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Step-by-Step Solution
Step 1: Write down the given relation
The force of interaction is given by
$F = \alpha \beta \exp\left(- \frac{x^2}{\alpha k T}\right)$.
Here,
• $x$ is the distance (with dimension $L$),
• $k$ is the Boltzmann constant (with dimension $M L^2 T^{-2} K^{-1}$, but for dimensional analysis regarding force, we primarily use it as having a factor $M L^2 T^{-2}$ when multiplied by temperature $T$),
• $T$ is the temperature (with dimension $K$),
• $\alpha$ and $\beta$ are constants whose dimensions we need to determine (eventually focusing on $\beta$).
Step 2: Recognize that the argument of the exponential must be dimensionless
In an exponential function $\exp(\dots)$, the argument must be dimensionless. Here, the argument is
$-\frac{x^2}{\alpha k T}$.
Thus, the combination
$\frac{x^2}{\alpha k T}$
must be dimensionless.
Step 3: Determine the dimension of $\alpha$
Let $[x] = L$ (distance) and $[kT] = M L^2 T^{-2}$ (since $kT$ has the dimension of energy). Therefore, from
$\frac{x^2}{\alpha \, (k T)}$ being dimensionless:
$\frac{L^2}{[\alpha] \cdot (M L^2 T^{-2})} = 1.$
Hence,
$[\alpha] \times M L^2 T^{-2} = L^2.$
$[\alpha] = \frac{L^2}{M L^2 T^{-2}} = M^{-1} T^2.$
Step 4: Write the dimensional equation for force
The force $F$ has dimensions $MLT^{-2}$. From the given expression
$F = \alpha \beta \exp(\dots),$
the exponential is dimensionless, so the dimensions of $F$ must match the product of dimensions of $\alpha$ and $\beta$:
$[F] = [\alpha] \, [\beta].$
Therefore,
$MLT^{-2} = (M^{-1} T^2)[\beta].$
Step 5: Solve for the dimension of $\beta$
Rearranging,
$[\beta] = (MLT^{-2}) \times (M \, T^{-2}) = M^2 L T^{-4}.$
Hence, the dimension of $\beta$ is $M^2 L T^{-4}.$
