Question

Suppose a star with radius 8.50 × 10⁸ m has a peak wavelength of 685 nm in the spectrum of its emitted radiation. Assume the star is a perfect blackbody. (A) What is the energy of a photon with this wavelength? (B) What is the surface temperature of the star? (C) At what rate is energy emitted from the star in the form of radiation?

Answer 1

Solution:

a)

Using Planck's equation;

E = hc/λ

=> E = 6.626^-34 x 3.0^8m/s / 685^-9m

=> E = 2.90^-19 J

b)

Using Wien's displacement law;

λ(max) x T= 2.898^-3 mK

=> T = 2.898^-3mK / 685^-9m

=> T = 4.23^3 K (4230 K)

c)

Using Stefan's black-body equation;

i.e,

(P/A) = εσT^4

emissivity ε = 1, Stefan constant = σ, (P/A) = emission intensity (W/m²) ..

Energy emission rate (J/s) = power

=> P = A εσT^4 .. (A = star surface area 4πR²)

=> P = {4π(8.50^8m)²}(1 x 5.70^-8)(4230K)^4

=> P = 2.10^27 W