The work done by a gas molecule in an isolated system is given by, $$W = \alpha \beta^2 e^{-\frac{x^2}{\alpha k T}}$$, where $$x$$ is the displacement, $$k$$ is the Boltzmann constant and $$T$$ is the temperature. $$\alpha$$ and $$\beta$$ are constants. Then the dimensions of $$\beta$$ will be:

We are given $$W = \alpha \beta^2 e^{-\frac{x^2}{\alpha k T}}$$, where $$x$$ is displacement, $$k$$ is Boltzmann constant, and $$T$$ is temperature.

Since the exponent of $$e$$ must be dimensionless, we need $$\frac{x^2}{\alpha k T}$$ to be dimensionless.

The dimensions of $$x^2$$ are $$[L^2]$$. The dimensions of $$kT$$ are $$[ML^2T^{-2}]$$ (energy).

So $$[\alpha] = \frac{[x^2]}{[kT]} = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^{2}$$.

Now, $$W$$ has dimensions of work, i.e., $$[ML^2T^{-2}]$$. The exponential is dimensionless, so $$[W] = [\alpha][\beta^2]$$.

This gives $$[\beta^2] = \frac{[W]}{[\alpha]} = \frac{ML^2T^{-2}}{M^{-1}T^{2}} = M^{2}L^{2}T^{-4}$$.

Taking the square root, $$[\beta] = MLT^{-2}$$.

Hence, the correct answer is Option C.