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Stefan-Boltzman's Law

Two spherical...

Question

Two spherical black bodies $A$ and $B$ , having radii $r_{A}$ and $r_{B},$ where $r_{B} = 2 r_{A}$ emit radiation with peak intensities at wavelength $400 nm$ and $800 nm$ respectively. If their temperatures are $T_{A}$ and $T_{B}$ respectively in Kelvin scale, their emissive powers are $E_{A}$ and $E_{B}$ and energies emitted per second are $P_{A}$ and $P_{B}$ respectively, then

A

\frac{T_{A}}{T_{B}} = 2

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B

\frac{P_{A}}{P_{B}} = 4

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C

\frac{E_{A}}{E_{B}} = 8

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D

\frac{E_{A}}{E_{B}} = 16

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Solution

The correct option is D $\frac{E_{A}}{E_{B}} = 16$
Given
$r_{B} = 2 r_{A}$

By Wein’s displacement law, wavelength corresponding to maximum intensity is
$λ_{A} \propto \frac{1}{T}$
$\frac{(λ_{m})_{A}}{(λ_{m})_{B}} = \frac{T_{B}}{T_{A}} = \frac{400}{800}$
$∴ \frac{T_{A}}{T_{B}} = 2$
By Stefan’s law,
$E = σ e T^{4}$
$\frac{E_{A}}{E_{B}} = \frac{T_{A}^{4}}{T_{B}^{4}} = 16$
$\frac{P_{A}}{P_{B}} = \frac{A_{A} T_{A}^{4}}{A_{B} T_{B}^{4}}$
$\frac{P_{A}}{P_{B}} = \frac{4 π r_{A}^{2} T_{A}^{4}}{4 π r_{B}^{2} T_{B}^{4}} = 4$

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