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EEJ MAIN Physics QUESTION #7041
Question 1
The force between two atoms is $F = \alpha \beta \exp \left( -\frac{x^2}{\alpha k T} \right)$, where $x$ is distance, $k$ is Boltzmann constant, and $T$ is temperature. Find the dimensions of $\beta$.
  • $M^0 L^2 T^{-4}$
  • $M^2 L T^{-4}$✔️
  • $M L T^{-2}$
  • $M^2 L^2 T^{-2}$
Correct Answer Explanation

The exponent must be dimensionless. Thus, $[\alpha] = \frac{[x^2]}{[kT]}$.

Since $[kT]$ has dimensions of energy $[ML^2T^{-2}]$, $[\alpha] = \frac{[L^2]}{[ML^2T^{-2}]} = [M^{-1}T^2]$.

Force $F$ has the same dimensions as $\alpha\beta$.

$[\beta] = \frac{[F]}{[\alpha]} = \frac{[MLT^{-2}]}{[M^{-1}T^2]} = [M^2 L T^{-4}]$.