Question

The sun as a blackbody:

a. The Sun is a 6000-K blackbody. At what characteristic wavelength does it radiate?

b. At what characteristic wavelength does a blackbody at room temperature radiate?

c. How much power per unit area is the Sun radiating?

Answer 1

Explanation:

(a) Using Wien's displacement law which states that the temperature and the wavelength of the blackbody radiation are inversely proportional.

So,

λmax= b/T

Where,  

λmax is the peak of wavelength

b is the Wien's displacement constant having value as $2.9\times 10^{-3}\  m K$

T is the Absolute Temperature in Kelvins = 6000 K

So,  

λmax= b/T  = $\frac{2.9\times 10^{-3}}{6000}$ = 4.8333 × 10⁻⁷ m

It lies in the visible region.

(b) Using Wien's displacement law:

λmax= b/T

So,  

λmax= b/T  = $\frac{2.9\times 10^{-3}}{298}$ = 9.73154 × 10⁻⁶ m

This wavelength corresponds to infrared region.

(c) The expression for power per unit area by using Stefan–Boltzmann law is:

Power per unit area = εσT⁴

Where, ε = 1 for blackbody

σ is Stefan–Boltzmann constant, having value as $5.67\times 10^{-8}\ W/m^2K^4$

Thus,

P = $5.67\times 10^{-8}\times 6000\ W/m^2=0.0003402\ W/m^2$