Assume that the sun is a sphere of radius `6.96 xx 10^(8) m` and that its surface temperature is ` 5.8 xx 10^(3) K`. If the Sun radiates at a rate of `3.90 xx 10^(26)`W and is a perfect emitter, at what rate is energy emitted per square meter at the Sun's surface ?

Options

A. `5.6 xx 10^(7) W//m^(2)`
B. `6.4 xx 10^(7) W//m^(2)`
C. `5.6 xx 10^(17) W//m^(2)`
D. `12.8 xx 10^(7) W//m^(2)`

Answer

To find the rate at which energy is emitted per square meter at the Sun's surface, we can follow these steps:

### Step 1: Calculate the Surface Area of the Sun
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]
where \( r \) is the radius of the sphere. Given that the radius of the Sun is \( r = 6.96 \times 10^8 \) m, we can substitute this value into the formula.

### Step 2: Substitute the Radius into the Surface Area Formula
Substituting the radius:
\[ A = 4 \pi (6.96 \times 10^8)^2 \]

### Step 3: Calculate the Surface Area
Calculating \( (6.96 \times 10^8)^2 \):
\[ (6.96 \times 10^8)^2 = 4.85 \times 10^{17} \, \text{m}^2 \]
Now, substituting this back into the area formula:
\[ A = 4 \pi (4.85 \times 10^{17}) \approx 6.08 \times 10^{18} \, \text{m}^2 \]

### Step 4: Calculate the Energy Emitted per Square Meter
The total power radiated by the Sun is given as \( P = 3.90 \times 10^{26} \) W. The energy emitted per square meter \( E \) can be calculated using the formula:
\[ E = \frac{P}{A} \]
Substituting the values we have:
\[ E = \frac{3.90 \times 10^{26}}{6.08 \times 10^{18}} \]

### Step 5: Perform the Division
Calculating the above expression:
\[ E \approx 6.42 \times 10^{7} \, \text{W/m}^2 \]

### Final Answer
The rate at which energy is emitted per square meter at the Sun's surface is approximately:
\[ E \approx 6.4 \times 10^{7} \, \text{W/m}^2 \]