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

Answer

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

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

we need to ensure that the terms in the exponent are dimensionless. Let's break down the steps to derive the dimensional formula for \(\beta\).

### Step 1: Understand the Terms
- \(P\) is pressure.
- \(Z\) is distance.
- \(k\) is the Boltzmann constant.
- \(\theta\) is temperature.
- \(\alpha\) and \(\beta\) are constants.

### Step 2: Identify the Dimensions
1. **Pressure (\(P\))**: The dimensional formula for pressure is given by:
\[ [P] = [M L^{-1} T^{-2}] \]

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

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

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

### Step 3: Ensure the Exponent is Dimensionless
For the expression \(-\frac{\alpha Z}{k \theta}\) to be dimensionless, the dimensions of \(\alpha Z\) must equal the dimensions of \(k \theta\). Therefore, we can write:

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

This implies:

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

Substituting the dimensions we have:

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

This simplifies to:

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

### Step 4: Solve for \(\alpha\)
From the above equation, we can isolate \([\alpha]\):

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

### Step 5: Relate \(\alpha\) and \(\beta\)
From the original equation, we have:

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

Rearranging gives:

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

### Step 6: Substitute Dimensions
Now substituting the dimensions we found:

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

This simplifies to:

\[ [\beta] = [M L T^{-2}] \cdot [M^{-1} L^{1} T^{2}] = [M^{0} L^{2} T^{0}] = [L^2] \]

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

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