The force of interaction between two atoms is | vedclass

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Question

$\begin{array}{l}
Power\,of\,e\,should\,be\,\dim ensionless.\
So,\left[ \lambda  ight] = \left( {\alpha Tk} ight)\
 \Rightarrow \,\

Vedclass · Accepted Answer

$M^2LT^{-4}$

Vedclass · Answer

$\begin{array}{l}
Power\,of\,e\,should\,be\,\dim ensionless.\
So,\left[ \lambda  ight] = \left( {\alpha Tk} ight)\
 \Rightarrow \,\,\,\,\,\,\,\,{L^2} = \left[ \alpha  ight]\left( {M{L^2}{T^{ - 2}}} ight)\
 \Rightarrow \,\,\,\,\,\,\,\,\,\left( \alpha  ight) = \left( {{M^{ - 1}}{T^2}} ight)\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,E = \frac{1}{2}KT\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {M{L^2}{T^{ - 2}}} ight)\,\,;\,\,\left( E ight) = \left[ {KT} ight]\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {\alpha \beta } ight) = \left( F ight)\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\left( {{M^{ - 1}}{T^2}} ight)\left( \beta  ight) = \left( {ML{T^{ - 2}}} ight)
\end{array}$

The force of interaction between two atoms is given by $F\, = \,\alpha \beta \,\exp \,\left( { - \frac{{{x^2}}}{{\alpha kt}}} \right);$ 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

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