The sun radiates electromagnetic energy at th | vedclass

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(D) The intensity $I$ is defined as the power radiated per unit surface area.$I = \frac{P}{A} = \frac{P}{4 \pi R^2}$Given:Power $P = 3.9 	imes 10^{26

Vedclass · Accepted Answer

$6.4 	imes 10^7$

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(D) The intensity $I$ is defined as the power radiated per unit surface area.$I = \frac{P}{A} = \frac{P}{4 \pi R^2}$Given:Power $P = 3.9 	imes 10^{26} \ W$Radius $R \approx 7.0 	imes 10^8 \ m$$I = \frac{3.9 	imes 10^{26}}{4 	imes 3.14 	imes (7.0 	imes 10^8)^2}$$I = \frac{3.9 	imes 10^{26}}{12.56 	imes 49 	imes 10^{16}}$$I = \frac{3.9 	imes 10^{10}}{615.44} \approx 0.00633 	imes 10^{10} \approx 6.33 	imes 10^7 \ W/m^2$Rounding to the nearest option,we get $6.4 	imes 10^7 \ W/m^2$.

The sun radiates electromagnetic energy at the rate of $3.9 \times 10^{26} \ W$. Its radius is $6.96 \times 10^8 \ m$. The intensity of sunlight at the solar surface will be (in $W/m^2$):

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