The force of interaction between two atoms is given by $F=\alpha \beta \exp \left(-\frac{x^{2}}{\alpha k…
The force of interaction between two atoms is given by $F=\alpha \beta \exp \left(-\frac{x^{2}}{\alpha k T}\right) ;$ where $x$ is the distance, $\mathrm{k}$ is the
Boltzmann constant and T is temperature and $\alpha$ and $\beta$ are two constants. The dimensions of $\beta$ is:
$\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-4}$
$\mathrm{M}^{2} \mathrm{LT}^{-4}$
$\mathrm{MLT}^{-2}$
$\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-2}$
Solution
Force of interaction between two atoms,
$\mathrm{F}=\alpha \beta \mathrm{e}^{\left(\frac{-\mathrm{x}^{2}}{\alpha \mathrm{kT}}\right)}$
Since exponential terms are dimensionless $\therefore\left[\frac{\mathrm{x}^{2}}{\alpha \mathrm{k} \mathrm{T}}\right]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
$\Rightarrow \frac{\mathrm{L}^{2}}{[\alpha] \mathrm{ML}^{2} \mathrm{~T}^{-2}}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
$\Rightarrow[\alpha]=\mathrm{M}^{-1} \mathrm{~T}^{2}$
$[\mathrm{F}]=[\alpha][\beta]$
$\mathrm{MLT}^{-2}=\mathrm{M}^{-1} \mathrm{~T}^{2}[\beta]$
$\Rightarrow[\beta]=M^{2} L T^{-4}$