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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
$M^0L^2T^{-4}$
$M^2LT^{-4}$
$MLT^{-2}$
$M^2L^2T^{-2}$
Solution
$\begin{array}{l}
Power\,of\,e\,should\,be\,\dim ensionless.\\
So,\left[ \lambda \right] = \left( {\alpha Tk} \right)\\
\Rightarrow \,\,\,\,\,\,\,\,{L^2} = \left[ \alpha \right]\left( {M{L^2}{T^{ – 2}}} \right)\\
\Rightarrow \,\,\,\,\,\,\,\,\,\left( \alpha \right) = \left( {{M^{ – 1}}{T^2}} \right)\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,E = \frac{1}{2}KT\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {M{L^2}{T^{ – 2}}} \right)\,\,;\,\,\left( E \right) = \left[ {KT} \right]\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {\alpha \beta } \right) = \left( F \right)\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\left( {{M^{ – 1}}{T^2}} \right)\left( \beta \right) = \left( {ML{T^{ – 2}}} \right)
\end{array}$
