`F=alphabetae^((-(x)/(alphakt)))`
k = Boltzmann constant
t = temperature
x = distance
The dimensions of B is

Options

A. `[MLT^(-4)]`
B. `[M^(2)L^(2)T^(-4)]`
C. `[MLT^(-2)]`
D. `[M^(2)LT^(-1)]`

Answer

To find the dimensions of \( \beta \) in the equation

\[ F = \alpha \beta e^{-\frac{x}{\alpha k t}} \]

we will follow these steps:

### Step 1: Identify the dimensions of force \( F \)
The dimension of force \( F \) is given by:

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

### Step 2: Analyze the exponential term
The term \( e^{-\frac{x}{\alpha k t}} \) must be dimensionless. This means that the exponent \( -\frac{x}{\alpha k t} \) must have dimensions of zero.

### Step 3: Determine the dimensions of \( x \), \( k \), and \( t \)
- The dimension of distance \( x \) is:

\[ [x] = L \]

- The dimension of Boltzmann constant \( k \) is:

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

where \( \Theta \) represents temperature.

- The dimension of temperature \( t \) is:

\[ [t] = \Theta \]

### Step 4: Combine the dimensions in the exponent
Now we need to find the dimensions of \( \alpha k t \):

\[ [\alpha k t] = [\alpha] \cdot [k] \cdot [t] \]

### Step 5: Set up the equation for dimensionless exponent
Since \( -\frac{x}{\alpha k t} \) is dimensionless, we can write:

\[ \frac{[x]}{[\alpha k t]} = 1 \]

This implies:

\[ [x] = [\alpha k t] \]

### Step 6: Substitute the known dimensions
Substituting the dimensions we have:

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

This simplifies to:

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

### Step 7: Solve for dimensions of \( \alpha \)
Rearranging gives:

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

Thus, the dimensions of \( \alpha \) are:

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

### Step 8: Relate \( F \) to \( \alpha \) and \( \beta \)
From the original equation, we can express \( F \) in terms of \( \alpha \) and \( \beta \):

\[ F = \alpha \beta \]

### Step 9: Solve for dimensions of \( \beta \)
Now, substituting the dimensions of \( F \) and \( \alpha \):

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

Substituting the dimension of \( \alpha \):

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

### Step 10: Rearranging to find \( \beta \)
Rearranging gives:

\[ [\beta] = [M L T^{-2}] \cdot (M L^{-1} T^{-2}) = M^{2} L^{2} T^{-4} \]

Thus, the dimensions of \( \beta \) are:

\[ [\beta] = M^{2} L^{2} T^{-4} \]

### Final Answer
The dimensions of \( \beta \) are \( M^{2} L^{2} T^{-4} \).

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