If the temperature of the Sun = 6000 , K,the | vedclass

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(A) The intensity of radiation $I$ at a distance $d$ from the Sun is given by the Stefan-Boltzmann Law:$I = \frac{P}{4 \pi d^2} = \frac{\sigma T^4 (4

Vedclass · Accepted Answer

$1.4 	imes 10^{3} \, W/m^2$

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(A) The intensity of radiation $I$ at a distance $d$ from the Sun is given by the Stefan-Boltzmann Law:$I = \frac{P}{4 \pi d^2} = \frac{\sigma T^4 (4 \pi R_s^2)}{4 \pi d^2} = \sigma T^4 \left( \frac{R_s}{d} ight)^2$Where $\sigma = 5.67 	imes 10^{-8} \, W/m^2K^4$,$T = 6000 \, K$,$R_s = 7.2 	imes 10^8 \, m$,and $d = 1.5 	imes 10^{11} \, m$.Substituting the values:$I = (5.67 	imes 10^{-8}) 	imes (6000)^4 	imes \left( \frac{7.2 	imes 10^8}{1.5 	imes 10^{11}} ight)^2$$I = (5.67 	imes 10^{-8}) 	imes (1.296 	imes 10^{15}) 	imes (4.8 	imes 10^{-3})^2$$I = 7.348 	imes 10^7 	imes 2.304 	imes 10^{-5} \approx 1.4 	imes 10^3 \, W/m^2$.

If the temperature of the $Sun = 6000 \, K$,the radius of the Sun is $7.2 \times 10^{5} \, km$,the radius of the Earth is $6000 \, km$,and the distance between the Earth and the Sun is $15 \times 10^{7} \, km$,find the intensity of light on Earth.

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