Animated Solution for Physics - Physics 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

When you first look at the equation $F=\alpha \beta \exp \left(-\frac{x^2}{\alpha kT}\right)$, it might seem intimidating. It describes the force of interaction between two atoms, but for this problem, we don't need to understand the deep physics of atomic interactions. We only need to analyze its dimensional structure.

The equation consists of a pre-exponential factor $\alpha \beta$ and an exponential term $\exp \left(-\frac{x^2}{\alpha kT}\right)$. Our ultimate goal is to find the dimension of the constant $\beta$.

There is a fundamental principle in dimensional analysis: any mathematical function like an exponential, logarithm, or trigonometric function must have a dimensionless argument. You cannot raise Euler's number $e$ to a physical quantity like "3 meters" or "5 seconds". The power must be a pure, dimensionless number.

Therefore, the entire argument of the exponential must have the dimensions of $[M^0{L}^0{T}^0]$. We can write this as:

Since the overall fraction is dimensionless, the dimensions of the numerator must perfectly cancel the dimensions of the denominator. This allows us to isolate $\alpha$ and determine its dimensions:

Now, let's break down the components. The variable $x$ represents distance, so its dimension squared is $[L^2]$. The term $kT$ is the product of the Boltzmann constant $k$ and temperature $T$. In thermodynamics, $kT$ represents thermal energy. The dimensional formula for any form of energy (like work done, which is force times displacement) is $[M{L}^2{T}^{-2}]$.

Substituting these known dimensions into our equation for $\alpha$:

The $[L^2]$ terms in the numerator and denominator cancel out beautifully. Bringing the remaining terms to the numerator, we get:

Finding Beta

Now that we have unlocked the dimension of $\alpha$, let's return to the main force equation. We established earlier that the entire exponential term $\exp (...)$ evaluates to a pure number, meaning it is completely dimensionless.

Because of this, the dimension of the Force $F$ on the left side of the equation must be exactly equal to the product of the dimensions of $\alpha$ and $\beta$ on the right side:

We want to find $\beta$, so we rearrange the equation:

Force is defined as mass times acceleration, so its dimensional formula is $[ML{T}^{-2}]$. We plug this in, along with our newly found dimension for $\alpha$:

When we simplify this expression by moving the denominator's powers to the numerator (changing their signs), we get:

And there we have it! By systematically applying the principle of dimensional homogeneity, we have successfully deduced the dimension of $\beta$.