Question

The intensity of solar radiation just outside the earth's atmosphere is measured to be $1.4 kW / m^{2}$ . If the radius of the sun $7 \times 10^{8} m$ , while the earth-sun distance is $W = 6.4 \times 10^{7} W / m^{2}$ , then find

the most probable wavelength in solar radiation.

Accepted Answer

Assuming the sun to be a "black body' at a temperature $T_{0}$ , we can write, $W =$ intensity of solar radiation on the sun's surface $= σ T_{0}^{4}$ , where $σ$ is the Stefan-Boltzmann constant.

The radiation emitted from the solar surface per unit time is spread over the surface of a sphere having a radius equal to earth-sun distance where it is received on the earth (just outside the atmosphere)

$W \times 4 π R_{S}^{2} = I_{0} \times 4 π D_{SE}^{2}$

where $D_{SE}$ is the distance between the sun and the earth. and $I_{0}$ is the intensity outside the earth's atmosphere.

$I_{0} = W \times {(\frac{R_{S}}{D_{SE}})}^{2}$

Now, $R_{S} = 7 \times 10^{8} m, D_{SE} = 150 \times 10^{9} m$

and $I_{0} = 1.4 \times 10^{3} W / m^{2}$

$∴ 1.4 \times 10^{3} = W \times {(\frac{7 \times 10^{8}}{150 \times 10^{9}})}^{2} = W \times \frac{49}{225} \times 10^{- 4}$ or $W = 6.4 \times 10^{7} W / m^{2}$

$6.4 \times 10^{7} = σ {T^{4}}_{0} = 5.67 \times 10^{- 8} {T_{0}}^{4} T_{0} \approx 5800 K$

Using Wien's displacement law,

$λ_{m p} T_{0} = 0.29 cm - K = 2.9 \times 10^{- 3} m - K$

or $λ_{mp} = \frac{2.0 \times 10^{- 3}}{5800} = 5 \times 10^{- 7} m = 5000 \overset{o}{A}$

Note: $λ_{m p}$ is also referred to as $λ_{\max}$ .

B M Sharma Solutions for Chapter: Transmission of Heat, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Transmission of Heat, Exercise 1: ILLUSTRATIONS

Questions from B M Sharma Solutions for Chapter: Transmission of Heat, Exercise 1: ILLUSTRATIONS with Hints & Solutions

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