The SI unit of Stefan's constant is :

Options

A. `Ws^(-1) m^(-2)K^(-4)`
B. `J s m^(-1)K^(-1)`
C. `J s^(-1)m^(-2)K^(-1)`
D. `W m^(-2)K^(-4)`

Answer

To find the SI unit of Stefan's constant (σ), we start with the formula that relates power radiated by a black body to its temperature:

\[ P = \sigma \cdot E \cdot A \cdot T^4 \]

Where:
- \( P \) is the power radiated (in watts, W)
- \( \sigma \) is Stefan's constant (which we want to find)
- \( E \) is the emissivity (dimensionless)
- \( A \) is the area (in square meters, m²)
- \( T \) is the absolute temperature (in Kelvin, K)

### Step 1: Rearranging the formula
To isolate σ, we rearrange the formula:

\[ \sigma = \frac{P}{E \cdot A \cdot T^4} \]

### Step 2: Identifying the units
Now we need to identify the units of each term in the equation:

- The SI unit of power \( P \) is watts (W), which is equivalent to joules per second (J/s).
- The emissivity \( E \) is dimensionless, so it has no units.
- The area \( A \) has units of square meters (m²).
- The temperature \( T \) is measured in Kelvin (K).

### Step 3: Substituting the units into the equation
Now we can substitute the units into the rearranged equation:

\[ \sigma = \frac{\text{W}}{\text{(dimensionless)} \cdot \text{m}^2 \cdot \text{K}^4} \]

### Step 4: Expressing the units of σ
Since \( \text{W} = \text{J/s} \) and \( \text{J} = \text{kg} \cdot \text{m}^2/\text{s}^2 \), we can express the units of power as:

\[ \text{W} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} \]

Now substituting this into the equation for σ:

\[ \sigma = \frac{\frac{\text{kg} \cdot \text{m}^2}{\text{s}^3}}{\text{m}^2 \cdot \text{K}^4} \]

### Step 5: Simplifying the units
Now we can simplify:

\[ \sigma = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3 \cdot \text{m}^2 \cdot \text{K}^4} \]

The \( \text{m}^2 \) cancels out:

\[ \sigma = \frac{\text{kg}}{\text{s}^3 \cdot \text{K}^4} \]

### Conclusion
Thus, the SI unit of Stefan's constant (σ) is:

\[ \sigma = \frac{\text{kg}}{\text{s}^3 \cdot \text{K}^4} \]