Question

The intensity of solar radiation just outside the earth's atmosphere is measured to be 1.4 kW/m². If the radius of the sun 7×10⁸m, while the earth-sun distance is 150×10⁶km, then find
i. the intensity of solar radiation at the surface of sun,
ii. the temperature at the surface of the sun assuming it to be a black body,
iii. the most probable wavelength in solar radiation.

Answer 1

Given :-

▪ The Intensity of solar radiation just outside the earth's atmosphere = 1.4 kW/m$^2$

▪ Radius of sun = 7×10$^8$m

▪ Distance between sun and earth = 150×10$^6$km

To Find :-

→ The intensity of radiation at the surface of sun.

→ Temperature at the surface of sun

→ Wavelength of solar radiation.

Please, see the attachment for better understanding.

Concept :-

☞ First question is completely based on law of energy conservation.

☞ Stefan's law : Energy emitted per second per unit area by a black body at absolute temperature T is given by

$\boxed{\bf{\pink{E=\sigma T^4}}}$

☞ Wien's displacement law : The wavelength corresponding to maximum energy emission by a black body at absolute temperature T is given by

$\boxed{\bf{\purple{\lambda=\dfrac{b}{T}}}}$

Calculation :-

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Answer : 1)

$\implies\bf\:I_1A_1=I_2A_2\\ \\ \implies\sf\:I_1(4\pi{R_s}^2)=I_2(4\pi{d}^2)\\ \\ \implies\sf\:I_1{R_s}^2=I_2d^2\\ \\ \implies\bf\:I_1=\dfrac{I_2\times d^2}{{R_s}^2}\\ \\ \implies\sf\:I_1=\dfrac{1.4\times 10^3\times (1.5\times 10^{11})^2}{(7\times 10^8)^2}\\ \\ \implies\sf\:I_1=\dfrac{3.15\times 10^{25}}{49\times 10^{16}}\\ \\ \implies\boxed{\bf{\red{I_1=6.428\times 10^7\:Wm^{-2}}}}$

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Answer : 2)

$\mapsto\bf\:E=\sigma T^4\\ \\ \mapsto\sf\:6.428\times 10^7=5.67\times 10^{-8}\times T^4\\ \\ \mapsto\sf\:T=(1.13\times 10^{15})^{\frac{1}{4}}\\ \\ \mapsto\boxed{\bf{\green{T=5800\:K}}}$

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Answer : 3)

$\leadsto\bf\:\lambda=\dfrac{b}{T}\\ \\ \leadsto\sf\:\lambda=\dfrac{0.002898}{5800}\\ \\ \leadsto\boxed{\bf{\gray{\lambda=0.5\: \mu{m}}}}$

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Nice question !!

Answer 2

Answer:

hope it is helpful to you.