In the relation, `P=alpha/beta e^((alphaZ)/(ktheta))P` is pressure, Z is distance, k is Boltzmann constant and `theta` is the temperature. 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 equation \( P = \frac{\alpha}{\beta} e^{\frac{\alpha Z}{k \theta}} P \), we can follow these steps:

### Step 1: Understand the dimensions of the variables involved
- Pressure \( P \) has the dimensions of \( [M L^{-1} T^{-2}] \).
- Distance \( Z \) has the dimensions of \( [L] \).
- Boltzmann constant \( k \) has the dimensions of \( [M L^2 T^{-2} K^{-1}] \).
- Temperature \( \theta \) has the dimensions of \( [K] \).

### Step 2: Analyze the exponent in the equation
The term \( \frac{\alpha Z}{k \theta} \) must be dimensionless because the exponent of an exponential function must be dimensionless. Therefore, we can write:

\[ \text{Dimensions of } \frac{\alpha Z}{k \theta} = 1 \]

### Step 3: Write down the dimensions of each component
- Dimensions of \( Z \) are \( [L] \).
- Dimensions of \( k \) are \( [M L^2 T^{-2} K^{-1}] \).
- Dimensions of \( \theta \) are \( [K] \).

### Step 4: Set up the equation for dimensions
Thus, we have:

\[ \text{Dimensions of } \alpha \cdot [L] = [M L^2 T^{-2} K^{-1}] \cdot [K] \]

This simplifies to:

\[ \text{Dimensions of } \alpha \cdot [L] = [M L^2 T^{-2}] \]

### Step 5: Solve for the dimensions of \( \alpha \)
From the equation above, we can isolate the dimensions of \( \alpha \):

\[ \text{Dimensions of } \alpha = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}] \]

### Step 6: Relate \( \alpha \) and \( \beta \) through pressure
Now, we can use the relationship given in the equation:

\[ P = \frac{\alpha}{\beta} \]

Rearranging gives us:

\[ \beta = \frac{\alpha}{P} \]

### Step 7: Substitute the dimensions of \( \alpha \) and \( P \)
We already found:

- Dimensions of \( \alpha = [M L T^{-2}] \)
- Dimensions of \( P = [M L^{-1} T^{-2}] \)

Thus, substituting these into the equation for \( \beta \):

\[ \text{Dimensions of } \beta = \frac{[M L T^{-2}]}{[M L^{-1} T^{-2}]} \]

### Step 8: Simplify the dimensions of \( \beta \)
This simplifies to:

\[ \text{Dimensions of } \beta = [L^2] \]

### Final Result
Thus, the dimensional formula of \( \beta \) is:

\[ \text{Dimensions of } \beta = [M^0 L^2 T^0] \]