# Q. 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 :

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Solution:

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

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