Question

The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the center of a star of radius r, whose outer surface radiates as a black body at a temperature T Kelvin is given by? (Where $\sigma$ is Stefan’s Constant)

Answer 1

Hint As stated in Stefan Boltzmann law, the radiation emitted per unit time from an area ‘a’ of a black body at absolute temperature ‘t’ is directly related to the temperature of the power four.
$\Rightarrow$ $u = sa{t^4}$
Where, s is Stefan’s Constant ($5.67 \times {10^{ - 8}}$)
A body that is non-black body absorbs radiation and hence emits it. And this emitted radiation can be represented by the following equation:
$\Rightarrow$ $u = e\sigma a{t^4}$
Where, e is emissivity (=absorptive power) lies between 0 – 1
Let the temperature of the surroundings be t’, then the total energy radiated by an area ‘a’ per unit time.
$\Rightarrow$ $\vartriangle u = u - u'$ $= e\sigma a[{t^4} - t{'^4}]$
Stefan’s Boltzmann Law establishes the correlation between the temperature of the black body to the amount of the energy it emits/radiates per unit area. The law is as follows:
“The total energy emitted/radiated per unit surface area of a blackbody across all wavelengths per unit time is directly proportional to the fourth power of the black body’s thermodynamic temperature.”

Complete step by step solution
According to Stefan’s Law,
$\Rightarrow$ $u = \sigma ea{t^4}$
For black body, e=1
Therefore,
$\Rightarrow$ $u = \sigma (4\pi {r^2}){t^4}$
Now the total energy per unit area is given by,
$\dfrac{{\sigma (4\pi {r^2}{t^4})}}{{4\pi {R^2}}}$ $= \dfrac{{\sigma {r^2}{t^4}}}{{{R^2}}}$

Therefore option a is correct

Note Stefan Boltzmann law relates the energy radiated by a blackbody to the temperature of the black body and this relation could be used to get the answer of the above question. With this it should also be kept in mind that the value of e=1 for the black body.