JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
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
- A \(M^0L^2T^{-4}\)
- B \(M^2LT^{-4}\)
- C \(MLT^{-2}\)
- D \(M^2L^2T^{-2}\)
Answer & Solution
Correct Answer
(B) \(M^2LT^{-4}\)
Step-by-step Solution
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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