Question

The force of interaction between two atoms is given by $F=\alpha \beta$ exp$(-\frac{x^2}{\alpha kt})$; 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. $M{L}^2{T}^{-4}$
• C. $M^{2}L{T}^{-4}$
• D. $ML{T}^{-2}$

Answer 1

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

Given:
$F=\alpha \beta e\left(\frac{-x^2}{\alpha KT}\right)\dots \left(i\right)$

Exponential terms are dimensionless

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

As we know, KT is thermal energy

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

Hence,

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

Therefore,

$\left[F\right]=\left[\alpha \right]\left[\beta \right]\dots \left(ii\right)$

By putting the dimensions in equation $\left(ii\right)$ we get,

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

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

Final answer :(b)