Question

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 dimension of $\beta$ is :

• A. $M^2 L^2 T^{-2}$
• B. $M^2 L T^{-4}$
• C. $M^0 L^2 T^{-4}$
• D. $MLT^{-2}$

Answer 1

Answer: (B)

$F =\alpha\beta e^{\left(\frac{-x^2}{\alpha KT}\right)} $
$ \left[\frac{x^{2}}{\alpha KT}\right] =M^{\circ}L^{\circ}T^{\circ} $
$ \frac{L^{2}}{\left[\alpha\right]ML^{2} T^{-2}} =M^{\circ}L^{\circ}T^{\circ} $
$ \Rightarrow \left[\alpha\right] =M^{-1} T^{2}$
$ \left[F\right] =\left[\alpha\right]\left[\beta\right]$
$ MLT^{-2} =M^{-1}T^{2}\left[\beta\right] $
$ \Rightarrow \left[\beta\right] =M^{2}LT^{-4} $