A dimensionless quantity is constructed in terms of electronic charge e, permittivity of free space

Q: Why do we use Ampere (A) as a fundamental dimension instead of Coulomb (C)?

In the SI system, electric current (Ampere) is chosen as the fundamental base quantity because it is easier to measure precisely in a laboratory (via magnetic forces) than static charge. Therefore, charge is a derived quantity: $Q = I \times t$, giving it the dimension $[A T]$.

Q: How can I quickly derive the dimensional formula for $\varepsilon_0$ if I forget it?

You can always derive it from Coulomb's Law: $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$. Rearranging for $\varepsilon_0$ gives $\varepsilon_0 = \frac{q^2}{F r^2}$. Substitute the dimensions of charge $[A T]$, force $[M L T^{-2}]$, and distance $[L]$ to get $[M^{-1} L^{-3} T^4 A^2]$.

Q: Is it necessary to check the equation for time (T) if I already found all variables?

While not strictly necessary if your math is flawless, checking the time equation acts as a perfect built-in verification. If the powers of T don't balance to zero with your found values, you know you made an algebraic mistake earlier.

Q: Why did the problem introduce the integer 'n'?

Because any power of a dimensionless quantity is still dimensionless! If $X$ is dimensionless, so is $X^2, X^{-1}$, or $X^{-n}$. The integer $n$ simply represents this infinite family of valid dimensionless combinations.

Q: What is the physical meaning of this specific dimensionless combination?

When $n=1$, the combination is proportional to the fine-structure constant, denoted by $\alpha$. It is a fundamental physical constant that characterizes the strength of the electromagnetic interaction between elementary charged particles.

Question

A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space $\varepsilon_0$, Planck's constant $h$, and speed of light $c$. If the dimensionless quantity is written as $e^\alpha \varepsilon_0^\beta h^\gamma c^\delta$ and $n$ is a non-zero integer, then $(\alpha, \beta, \gamma, \delta)$ is given by

Options:

(A) $(2n, -n, -n, -n)$

(B) $(n, -n, -2n, -n)$

(C) $(n, -n, -n, -2n)$

(D) $(2n, -n, -2n, -2n)$

Sigma Percentile · Accepted Answer

### The Power of Dimensional Analysis

Imagine you are an explorer in the vast universe of physics, trying to uncover the hidden connections between fundamental constants. Dimensional analysis is your ultimate compass.

It allows us to strip away the numerical values and look purely at the "DNA" of physical quantities—their fundamental dimensions of mass, length, time, and electric current.

In this thrilling problem, we are tasked with combining four fundamental constants of nature to create a dimensionless quantity. This isn't just a random math exercise; it's how physicists discover profound truths about the universe!

### Deconstructing the Physical Constants

Before we can build our dimensionless quantity, we need to gather our building blocks. We must express each given constant in terms of the fundamental SI dimensions: Mass ($M$), Length ($L$), Time ($T$), and Electric Current ($A$).

First, we have the elementary charge, $e$. Since current is the rate of flow of charge ($I = Q/t$), charge is simply current multiplied by time. Thus, its dimensional formula is $[A T]$.

Next is the speed of light, $c$. Like any speed, it is distance divided by time, giving us $[L T^{-1}]$.

Then we have Planck's constant, $h$. We can derive its dimensions from the energy equation $E = h
u$. Since energy is $[M L^2 T^{-2}]$ and frequency $
u$ is $[T^{-1}]$, dividing them gives $h = [M L^2 T^{-1}]$.

Finally, we need the permittivity of free space, $\varepsilon_0$. We can extract this from Coulomb's Law: $F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2}$. Rearranging for $\varepsilon_0$ and substituting the dimensions of force, charge, and distance yields $[M^{-1} L^{-3} T^4 A^2]$.

### The Master Equation

Now that we have our building blocks, let's construct the master equation. We are given the expression $e^\alpha \varepsilon_0^\beta h^\gamma c^\delta$.

Because this combination is stated to be **dimensionless**, its net dimensional formula must be exactly $[M^0 L^0 T^0 A^0]$.

Let's substitute our derived dimensional formulas into the expression:

$$ [A T]^\alpha [M^{-1} L^{-3} T^4 A^2]^\beta [M L^2 T^{-1}]^\gamma [L T^{-1}]^\delta = [M^0 L^0 T^0 A^0] $$

To make sense of this, we need to group the like base quantities together. By carefully applying the laws of exponents, we consolidate the powers for $M, L, T,$ and $A$:

$$ M^{-\beta + \gamma} L^{-3\beta + 2\gamma + \delta} T^{\alpha + 4\beta - \gamma - \delta} A^{\alpha + 2\beta} = M^0 L^0 T^0 A^0 $$

### Solving the Matrix of Powers

We have now arrived at a beautiful system of linear equations. For the entire expression to be dimensionless, the exponent of every single base quantity must independently equal zero.

Let's start with the easiest ones. Equating the powers of Mass ($M$) to zero gives us:

$$ -\beta + \gamma = 0 \implies \gamma = \beta $$

Next, let's look at the powers of Electric Current ($A$):

$$ \alpha + 2\beta = 0 \implies \alpha = -2\beta $$

We are making great progress! Now, let's tackle the powers of Length ($L$):

$$ -3\beta + 2\gamma + \delta = 0 $$

Since we already know that $\gamma = \beta$, we can substitute this directly into the length equation:

$$ -3\beta + 2(\beta) + \delta = 0 $$

Simplifying this yields $-\beta + \delta = 0$, which means **$\delta = \beta$**.

### The Grand Reveal

We have successfully expressed all the unknown powers in terms of a single variable, $\beta$. Our solution tuple $(\alpha, \beta, \gamma, \delta)$ can be written generally as $(-2\beta, \beta, \beta, \beta)$.

To ensure our algebra is flawless, we can use the Time ($T$) equation as a built-in verification check:

$$ \alpha + 4\beta - \gamma - \delta = 0 $$

Substituting our derived values gives $(-2\beta) + 4\beta - (\beta) - (\beta) = 0$, which simplifies perfectly to $0 = 0$. Our math is rock solid!

Now, we look at the given options. The problem introduces an integer $n$. If we strategically set $\beta = -n$, our general tuple transforms into:

$$ (2n, -n, -n, -n) $$

This perfectly matches Option (A). We have conquered the problem!

### Beyond the Problem: The Fine-Structure Constant

Why did the examiners choose this specific combination of constants? It is not a coincidence.

If we set $n=1$, our dimensionless quantity becomes $\frac{e^2}{\varepsilon_0 h c}$. In the realm of quantum electrodynamics, this combination is intimately related to the **fine-structure constant**, denoted by $\alpha$.

The fine-structure constant (approximately equal to $1/137$) is a fundamental physical constant that characterizes the strength of the electromagnetic interaction between elementary charged particles.

By solving this problem, you haven't just performed an algebraic exercise; you have mathematically reconstructed one of the most important numbers in the universe!

Select Answer:

Visualized Solution

The Goal

Dimensional Formulas

The Dimensional Equation

Equating Powers of $M$ and $A$

Equating Powers of $L$

Verifying with $T$

Final Answer

The Fine-Structure Constant

The Sigma Insight: Dimensional Analysis

The Power of Dimensional Analysis

Deconstructing the Physical Constants

The Master Equation

Solving the Matrix of Powers

The Grand Reveal

Beyond the Problem

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