Question

If $T$ is surface temperature of sun, $R$ is the radius of sun, $r$ is radius of earth’s orbit and $S$ is solar constant, then total radiant energy of sun per unit time from the sphere of radius $r,$ then
(A) $\pi {{r}^{2}}S$
( B) $4\pi {{r}^{2}}S$
(C) $\sigma \dfrac{4}{3}\pi {{R}^{3}}{{T}^{4}}$
(D) $\sigma 4\pi {{r}^{2}}{{T}^{4}}$

Answer 1

Hint to calculate here, the total radiant energy of sun per unit time(from sphere of radius $r$ ) ,we can use here the Stefan’s-Boltzmann law.
It states that total energy radiated per unit surface area of the black body across all wavelengths per unit time, is directly proportional to the fourth power of the blackbody’s thermodynamic temperature $T$.
Formula used
$\text{Energy radiated per unit surface area per unit time}=\sigma {{T}^{4}}\times 4\pi {{r}^{2}}$

Complete step by step solution : Using Stefan’s Boltzmann law.
$\text{Energy radiated per unit surface area per unit time}=\sigma {{T}^{4}}$
Where $\sigma$ is Stefan’s Boltzmann constant
Here, surface area of earth$=4\pi {{r}^{2}}$
Thus, total energy radiated per unit time,
$L=\sigma {{T}^{4}}\times 4\pi {{r}^{2}}$

Thus, option (d) is correct.

Additional information

Blackbody, in physics, a surface that absorbs all radiant energy falling on it. The term arises because incident visible light will be absorbed rather than reflected, and therefore the surface will appear black. The concept of such a perfect absorber of energy is extremely useful in the study of radiation phenomena.

Note: Learn the formula and statement of Stefan’s Boltzmann law. Learn the value of the Stefan’s Boltzmann constant as well, to be used in numerical problems.
It is derived from other known constants. The value of constant is:

& \sigma =\dfrac{2{{\pi }^{5}}{{k}^{4}}}{15{{c}^{2}}{{h}^{3}}}=5\cdot 670373\times {{10}^{-8}}\text{ }W{{m}^{-2}}{{k}^{-4}} \\\ & k:\text{Boltzmann constant} \\\ & h:\text{Plank }\\!\\!'\\!\\!\text{ s constant} \\\ & c:\text{speed of light in a vacuum} \\\ \end{aligned}$$