Q113 / 200 · NEET 2010 (Screening) · Physics

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

  • Correct answer
    A
    $\sigma r^2 \mathrm{~T}^4 / \mathrm{R}^2$
  • B
    $\sigma r^2 T^4 / 4 \pi r^2$
  • C
    $\sigma r^4 T^4 / r^4$
  • D
    $4 \pi \sigma r^2 T^4 / R^2$

Solution

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

Hence, the correct answer is option: 'A'.

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