In a typical combustion engine the work done by a gas molecule is given `W=alpha^2betae^((-betax^2)/(kT))` where x is the displacement, k is the Boltzmann constant and T is the temperature. If `alpha and beta` are constants, dimensions of `alpha` will be:

Options

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

Answer

To determine the dimensions of the constant \( \alpha \) in the given equation for work done by a gas molecule, we start with the expression:

\[ W = \alpha^2 \beta e^{\left(-\frac{\beta x^2}{kT}\right)} \]

Where:
- \( W \) is the work done,
- \( \alpha \) and \( \beta \) are constants,
- \( x \) is the displacement,
- \( k \) is the Boltzmann constant,
- \( T \) is the temperature.

### Step 1: Identify the dimensions of work done \( W \)

The dimension of work done \( W \) is given by:

\[ [W] = [\text{Force}] \times [\text{Distance}] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}] \]

### Step 2: Analyze the exponential term

The term inside the exponential \( e^{\left(-\frac{\beta x^2}{kT}\right)} \) must be dimensionless. Therefore, we need to ensure that the dimensions of \( \frac{\beta x^2}{kT} \) are dimensionless:

\[ \left[\frac{\beta x^2}{kT}\right] = 1 \]

### Step 3: Determine the dimensions of \( x^2 \)

The dimension of displacement \( x \) is:

\[ [x] = [L] \]

Thus, the dimension of \( x^2 \) is:

\[ [x^2] = [L^2] \]

### Step 4: Determine the dimensions of \( k \) and \( T \)

The Boltzmann constant \( k \) has the dimensions:

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

The temperature \( T \) has the dimension:

\[ [T] = [K] \]

### Step 5: Combine the dimensions in the exponential

Now, we can express the dimensions of \( kT \):

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

### Step 6: Set up the equation for \( \beta \)

From the dimensional analysis of the exponential term, we have:

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

This implies:

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

### Step 7: Substitute back to find dimensions of \( \alpha \)

Now, substituting the dimensions of \( \beta \) back into the work done equation:

\[ [W] = [\alpha^2] \cdot [\beta] \]

This gives us:

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

Dividing both sides by \( [M T^{-2}] \):

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

This simplifies to:

\[ [L^2] = [\alpha^2] \]

### Step 8: Solve for \( \alpha \)

Taking the square root of both sides:

\[ [\alpha] = [L] \]

### Conclusion

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

\[ [\alpha] = [L^1] = [M^0 L^1 T^0] \]