The force of interaction between two atoms is given by $$F = \alpha\beta \exp\left(-\frac{x^2}{\alpha kT}\right)$$; where $$x$$ is the distance, k is the Boltzmann constant and T is temperature and $$\alpha$$ and $$\beta$$ are two constants. The dimensions of $$\beta$$ is:

We have the given force law $$F=\alpha\beta\exp\!\left(-\dfrac{x^{2}}{\alpha kT}\right).$$

The dimensions of the exponential function must be unity, because an exponential of a dimensional quantity is not physically meaningful. Hence the entire argument of the exponential must be dimensionless:

$$\dfrac{x^{2}}{\alpha kT}\; \text{is dimensionless.}$$

So the numerator and denominator must possess identical dimensions. The distance $$x$$ has the dimension $$[x]=L,$$ therefore $$[x^{2}]=L^{2}.$$

The Boltzmann constant $$k$$ has the well-known dimension “energy per kelvin,” that is

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

and temperature $$T$$ carries the dimension $$[T]=\Theta.$$ Multiplying these,

$$[kT]=(M\,L^{2}T^{-2}\Theta^{-1})(\Theta)=M\,L^{2}T^{-2}.$$

Because $$\alpha kT$$ must match $$x^{2},$$ we write

$$[\alpha kT]=L^{2}.$$

Substituting the dimension of $$kT,$$ we obtain

$$[\alpha]\,(M\,L^{2}T^{-2})=L^{2}.$$

Dividing both sides by $$M\,L^{2}T^{-2},$$

$$[\alpha]=\dfrac{L^{2}}{M\,L^{2}T^{-2}}=M^{-1}T^{2}.$$

Now we return to the original force equation. Force has the fundamental dimension

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

This force equals the product $$\alpha\beta$$ (the exponential being dimensionless), hence

$$[F]=[\alpha][\beta].$$

We already found $$[\alpha]=M^{-1}T^{2},$$ so

$$[\beta]=\dfrac{[F]}{[\alpha]}=\dfrac{M\,L\,T^{-2}}{M^{-1}T^{2}}.$$

Carrying out the division means multiplying by the reciprocal of $$[\alpha]:$$

$$[\beta]=(M\,L\,T^{-2})(M\,T^{-2})=M^{2}L\,T^{-4}.$$

Therefore the dimensions of $$\beta$$ are $$M^{2}L\,T^{-4}.$$

Looking at the options, this matches option B.

Hence, the correct answer is Option B.