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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement

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।

  1. A \(M^0L^2T^{-4}\)
  2. B \(M^2LT^{-4}\)
  3. C \(MLT^{-2}\)
  4. D \(M^2L^2T^{-2}\)
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Answer & Solution

Correct Answer

(B) \(M^2LT^{-4}\)

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Detailed explanation

\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^{…

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