In the relation: `P=(alpha)/(beta)e^(-(alphaZ)/(ktheta)),P` is pressure `Z` is distance `k` is Boltzmann constant and `theta` ils the temperature. The dimensional formula of `beta` will bes

Options

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

Explanation

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

\[ P = \frac{\alpha}{\beta} e^{-\frac{\alpha Z}{k \theta}} \]

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

### Step 1: Understand the Equation
The equation states that pressure \( P \) is equal to \( \frac{\alpha}{\beta} \) multiplied by a dimensionless exponential term. For the entire expression to be dimensionally consistent, \( \frac{\alpha}{\beta} \) must have the same dimensions as pressure \( P \).

### Step 2: Identify the Dimensions of Pressure
The dimensional formula for pressure \( P \) is given by:

\[ [P] = \frac{\text{Force}}{\text{Area}} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 3: Identify the Dimensions of \( \alpha \)
To find the dimensions of \( \beta \), we first need to determine the dimensions of \( \alpha \). The term \( e^{-\frac{\alpha Z}{k \theta}} \) is dimensionless, which means that \( \frac{\alpha Z}{k \theta} \) must also be dimensionless.

### Step 4: Write the Dimensions of Each Component
- \( Z \) (distance) has dimensions: \( [Z] = L \)
- \( k \) (Boltzmann constant) has dimensions:

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

where \( \Theta \) represents temperature.

- \( \theta \) (temperature) has dimensions: \( [\theta] = \Theta \)

### Step 5: Set Up the Dimensionless Condition
Since \( \frac{\alpha Z}{k \theta} \) is dimensionless, we can write:

\[ [\alpha] \cdot [Z] = [k] \cdot [\theta] \]

Substituting the dimensions we have:

\[ [\alpha] \cdot L = (M L^2 T^{-2} \Theta^{-1}) \cdot \Theta \]

This simplifies to:

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

### Step 6: Solve for the Dimensions of \( \alpha \)
Rearranging gives:

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

### Step 7: Find the Dimensions of \( \beta \)
Now, since \( \frac{\alpha}{\beta} \) must have the same dimensions as pressure \( P \):

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

Substituting the known dimensions:

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

### Step 8: Solve for \( [\beta] \)
Cross-multiplying gives:

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

Thus, the dimensional formula for \( \beta \) is:

\[ [\beta] = L^2 \]

### Final Answer
The dimensional formula of \( \beta \) is \( L^2 \).

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