The amount of radiations emitted per second by unit area of a hollow container at `10^(3^(@))` K will be-

Options

A. 567 J
B. 5670 J
C. 56700 J
D. 567000 J

Answer

To solve the problem of finding the amount of radiation emitted per second by unit area of a hollow container at a temperature of \(10^3\) K, we will use Stefan's Law. Here’s a step-by-step solution:

### Step 1: Understand Stefan's Law
Stefan's Law states that the power emitted per unit area (E) of a black body is proportional to the fourth power of its absolute temperature (T). The formula is given by:

\[ E = \sigma T^4 \]

where:
- \(E\) is the power emitted per unit area (in Watts per square meter, W/m²),
- \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\),
- \(T\) is the absolute temperature in Kelvin (K).

### Step 2: Substitute the Values
Given that the temperature \(T = 10^3 \, \text{K}\), we can substitute this value into the equation:

\[ E = \sigma (10^3)^4 \]

### Step 3: Calculate \(T^4\)
Calculate \( (10^3)^4 \):

\[ (10^3)^4 = 10^{12} \]

### Step 4: Substitute and Calculate \(E\)
Now substitute \(T^4\) back into the equation for \(E\):

\[ E = 5.67 \times 10^{-8} \times 10^{12} \]

### Step 5: Simplify the Expression
Now, simplify the multiplication:

\[ E = 5.67 \times 10^{4} \, \text{W/m}^2 \]

### Step 6: Final Result
Thus, the amount of radiation emitted per second by unit area of the hollow container at \(10^3\) K is:

\[ E = 56700 \, \text{W/m}^2 \]

### Conclusion
The correct answer is \(56700 \, \text{W/m}^2\).

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