The dimension of `sigma b^4` (where `sigma` is Stefan's constant and b is Wien's constant) are `[ML^4 T^(-3)]` is it true.

Options

A. `[M^(0)L^(0)T^(0)]`
B. `[ML^(4)T^(-3)]`
C. `[ML^(-2)T]`
D. `[ML^(6)T^(-3)]`

Answer

To determine if the dimension of \( \sigma b^4 \) is indeed \( [ML^4 T^{-3}] \), we will calculate the dimensions of both Stefan's constant (\( \sigma \)) and Wien's constant (\( b \)) step by step.

### Step 1: Determine the Dimensions of Stefan's Constant (\( \sigma \))

Stefan's constant (\( \sigma \)) is defined as:

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

Where:
- \( E \) is energy,
- \( A \) is area,
- \( T \) is temperature.

#### Dimensions of Energy (\( E \)):
The dimension of energy is given by:

\[ [E] = [ML^2T^{-2}] \]

#### Dimensions of Area (\( A \)):
The dimension of area is:

\[ [A] = [L^2] \]

#### Dimensions of Temperature (\( T \)):
The dimension of temperature is:

\[ [T] = [K] \]

#### Putting it all together:
Now substituting these into the equation for \( \sigma \):

\[ [\sigma] = \frac{[E]}{[A] \cdot [T^4]} = \frac{[ML^2T^{-2}]}{[L^2] \cdot [K^4]} = \frac{[ML^2T^{-2}]}{[L^2K^4]} \]

Simplifying this gives:

\[ [\sigma] = [M][L^2][T^{-2}][L^{-2}][K^{-4}] = [M][T^{-2}][K^{-4}] \]

Thus, the dimensions of \( \sigma \) are:

\[ [\sigma] = [MT^{-3}K^{-4}] \]

### Step 2: Determine the Dimensions of Wien's Constant (\( b \))

Wien's constant (\( b \)) is defined in relation to the wavelength (\( \lambda \)) and temperature (\( T \)) as:

\[ \lambda = \frac{b}{T} \]

Rearranging gives:

\[ b = \lambda \cdot T \]

#### Dimensions of Wavelength (\( \lambda \)):
The dimension of wavelength is:

\[ [\lambda] = [L] \]

#### Putting it all together:
Now substituting into the equation for \( b \):

\[ [b] = [L][K] = [LK] \]

### Step 3: Calculate the Dimensions of \( \sigma b^4 \)

Now we can calculate the dimensions of \( \sigma b^4 \):

\[ [\sigma b^4] = [\sigma] \cdot [b^4] = [MT^{-3}K^{-4}] \cdot [L^4K^4] \]

#### Simplifying this expression:
\[ [\sigma b^4] = [M][T^{-3}][K^{-4}][L^4][K^4] = [M][L^4][T^{-3}][K^{0}] \]

Since \( K^0 = 1 \), we can ignore it:

\[ [\sigma b^4] = [M][L^4][T^{-3}] \]

### Conclusion

Thus, the final dimensions of \( \sigma b^4 \) are:

\[ [\sigma b^4] = [ML^4T^{-3}] \]

This confirms that the statement in the question is **true**.

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