bodies A (radius $6 c m$ ) and B (radius $18 c m$ ) are at temperatures $T_{A} a n d T_{B},$ respectively. The maximum intensity in the emission spectrum of A is at $500 m m$ and in that of B is at $1500 m m .$ Considering them to be black bodies, the ratio of the rate of total energy radiated by A to that of B will be______.

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answer is 9.

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Detailed Solution

According to Wien's displacement law, $λ_{m} T = C o n s \tan t$

$∴ {(λ_{m})}_{A} T_{A} = {(λ_{m})}_{B} T_{B}, o r \frac{T_{A}}{T_{B}} = \frac{{(λ_{m})}_{B}}{{(λ_{m})}_{A}} = \frac{1500 n m}{500 n m} o r \frac{T_{A}}{T_{B}} = 3$

According to Stefan's Boltzmann law, rate of energy, radiated by a black body

$E = σ A T^{4} = σ 4 π R^{2} T^{4} [H e r e, A = 4 π R^{2}]$

$∴ \frac{E_{A}}{E_{B}} = {(\frac{R_{A}}{R_{B}})}^{2} {(\frac{T_{A}}{T_{B}})}^{4} = {(\frac{6 c m}{18 c m})}^{2} \times {(3)}^{4} = 9$

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