The dimensions of Stefan-Boltzman constant `sigma` can be written in terms of Plank's constant h. Boltzmann constant `k_(B)` and the speed of light c as `sigma = h^(alpha) K_(B)^(b) c^(Y)`. Here

Options

A. `alpha = 3,beta =4 and gamma = -3`
B. `alpha =3,beta = -4 and gamma =2`
C. `alpha =-3,beta =4 and gamma = -2`
D. `alpha =2,beta =-3 and gamma =-1`

Explanation

To find the values of \( \alpha \), \( \beta \), and \( \gamma \) in the equation for the Stefan-Boltzmann constant \( \sigma = h^{\alpha} k_B^{\beta} c^{\gamma} \), we will start by writing down the dimensions of each constant involved.

### Step 1: Write down the dimensions of the Stefan-Boltzmann constant \( \sigma \)

The Stefan-Boltzmann constant \( \sigma \) has the units of power per unit area per unit temperature to the fourth power. In SI units, it can be expressed as:

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

Since \( \text{W} = \text{J/s} \) and \( \text{J} = \text{kg} \cdot \text{m}^2/\text{s}^2 \), we can rewrite this as:

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

Thus, the dimensions of \( \sigma \) can be expressed as:

\[ [\sigma] = M^1 L^0 T^{-3} \Theta^{-4} \]

### Step 2: Write down the dimensions of Planck's constant \( h \)

Planck's constant \( h \) has the units of energy multiplied by time:

\[ h = \text{J} \cdot \text{s} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^2} \cdot \text{s} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1} \]

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

\[ [h] = M^1 L^2 T^{-1} \]

### Step 3: Write down the dimensions of the Boltzmann constant \( k_B \)

The Boltzmann constant \( k_B \) has the units of energy per temperature:

\[ k_B = \frac{\text{J}}{\text{K}} = \frac{\text{kg} \cdot \text{m}^2/\text{s}^2}{\Theta} \]

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

\[ [k_B] = M^1 L^2 T^{-2} \Theta^{-1} \]

### Step 4: Write down the dimensions of the speed of light \( c \)

The speed of light \( c \) has the units of distance per time:

\[ c = \frac{\text{m}}{\text{s}} = L^1 T^{-1} \]

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

\[ [c] = L^1 T^{-1} \]

### Step 5: Combine the dimensions

Now we can express the dimensions of \( \sigma \) in terms of \( h \), \( k_B \), and \( c \):

\[ [\sigma] = [h]^{\alpha} [k_B]^{\beta} [c]^{\gamma} \]

Substituting the dimensions we found:

\[ M^1 L^0 T^{-3} \Theta^{-4} = (M^1 L^2 T^{-1})^{\alpha} (M^1 L^2 T^{-2} \Theta^{-1})^{\beta} (L^1 T^{-1})^{\gamma} \]

### Step 6: Equate the powers of each dimension

Now we will equate the powers of \( M \), \( L \), \( T \), and \( \Theta \):

1. For \( M \):
\[ 1 = \alpha + \beta \quad (1) \]

2. For \( L \):
\[ 0 = 2\alpha + 2\beta + \gamma \quad (2) \]

3. For \( T \):
\[ -3 = -\alpha - 2\beta - \gamma \quad (3) \]

4. For \( \Theta \):
\[ -4 = -\beta \quad (4) \]

### Step 7: Solve the equations

From equation (4):
\[ \beta = 4 \]

Substituting \( \beta \) into equation (1):
\[ 1 = \alpha + 4 \implies \alpha = -3 \]

Now substituting \( \alpha \) and \( \beta \) into equation (2):
\[ 0 = 2(-3) + 2(4) + \gamma \implies 0 = -6 + 8 + \gamma \implies \gamma = -2 \]

### Final Result

Thus, we have:
\[ \alpha = -3, \quad \beta = 4, \quad \gamma = -2 \]