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 Boltzman 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. $ML{T}^{-2}$

Answer 1

Answer: (B)

$M^{2}L{T}^{-4}$

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