Question

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

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

Answer 1

The correct option is B $M^{2}L{T}^{-4}$

$\text{The dimension of force,}F=\left[ML{T}^{-2}\right]$

$\text{The dimension of distance,}x=\left[L\right]$

$\text{The dimension of Energy,}kT=\left[M{L}^2{T}^{-2}\right]$


Given that, the Force of interaction between two atoms,
$F=\alpha \beta e^{\left(\frac{-x^2}{\alpha kT}\right)}$

Since exponential terms are dimensionless,

$\therefore \left[\frac{x^2}{\alpha kT}\right]=M^0{L}^0{T}^0$

$\Rightarrow \frac{L^2}{\left[\alpha \right]M{L}^2{T}^{-2}}=M^0{L}^0{T}^0$

$\Rightarrow \left[\alpha \right]=M^{-1}{T}^2$

$\text{Also},\left[F\right]=\left[\alpha \right]\left[\beta \right]$

$\Rightarrow ML{T}^{-2}=M^{-1}{T}^2\left[\beta \right]$

$\Rightarrow \left[\beta \right]=M^{2}L{T}^{-4}$

Hence, option $\left(B\right)$ is correct.