Radiation: Betelgeuse is a red supergiant star in the constellation Orion. It radiates heat at the rate of 2.70 × 10³⁰ W and has a surface temperature of 3000 K. Assuming that it is a perfect emitter, what is the radius of Betelgeuse? The Stefan-Boltzmann constant is 5.670 × 10^-8 W/m² · K⁴.

Question

Quizplus · Accepted Answer

The luminosity of Betelgeuse can be calculated using the Stefan-Boltzmann law:

L = 4πR²σT⁴

where R is the radius of the star, σ is the Stefan-Boltzmann constant, and T is the surface temperature. Substituting the given values:

2.70 × 10^31 W = 4πR²(5.670 × 10^-8 W/m²K⁴)(3000 K)⁴

Solving for R gives:

R = (L/4πσT⁴)^(1/2)
  = [(2.70 x 10^31 W)/(4π(5.670 x 10^-8 W/m²K⁴)(3000 K)⁴)]^(1/2)
  ≈ 2.16 x 10^11 m

Therefore, the answer is choice E: 2.16 × 10^11 m.

Radiation: Betelgeuse Is a Red Supergiant Star in the Constellation