According to Maxwell`-`distribution law, the probability function representing the ratio of molecules at a particular velocity to the total number of molecules is given by
`f(v)=k_(1)sqrt(((m)/(2piKT^(2))))4piv^(2)e^(-(mv^(2))/(2KT))`
Where `m` is the mass of the molecule, `v` is the velocity of the molecule, `T` is the temperature `k` and `k_(1)` are constant. The dimensional formulae of `k_(1)` is

Options

A. `L^(2)T^(-2)`
B. `L^(1)T^(-1)K^(-3//2)`
C. `L^(1)T^(-1)K^(+3//2)`
D. `L^(2)T^(-1)K^(+3//2)`

Explanation

To find the dimensional formula of the constant \( k_1 \) in the given probability function according to Maxwell's distribution law, we start by analyzing the equation:

\[ f(v) = k_1 \sqrt{\frac{m}{2\pi k T^2}} 4\pi v^2 e^{-\frac{mv^2}{2kT}} \]

### Step 1: Identify the dimensions of each component in the equation

1. **Dimensions of \( m \)** (mass of the molecule):
- \( [m] = M \)

2. **Dimensions of \( k \)** (Boltzmann constant):
- \( k \) has dimensions of energy per temperature, which is \( [k] = \frac{[E]}{[T]} = \frac{ML^2T^{-2}}{Θ} = ML^2T^{-2}Θ^{-1} \)

3. **Dimensions of \( T \)** (temperature):
- \( [T] = Θ \)

4. **Dimensions of \( v \)** (velocity):
- \( [v] = LT^{-1} \)

### Step 2: Analyze the term \( \sqrt{\frac{m}{2\pi k T^2}} \)

The term inside the square root is:

\[ \frac{m}{2\pi k T^2} \]

Ignoring the constant \( 2\pi \) for dimensional analysis, we have:

\[ \frac{m}{k T^2} = \frac{M}{ML^2T^{-2}Θ^{-1} \cdot Θ^2} = \frac{M}{ML^2T^{-2}Θ} = \frac{1}{L^2T^{-2}Θ} \]

Thus, the dimensions of this term are:

\[ [L^{-2}T^2Θ^{-1}] \]

Taking the square root gives:

\[ \sqrt{\frac{m}{2\pi k T^2}} = [L^{-1}TΘ^{-1}] \]

### Step 3: Analyze the term \( 4\pi v^2 \)

The term \( v^2 \) has dimensions:

\[ [v^2] = [L^2T^{-2}] \]

Thus, the dimensions of \( 4\pi v^2 \) are:

\[ [L^2T^{-2}] \]

### Step 4: Combine the dimensions

Now, we combine the dimensions from the two components:

\[ \sqrt{\frac{m}{2\pi k T^2}} \cdot 4\pi v^2 = [L^{-1}TΘ^{-1}] \cdot [L^2T^{-2}] = [L^{1}T^{-1}Θ^{-1}] \]

### Step 5: Determine the dimensions of \( k_1 \)

The function \( f(v) \) represents a probability density function, which is dimensionless. Therefore, to make the entire expression dimensionless, the dimensions of \( k_1 \) must cancel out the dimensions of the combined terms:

\[ [k_1] \cdot [L^{1}T^{-1}Θ^{-1}] = [1] \]

Thus, we have:

\[ [k_1] = [L^{-1}T^{1}Θ] \]

### Final Answer

The dimensional formula of \( k_1 \) is:

\[ \boxed{L^{-1}T^{1}Θ} \]