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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Thermal Properties of Matter

NEET 2010 · Physics · Thermal Properties of Matter

The total radiant energy per unit area, normal to the direction of...

The total radiant energy per unit area, normal to the direction of incidence, received at a distance R\mathrm{R} from the centre of a star of radius rr, whose outer surface radiates as a black body at a temperature T KT \mathrm{~K} is given by (where σ\sigma is Stefan's constant)

Options

  1. Aσr2 T4/R2\sigma r^2 \mathrm{~T}^4 / \mathrm{R}^2
  2. Bσr2T4/4πr2\sigma r^2 T^4 / 4 \pi r^2
  3. Cσr4T4/r4\sigma r^4 T^4 / r^4
  4. D4πσr2T4/R24 \pi \sigma r^2 T^4 / R^2

Correct answer

A. σr2 T4/R2\sigma r^2 \mathrm{~T}^4 / \mathrm{R}^2

Step-by-step solution

If r\mathrm{r} is the radius of the star and TT its temperature, then the energy emitted by the star per second through radiation in accordance with Stefan's law will be given by AσT4=4πr2σT4A \sigma \mathrm{T}^4=4 \pi \mathrm{r}^2 \sigma \mathrm{T}^4 In reaching a distance RR this energy will spread over a sphere of radius RR; so the intensity of radiation will be given by S=P4πR2=4πr2σT44πR2\frac{P}{4 \pi R^2}=\frac{4 \pi r^2 \sigma T^4}{4 \pi \mathrm{R}^2}
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