Pressure depends on distance as, `P = (alpha)/(beta)exp((-alpha z)/(k theta))`, where `alpha, beta` are constants, z is distance, k is Boltzmann's constant and `theta` is temperature. The dimensions of `beta` are :

Options

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

Answer

To find the dimensions of the constant \( \beta \) in the equation \( P = \frac{\alpha}{\beta} \exp\left(-\frac{\alpha z}{k \theta}\right) \), we can follow these steps:

### Step 1: Understand the Equation
The equation relates pressure \( P \) to constants \( \alpha \) and \( \beta \), as well as the distance \( z \), Boltzmann's constant \( k \), and temperature \( \theta \). We need to ensure that the exponent is dimensionless.

### Step 2: Analyze the Exponent
The term in the exponent is \( -\frac{\alpha z}{k \theta} \). For this to be dimensionless, the dimensions of \( \alpha z \) must equal the dimensions of \( k \theta \).

### Step 3: Identify Dimensions
- **Distance \( z \)** has dimensions of length: \( [z] = L \).
- **Boltzmann's constant \( k \)** has dimensions of energy per temperature, which is \( [k] = \frac{ML^2}{T^2 \Theta} \) (where \( \Theta \) represents temperature).
- **Temperature \( \theta \)** has dimensions of temperature: \( [\theta] = \Theta \).

### Step 4: Set Up the Dimension Equation
From the dimensionless condition:
\[ [\alpha z] = [k \theta] \]
This gives us:
\[ [\alpha] \cdot [z] = [k] \cdot [\theta] \]
Substituting the dimensions:
\[ [\alpha] \cdot L = \left(\frac{ML^2}{T^2 \Theta}\right) \cdot \Theta \]
This simplifies to:
\[ [\alpha] \cdot L = \frac{ML^2}{T^2} \]

### Step 5: Solve for Dimensions of \( \alpha \)
Rearranging gives:
\[ [\alpha] = \frac{ML^2}{T^2} \cdot \frac{1}{L} = \frac{ML}{T^2} \]
Thus, the dimensions of \( \alpha \) are:
\[ [\alpha] = M L T^{-2} \]

### Step 6: Determine Dimensions of Pressure
Pressure \( P \) has dimensions of force per unit area:
\[ [P] = \frac{F}{A} = \frac{MLT^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 7: Relate Dimensions of \( \beta \)
From the equation \( P = \frac{\alpha}{\beta} \), we can express the dimensions of \( \beta \):
\[ [P] = \frac{[\alpha]}{[\beta]} \implies [\beta] = \frac{[\alpha]}{[P]} \]
Substituting the dimensions we found:
\[ [\beta] = \frac{M L T^{-2}}{M L^{-1} T^{-2}} = L^2 \]

### Final Answer
Thus, the dimensions of \( \beta \) are:
\[ [\beta] = L^2 \]