Final answer:
The Sun, as a blackbody, radiates at a characteristic wavelength in the yellow-orange part of the spectrum. It radiates approximately 6.31 x 10^7 W/m^2 of power per unit surface area. If the Sun had a radius twice as large but emitted the same total energy, it would have a temperature of approximately 4247 K.
Step-by-step explanation:
The Sun, as a blackbody, radiates with a temperature of 6000 K. The characteristic wavelength at which it radiates can be determined using Wien's displacement law. According to this law, the wavelength at which a blackbody radiates most intensely is inversely proportional to its temperature. So, we can use the formula:
λ(max) = 2.898 × 10^(-3) / T
Plugging in the temperature of 6000 K, we get:
λ(max) = 2.898 × 10^(-3) / 6000
λ(max) ≈ 4.83 × 10^(-7) meters, which falls in the visible range of the spectrum, specifically in the yellow-orange part.
As for the power per unit surface area that the Sun is radiating, we can use the Stefan-Boltzmann law. This law states that the energy flux from a blackbody at temperature T is proportional to the fourth power of its absolute temperature. The formula is:
F = σT^4
where F is the energy flux and σ is the Stefan-Boltzmann constant (approximately 5.67 × 10^(-8) W/m^2K^4). Plugging in the temperature of 6000 K, we get:
F = 5.67 × 10^(-8) × (6000)^4
F ≈ 6.31 × 10^7 W/m^2
Now, if the Sun had a radius twice as large as its current radius but emitted the same total amount of energy, we can use the formula for the total power output of a blackbody:
P = 4πR^2σT^4
With the new radius being 2R, we can express the total power output as:
P = 4π(2R)^2σT^4
If P remains the same, we can set the two equations equal to each other and solve for the new temperature:
4πR^2σT^4 = 4π(2R)^2σT'^4
Rearranging and simplifying, we get:
T'^4 = T^4 / 4
T' ≈ T / √2
Substituting the value of T = 6000 K, we get:
T' ≈ 4247 K.