In a typical combustion engine the workdone by a gas molecule is given by $$W = \alpha^2 \beta e^{\frac{-\beta x^2}{kT}}$$, 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:

The work done is given by $$W = \alpha^2 \beta e^{-\beta x^2/(kT)}$$.

Since the exponent must be dimensionless, we analyze $$\frac{\beta x^2}{kT}$$. Here $$x$$ is displacement with dimension $$[L]$$, $$k$$ is Boltzmann constant with dimension $$[ML^2T^{-2}K^{-1}]$$, and $$T$$ is temperature with dimension $$[K]$$. So $$kT$$ has dimension $$[ML^2T^{-2}]$$.

For the exponent to be dimensionless: $$[\beta] \cdot [L^2] = [ML^2T^{-2}]$$, giving $$[\beta] = [MT^{-2}]$$.

Now, the exponential is dimensionless, so $$[W] = [\alpha^2][\beta]$$. Since $$W$$ has dimension $$[ML^2T^{-2}]$$:

$$[\alpha^2][\beta] = [ML^2T^{-2}]$$

$$[\alpha^2][MT^{-2}] = [ML^2T^{-2}]$$

$$[\alpha^2] = \frac{[ML^2T^{-2}]}{[MT^{-2}]} = [L^2]$$

Therefore $$[\alpha] = [L] = [M^0LT^0]$$.

The answer is Option 4: $$[M^0LT^0]$$.