77.9k views
2 votes
Problems related to radiation. (a) The temperature of the Sun’s photosphere is 5700 K. Assume it is a blackbody. What is the peak wavelength (the wavelength of the most typical photon emitted)? (b) The solar radius is 695,000 km. How much energy does the Sun emit per unit time, given the radius and temperature? (c) Suppose you double the temperature of the Sun’s photosphere. How much more energy would it emit per unit time compared to the value you found in part (b)? (d) With the increased temperature, where would be the peak wavelength? Would you see that light with your eyes?

1 Answer

3 votes

Answer:

a) λ = 5,084 10⁻⁷ m , b) P = 3.63 10²⁶ W , c) P = 5.8 10²⁷ W and d) λ = 2.54 10⁻⁷ m

Step-by-step explanation:

a) The maximum emission of the sun can be calculated using the Win equation

λ T = 2,898 10⁻³ m.K

λ = 2,898 10⁻³ / T

λ = 2,898 10⁻³ / 5700

λ = 5,084 10⁻⁷ m

λ = 5,084 10⁻⁷ m (1 10⁹ nm / 1m) =

λ = 5,084 10² nm = 508.4 nm

photon in the visible range

b) The emission of the Sun, is described by the Stefan equation

P = σ A e T⁴

Where σ is the Stefan-Boltzmann constant that vslue is 5,670 10-8 W/m²K⁴, A area of ​​the Sun, and e the emissivity that for a perfect black body is 1

In order to use this equation, we must calculate the area of ​​the sun, we consider it a perfect sphere

r = 695,000 km (1000m / 1 km) = 6.95 10⁸ m

Area of ​​a sphere

A = 4π R²

A = 4π (6.95 10⁸8)²

A = 6.07 10¹⁸ m²

P = 5,670 10⁻⁸ 6.07 10¹⁸ 1 5700⁴

P = 3.63 10²⁶ W

c) The new temperature is double the previous one

T = 2 To

Let's substitute in the formula and calculate

P = σ A e (2To)⁴

P = σ A e T⁴ 2⁴

Po = σ A e T4 = 3.63 10 26 W

P = 16 Po

P= 16 (3.63 10²⁶)

P = 5.8 10²⁷ W

d) Let's calculate the explicit value of the temperature and use the Win equation

T = 2 5700

T = 11400K

λ = 2,898 10⁻³ / 11400

λ = 2.54 10⁻⁷ m

λ = 2.54 10²nm = 254 nm

photon in the UV range

User Promit
by
6.2k points