# 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;

Question:

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;

1. $\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-4}$

2. $\mathrm{M}^{2} \mathrm{LT}^{-4}$

3. $\mathrm{MLT}^{-2}$

4. $\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-2}$

JEE Main Previous Year Single Correct Question of JEE Main from Chemistry Laws of Motion chapter.

JEE Main Previous Year 2019

Correct Option: 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

\begin{aligned} &\therefore\left[\frac{x^{2}}{\alpha k T}\right]=M^{0} L^{0} T^{0} \\ &\Rightarrow \frac{L^{2}}{[\alpha] M L^{2} T^{-2}}=M^{0} L^{0} T^{0} \\ &\Rightarrow[\alpha]=M^{-1} T^{2} \\ &{[F]=[\alpha][\beta]} \\ &M L T^{-2}=M^{-1} T^{2}[\beta] \\ &\Rightarrow[\beta]=M^{2} L T^{-4} \end{aligned}

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