A star (P) behaves like a perfectly black bod | vedclass

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(C) Concept: According to the Stefan-Boltzmann law,the total radiant energy $E$ emitted by a black body of surface area $A$ at temperature $T$ is give

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$4$:$1$

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(C) Concept: According to the Stefan-Boltzmann law,the total radiant energy $E$ emitted by a black body of surface area $A$ at temperature $T$ is given by $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant.For a spherical star of radius $R$,the surface area is $A = 4 \pi R^2$.Thus,$E = \sigma (4 \pi R^2) T^4$.Given for star $(P)$: Radius $= R$,Temperature $= T$.So,$E_P = \sigma (4 \pi R^2) T^4$.Given for star $(Q)$: Radius $= 8R$,Temperature $= T/4$.So,$E_Q = \sigma (4 \pi (8R)^2) (T/4)^4$.Calculating the ratio:$\frac{E_P}{E_Q} = \frac{\sigma (4 \pi R^2) T^4}{\sigma (4 \pi (64 R^2)) (T^4 / 256)} = \frac{T^4}{64 R^2 \cdot (T^4 / 256)} = \frac{256}{64} = 4$.Wait,re-evaluating: $\frac{E_P}{E_Q} = \frac{R^2 T^4}{(8R)^2 (T/4)^4} = \frac{R^2 T^4}{64 R^2 (T^4 / 256)} = \frac{256}{64} = 4$.The ratio of radiant energy emitted by $(P)$ to that by $(Q)$ is $4:1$.

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