The work done by a gas molecule in an isolated system is given by, W = alpha * beta^2 * e^(-x^2 / (a

Q: Why must the exponent be dimensionless?

Exponential functions can be expanded as an infinite series (Maclaurin series): $e^y = 1 + y + y^2/2! + ...$. If $y$ had a dimension like length, you would be adding a pure number (1) to a length ($y$) and an area ($y^2$). This violates the principle of dimensional homogeneity, which states you can only add quantities with the same dimensions.

Q: I always forget the dimensions of the Boltzmann constant $k$. How can I derive it quickly?

Think of a simple formula involving $k$. The kinetic energy of a gas molecule is $E = \frac{3}{2} kT$. Therefore, $k = \frac{2E}{3T}$. The dimensions of $k$ are simply the dimensions of energy divided by temperature: $[M L^2 T^{-2}] / [K] = [M L^2 T^{-2} K^{-1}]$.

Q: Why did we ignore the negative sign in the exponent $-\frac{x^2}{\alpha kT}$?

Dimensions represent the physical nature of a quantity, not its magnitude or direction. A negative sign is just a dimensionless constant ($-1$). It does not affect the dimensional formula at all.

Q: Does the exponential term $e^{...}$ have any dimensions?

No. Since the power it is raised to is a pure number, the result of the exponential function is also a pure number. It is completely dimensionless, $[M^0 L^0 T^0]$.

Q: Is it possible for $\beta$ to have different dimensions if the equation was slightly different?

Absolutely! The dimensions of constants like $\alpha$ and $\beta$ are entirely dictated by the structure of the equation they are in. They are 'fitted' to ensure the equation is dimensionally consistent. If the equation changed, their dimensions would likely change too.

Question

The work done by a gas molecule in an isolated system is given by, $W = \alpha \beta^2 e^{-\frac{x^2}{\alpha kT}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature, $\alpha$ and $\beta$ are constants. Then, the dimensions of $\beta$ will be

Options:

(A) $[M^2 L T^2]$

(B) $[M^0 L T^0]$

(C) $[M L T^{-2}]$

(D) $[M L^2 T^{-2}]$

Sigma Percentile · Accepted Answer

### The Beauty of Dimensional Homogeneity

Imagine you are trying to add three apples and four oranges. It doesn't make sense, right? In physics, this simple logic is elevated to a powerful tool called the **Principle of Dimensional Homogeneity**. It states that you can only add, subtract, or equate quantities that have the exact same physical dimensions.

When we look at the intimidating equation given in the problem:

$$W = \alpha \beta^2 e^{-\frac{x^2}{\alpha kT}}$$

It might seem like a tangled mess of variables. But dimensional analysis gives us a pair of X-ray glasses to see right through the complexity.

### The Secret of the Exponent

Here is a golden rule that will save you countless times in JEE physics: **The argument of any transcendental function (like exponentials, logarithms, or trigonometric functions) must be a pure, dimensionless number.**

Why? Because functions like $e^y$ are defined by infinite series:

$$e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$$

If $y$ had dimensions of, say, length, you would be adding a pure number ($1$) to a length ($y$), to an area ($y^2$), and so on. That violates our apple-and-orange rule!

Therefore, the entire exponent in our equation must be dimensionless:

$$\left[ \frac{x^2}{\alpha kT} \right] = [M^0 L^0 T^0]$$

### Unlocking the Dimensions of Alpha

Since the exponent is dimensionless, we can easily isolate the dimensions of the unknown constant $\alpha$.

$$[\alpha] = \frac{[x^2]}{[k][T]}$$

Now, let's gather the dimensions of the known players. The displacement $x$ is simply length:

$$[x] = [L]$$

The Boltzmann constant $k$ is a bridge between temperature and energy. Its unit is Joules per Kelvin, so its dimensions are energy divided by temperature:

$$[k] = [M L^2 T^{-2} K^{-1}]$$

And temperature $T$ is a fundamental quantity:

$$[T] = [K]$$

Substituting these into our equation for $\alpha$:

$$[\alpha] = \frac{[L^2]}{[M L^2 T^{-2} K^{-1}] [K]}$$

Notice how beautifully the temperature dimensions ($K^{-1}$ and $K$) cancel each other out. The $L^2$ in the numerator and denominator also vanish. We are left with:

$$[\alpha] = [M^{-1} T^2]$$

### The Final Stretch: Finding Beta

Now that we have unmasked $\alpha$, let's return to the main equation. Since the exponential term $e^{-\frac{x^2}{\alpha kT}}$ is just a pure number, it contributes nothing to the overall dimensions.

This means the dimensions of the left side (Work) must perfectly match the dimensions of the remaining terms on the right side:

$$[W] = [\alpha] [\beta^2]$$

We want to find $\beta$, so let's rearrange this:

$$[\beta^2] = \frac{[W]}{[\alpha]}$$

Work is a form of energy, so its dimensions are:

$$[W] = [M L^2 T^{-2}]$$

Plugging in the dimensions of Work and $\alpha$:

$$[\beta^2] = \frac{[M L^2 T^{-2}]}{[M^{-1} T^2]}$$

When we bring the terms from the denominator up, the signs of their powers flip:

$$[\beta^2] = [M^2 L^2 T^{-4}]$$

Finally, taking the square root of both sides gives us the dimensions of $\beta$:

$$[\beta] = [M L T^{-2}]$$

### The Physical Revelation

We have our answer, which matches option (c). But let's take a moment to appreciate what we just found.

The dimensions $[M L T^{-2}]$ are the exact dimensions of **Force**! In the context of this specific physical model, the constant $\beta$ isn't just a random mathematical placeholder; it represents a physical force acting within the system.

This is the true magic of dimensional analysis. It doesn't just help you solve equations; it helps you understand the physical reality hiding behind the math. Keep practicing, and soon you'll be reading equations like a seasoned physicist!

Select Answer:

Visualized Solution

$Analyzing the Equation$

$Dimensionless Exponent$

$Dimensions of α$

$Substituting Known Dimensions$

$Calculating [α]$

$Homogeneity of the Main Equation$

$Solving for [β^{2}]$

$Calculating [β]$

$Physical Significance$

The Sigma Insight: Dimensional Analysis

The Beauty of Dimensional Homogeneity

The Secret of the Exponent

Unlocking the Dimensions of Alpha

The Final Stretch

The Physical Revelation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensionless Exponent

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensions of α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Known Dimensions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating [α]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Homogeneity of the Main Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for [β2]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating [β]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Significance