In the relaction ` p = (a)/(beta) e ^((aZ)/(k theta )`, p is pressure Z is distance .k is Boltamann constant and `theta` is the teperations . The dimensional formula of `beta` will be

Options

A. `[M^(0)L^(2)T^(0)]`
B. `[ML^(2)T]`
C. `[ML^(0)T^(-1)]`
D. `[M^(0)L^(2)T^(-1)]`

Explanation

To find the dimensional formula of \( \beta \) in the relation

\[ p = \frac{A}{\beta} e^{\frac{aZ}{k\theta}}, \]

where \( p \) is pressure, \( Z \) is distance, \( k \) is the Boltzmann constant, and \( \theta \) is temperature, we will follow these steps:

### Step 1: Identify the dimensions of the known quantities.

1. **Pressure \( p \)**: Pressure is defined as force per unit area. The dimensional formula for force is \( [F] = [M][L][T^{-2}] \). Therefore, the dimensional formula for pressure is:
\[ [p] = \frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}]. \]

2. **Distance \( Z \)**: The dimensional formula for distance is:
\[ [Z] = [L]. \]

3. **Boltzmann constant \( k \)**: The dimensional formula for the Boltzmann constant is:
\[ [k] = [M][L^2][T^{-2}][K^{-1}]. \]

4. **Temperature \( \theta \)**: The dimensional formula for temperature is:
\[ [\theta] = [K]. \]

### Step 2: Analyze the exponential term.

The term \( e^{\frac{aZ}{k\theta}} \) must be dimensionless. Therefore, the dimensions of \( \frac{aZ}{k\theta} \) must equal \( [M^0][L^0][T^0] \).

### Step 3: Determine the dimensions of \( a \).

From the relation, we have:
\[ \frac{aZ}{k\theta} \text{ is dimensionless}. \]
Substituting the dimensions:
\[ \frac{[a][L]}{[M][L^2][T^{-2}][K^{-1}][K]} = \frac{[a][L]}{[M][L^2][T^{-2}]}. \]
This implies:
\[ [a][L] = [M][L^2][T^{-2}]. \]
Thus,
\[ [a] = \frac{[M][L^2][T^{-2}]}{[L]} = [M][L][T^{-2}]. \]

### Step 4: Find the dimensions of \( \beta \).

From the original equation:
\[ p = \frac{A}{\beta} e^{\frac{aZ}{k\theta}}. \]
Since the exponential term is dimensionless, we can write:
\[ p = \frac{A}{\beta}. \]
Rearranging gives:
\[ \beta = \frac{A}{p}. \]

Substituting the dimensions we found:
- Dimensions of \( A \) are \( [M][L][T^{-2}] \).
- Dimensions of \( p \) are \( [M][L^{-1}][T^{-2}] \).

Thus,
\[ [\beta] = \frac{[M][L][T^{-2}]}{[M][L^{-1}][T^{-2}]} = [L^2]. \]

### Conclusion

The dimensional formula for \( \beta \) is:
\[ [\beta] = [L^2]. \]