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 will follow these steps:

### Step 1: Identify the dimensions of pressure \( P \)
Pressure is defined as force per unit area. The dimensional formula for force is given by:
\[ \text{Force} = \text{mass} \times \text{acceleration} = M \cdot L \cdot T^{-2} \]
Since pressure is force per unit area, we have:
\[ P = \frac{\text{Force}}{\text{Area}} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 2: Analyze the equation
The equation can be rearranged to isolate \( \beta \):
\[ P = \frac{\alpha}{\beta} \]
This implies:
\[ \beta = \frac{\alpha}{P} \]

### Step 3: Find the dimensions of \( \alpha \)
From the equation, we can see that \( \alpha \) must have the same dimensions as \( P \) multiplied by \( \beta \). However, we need to express \( \beta \) in terms of \( P \) and \( \alpha \).

### Step 4: Substitute the dimensions
Assuming \( \alpha \) has some dimensions \( [\alpha] \), we can express \( \beta \) as:
\[ [\beta] = \frac{[\alpha]}{[P]} \]

### Step 5: Substitute the known dimensions
We know:
\[ [P] = M L^{-1} T^{-2} \]
Thus, substituting this into our equation gives:
\[ [\beta] = \frac{[\alpha]}{M L^{-1} T^{-2}} \]

### Step 6: Determine the dimensions of \( \beta \)
Assuming \( \alpha \) has dimensions \( [\alpha] = M^a L^b T^c \), we can express \( \beta \) as:
\[ [\beta] = \frac{M^a L^b T^c}{M L^{-1} T^{-2}} = M^{a-1} L^{b+1} T^{c+2} \]

### Step 7: Conclusion
To find the specific dimensional formula for \( \beta \), we need to know the dimensions of \( \alpha \). If we assume \( \alpha \) has dimensions of pressure (for example), then \( [\alpha] = M L^{-1} T^{-2} \):
\[ [\beta] = \frac{M L^{-1} T^{-2}}{M L^{-1} T^{-2}} = M^0 L^0 T^0 = 1 \]
Thus, the dimensional formula of \( \beta \) is dimensionless.

### Final Answer
The dimensional formula of \( \beta \) is \( M^0 L^0 T^0 \) (dimensionless).

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