Back to Directory
NEET PHYSICSMedium

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

A

σr2T4R2\frac{\sigma r^2 T^4}{R^2}

B

σr2T44πr2\frac{\sigma r^2 T^4}{4\pi r^2}

C

σr4T4r4\frac{\sigma r^4 T^4}{r^4}

D

4πσr2T4R2\frac{4\pi \sigma r^2 T^4}{R^2}

Step-by-Step Solution

According to the Stefan-Boltzmann law, the total radiant power PP emitted by a star of radius rr and temperature TT is P=σAT4=σ(4πr2)T4P = \sigma A T^4 = \sigma (4\pi r^2) T^4. This energy spreads uniformly over a spherical region as it travels outwards. At a distance RR from the center of the star, the energy crosses a spherical surface of area 4πR24\pi R^2. Therefore, the radiant energy received per unit area per unit time (intensity II) at distance RR is given by: I=P4πR2=σ(4πr2)T44πR2=σr2T4R2I = \frac{P}{4\pi R^2} = \frac{\sigma (4\pi r^2) T^4}{4\pi R^2} = \frac{\sigma r^2 T^4}{R^2}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started