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Assuming the sun to be a black body at a temperatenc of 5700∘C, calculati emissive pouser of the surtace of the sun, wavelength for maximum spectral intensity and Leat Energy emitted by sun per unit time assuming its diametir as 1.391×109 m.

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Answer:

emissive power

E ≈ 3.87 x 10^26 W/m^2

Maximum wavelength

λmax ≈ 4.85 x 10^-7 m

total energy emitted by sun

Energy ≈ 2.36 x 10^45 Joules/sec

Step-by-step explanation:

To calculate the emissive power of the surface of the Sun, we will use Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its temperature.

E = σ * T^4

where E is the emissive power, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2K^4)), and T is the temperature in Kelvin.

First, we need to convert the temperature of the Sun from Celsius to Kelvin:

5700°C + 273.15 = 5973.15 K

Now we can calculate the emissive power:

E = 5.67 x 10^-8 * (5973.15)^4

E ≈ 3.87 x 10^26 W/m^2

To calculate the wavelength for the maximum spectral intensity (λmax), we can use Wien's displacement law:

λmax = b / T

where b is Wien's displacement constant (approximately 2.898 x 10^-3 m·K).

λmax = 2.898 x 10^-3 / 5973.15

λmax ≈ 4.85 x 10^-7 m

Finally, to calculate the total energy emitted by the Sun per unit time, we need to consider its surface area. The formula is:

Energy = E * A

where E is the emissive power and A is the surface area.

The surface area of a sphere is given by:

A = 4πr^2

where r is the radius of the sphere.

Given the diameter of the Sun as 1.391 x 10^9 m, we can calculate the radius (r) as half of the diameter:

r = 1.391 x 10^9 / 2 = 6.955 x 10^8 m

Now we can calculate the surface area:

A = 4π(6.955 x 10^8)^2

A ≈ 6.087 x 10^18 m^2

Finally, we can calculate the total energy emitted by the Sun per unit time:

Energy = 3.87 x 10^26 * 6.087 x 10^18

Energy ≈ 2.36 x 10^45 Joules/sec

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