Animated Solution for Physics - Physics and Measurement: In a typical combustion engine, the work done by a gas molecule is given $W={\alpha }^2\beta e^{-\frac{\beta x^2}{kT}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be

## Dimensional Analysis of a Complex Exponential Function

When you first encounter an equation like $W={\alpha }^2\beta e^{-\frac{\beta x^2}{kT}}$, it is completely normal to feel a bit intimidated. It looks like a dense forest of Greek letters, constants, and exponential functions. However, dimensional analysis acts as a powerful machete, allowing us to clear the path and reveal the underlying simplicity of the physics.

Let's embark on a step-by-step journey to decode this equation and find the dimensions of the constant $\alpha$.

The most crucial rule to remember in dimensional analysis is that the argument of any exponential function must be a pure, dimensionless number. An exponent is simply a power to which a base is raised; it cannot possess physical units like meters or seconds.

Therefore, the entire term $-\frac{\beta x^2}{kT}$ must be dimensionless. We can write this mathematically as:

To find the dimensions of $\beta$, we need to understand the dimensions of the other variables in this term. We are given that $x$ is displacement, so its dimension is simply $[L]$. The denominator contains $kT$, where $k$ is the Boltzmann constant and $T$ is temperature.

If you recall from the kinetic theory of gases, the term $kT$ is a measure of thermal energy (for example, the average kinetic energy of a gas molecule is $\frac{3}{2}kT$). Since $kT$ represents energy, its dimensions must be identical to the dimensions of work or energy:

Now, we can isolate the dimensions of $\beta$:

Notice how beautifully the $[L^2]$ terms in the numerator and denominator cancel each other out. This leaves us with the dimensions for $\beta$:

Now that we have successfully found the dimensions of $\beta$, let's shift our focus back to the main equation: $W={\alpha }^2\beta e^{-\frac{\beta x^2}{kT}}$.

We have already established that the entire exponential part, $e^{-\frac{\beta x^2}{kT}}$, is just a dimensionless number. In the realm of dimensional analysis, dimensionless numbers effectively 'disappear' because they do not contribute to the overall units of the equation.

This means that the dimensions of the Work done ($W$) on the left-hand side must be entirely provided by the pre-factor on the right-hand side, which is ${\alpha }^2\beta$. We can express this as:

We know that Work has the same dimensions as Energy, which is $[M{L}^2{T}^{-2}]$. We also just calculated the dimensions of $\beta$ to be $[M{T}^{-2}]$. Let's substitute these known dimensions into our equation to solve for ${\alpha }^2$:

Observe the elegance of the cancellation here. Both the mass dimension $[M]$ and the time dimension $[T^{-2}]$ are present in the numerator and the denominator, so they cancel out completely. We are left with:

To find the dimensions of $\alpha$, we simply take the square root of both sides. This yields:

In standard dimensional notation, where we explicitly state the zero powers of mass and time, this is written as $[M^{0}L{T}^0]$. Therefore, the correct option is (b).

This problem is a fantastic reminder that no matter how complex an equation appears, applying the fundamental rules of dimensional analysis systematically will always lead you to the correct answer.