The earth receives its surface radiation from the sun at the rate of 1400 `W//m^(2)`. The distance of the centre of the sun from the surface of the earth is `1.5 xx 10^(11)` m and the radius of the sun is `7.0 xx 10^(8)` m. Treating sun as a black body, it follows from the above data that its surface temeperature is

Options

A. 5810 K
B. `10^(6)` K
C. 50.1 K
D. `5801^(@)C`

Explanation

To find the surface temperature of the Sun based on the given data, we can follow these steps:

### Step 1: Understand the relationship between intensity, power, and area
The intensity \( I \) of radiation received at the Earth from the Sun is given by the formula:
\[ I = \frac{P}{4\pi d^2} \]
where \( P \) is the total power emitted by the Sun and \( d \) is the distance from the Sun to the Earth.

### Step 2: Rearrange the formula to find the power emitted by the Sun
From the intensity formula, we can rearrange it to find the power \( P \):
\[ P = I \cdot 4\pi d^2 \]
Substituting the values:
- \( I = 1400 \, \text{W/m}^2 \)
- \( d = 1.5 \times 10^{11} \, \text{m} \)

Calculating \( P \):
\[ P = 1400 \cdot 4\pi (1.5 \times 10^{11})^2 \]

### Step 3: Calculate the distance squared
Calculating \( (1.5 \times 10^{11})^2 \):
\[ (1.5 \times 10^{11})^2 = 2.25 \times 10^{22} \, \text{m}^2 \]

### Step 4: Substitute back into the power equation
Now substituting back:
\[ P = 1400 \cdot 4\pi \cdot 2.25 \times 10^{22} \]

### Step 5: Calculate \( P \)
Using \( \pi \approx 3.14 \):
\[ P \approx 1400 \cdot 4 \cdot 3.14 \cdot 2.25 \times 10^{22} \]
Calculating this gives:
\[ P \approx 1400 \cdot 12.56 \cdot 2.25 \times 10^{22} \approx 1400 \cdot 28.26 \times 10^{22} \approx 3.95 \times 10^{25} \, \text{W} \]

### Step 6: Use Stefan-Boltzmann Law to find the temperature
The power emitted by a black body is given by:
\[ P = \sigma A T^4 \]
where \( A \) is the surface area of the Sun and \( T \) is its surface temperature. The surface area \( A \) of the Sun is:
\[ A = 4\pi R^2 \]
where \( R = 7.0 \times 10^8 \, \text{m} \).

### Step 7: Calculate the surface area of the Sun
Calculating \( A \):
\[ A = 4\pi (7.0 \times 10^8)^2 = 4\pi (4.9 \times 10^{17}) \approx 61.57 \times 10^{17} \, \text{m}^2 \]

### Step 8: Substitute into the Stefan-Boltzmann equation
Now substituting \( A \) into the power equation:
\[ P = \sigma (4\pi R^2) T^4 \]
Rearranging for \( T^4 \):
\[ T^4 = \frac{P}{\sigma A} \]

### Step 9: Substitute values and solve for \( T \)
Using \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4 \):
\[ T^4 = \frac{3.95 \times 10^{25}}{5.67 \times 10^{-8} \cdot 61.57 \times 10^{17}} \]

### Step 10: Calculate \( T \)
Calculating the denominator:
\[ 5.67 \times 10^{-8} \cdot 61.57 \times 10^{17} \approx 3.49 \times 10^{10} \]
Thus,
\[ T^4 \approx \frac{3.95 \times 10^{25}}{3.49 \times 10^{10}} \approx 1.13 \times 10^{15} \]
Taking the fourth root gives:
\[ T \approx (1.13 \times 10^{15})^{1/4} \approx 5810 \, \text{K} \]

### Final Answer
The surface temperature of the Sun is approximately \( 5810 \, \text{K} \).

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