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दो परमाणुओं के मध्य अन्योन्यक्रिया बल सम्बन्ध $F =\alpha \beta \exp \left(-\frac{ x ^{2}}{\alpha kt }\right)$ से दिया जाता है जहाँ $x$ दूरी है, $k$ बोल्ट्जमैन नियतांक तथा $T$ तापमान है और $\alpha$ तथा $\beta$ दो स्थिरांक हैं। $\beta$ की विमा होगी।
$M^0L^2T^{-4}$
$M^2LT^{-4}$
$MLT^{-2}$
$M^2L^2T^{-2}$
Solution
$\begin{array}{l}
Power\,of\,e\,should\,be\,\dim ensionless.\\
So,\left[ \lambda \right] = \left( {\alpha Tk} \right)\\
\Rightarrow \,\,\,\,\,\,\,\,{L^2} = \left[ \alpha \right]\left( {M{L^2}{T^{ – 2}}} \right)\\
\Rightarrow \,\,\,\,\,\,\,\,\,\left( \alpha \right) = \left( {{M^{ – 1}}{T^2}} \right)\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,E = \frac{1}{2}KT\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {M{L^2}{T^{ – 2}}} \right)\,\,;\,\,\left( E \right) = \left[ {KT} \right]\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {\alpha \beta } \right) = \left( F \right)\\
\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\left( {{M^{ – 1}}{T^2}} \right)\left( \beta \right) = \left( {ML{T^{ – 2}}} \right)
\end{array}$
