Q 11-01-223JEE MainJEE Main 2019 (11 Jan, Shift 1)Medium
The force of interaction between two atoms is given by $F = \alpha\beta\exp\left(-\dfrac{x^2}{\alpha kT}\right)$, where $x$ is the distance, $k$ is the Boltzmann constant, $T$ is temperature and $\alpha$ and $\beta$ are two constants. The dimensions of $\beta$ are:
Answer: (B) $M^2LT^{-4}$
The exponent must be dimensionless, so $\alpha$ has the dimensions of $\dfrac{x^2}{kT}$. Since $kT$ is an energy:
$$[\alpha] = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^{2}$$
The exponential is a pure number, so $\alpha\beta$ has the dimensions of force:
$$[\beta] = \frac{MLT^{-2}}{M^{-1}T^{2}} = M^2LT^{-4}$$
Solution by Sreeraj P, M.Sc Physics