Question:
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 :
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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The exponent must be dimensionless, so \([\alpha] = \frac{[x^2]}{[kT]}\) where \(kT\) has dimensions of energy; then \([\beta] = \frac{[F]}{[\alpha]}\).
Updated On: Nov 5, 2026
- \(M^2L^2T^{-2}\)
- \(M^2LT^{-4}\)
- \(MLT^{-4}\)
- \(M^0L^2LT^{-4}\)
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The Correct Option is B
Solution and Explanation
The argument of the exponential is dimensionless: \(\frac{x^2}{\alpha kT}\) has no dimensions.
\([kT] = \) energy \(= ML^2T^{-2}\), \([x^2] = L^2\).
\([\alpha] = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^{2}\).
Exp is dimensionless, so \([F] = [\alpha][\beta]\Rightarrow [\beta] = \frac{[F]}{[\alpha]} = \frac{MLT^{-2}}{M^{-1}T^{2}} = M^2LT^{-4}\).
Answer: (2) \(M^2LT^{-4}\)
\([kT] = \) energy \(= ML^2T^{-2}\), \([x^2] = L^2\).
\([\alpha] = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^{2}\).
Exp is dimensionless, so \([F] = [\alpha][\beta]\Rightarrow [\beta] = \frac{[F]}{[\alpha]} = \frac{MLT^{-2}}{M^{-1}T^{2}} = M^2LT^{-4}\).
Answer: (2) \(M^2LT^{-4}\)
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