The total radiant energy per unit area, normal to the direction o
The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by
(Where σ is Stefan's Constant)
A. <span class="math-tex">\(\frac{4\pi \sigma {{r}^{2}}{{T}^{4}}}{{{R}^{2}}}\)</span>
B. <span class="math-tex">\(\frac{\sigma {{r}^{2}}{{T}^{4}}}{{{R}^{2}}}\)</span>
C. <span class="math-tex">\(\frac{\sigma {{r}^{2}}{{T}^{4}}}{4\pi {{r}^{2}}}\)</span>
D. <span class="math-tex">\(\frac{\sigma {{r}^{4}}{{T}^{4}}}{{{r}^{4}}}\)</span>
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
CONCEPT:
- Stefan-Boltzmann law states that the power radiated from a blackbody is directly proportional to the fourth power of absolute temperature and is expressed as
E =σ eAT4
Where σ = Stefan-Boltzmann constant =5.67× 108 JS-1m-2K-4, e = Emissivity, T =Temperature
- A black body absorbs all kinds of radiation incident upon it regardless of frequency of radiation.
- The emissivity of a blackbody is one, e =1.
CALCULATION:
Given - A = 4π R2, e =1 (for a blackbody)
- According to Stefan's law
⇒ E =σ eAT4 -------(1)
Applying the given values equation on becomes
⇒ E =σ 4πr2T4 --------(2)
- The intensity of the total energy is given by
\(\Rightarrow I= \frac{E}{A}\)
\(\Rightarrow I= \frac{σ 4\pi r^{2}T^{4}}{4 \pi R^{2}}\)
\(\Rightarrow I= \frac{σ r^{2}T^{4}}{ R^{2}}\)
