JEE PYQ: Units & Measurements - Question ID 59359e97e57e (JEE Main 2021)

In dimensional analysis, every physical quantity can be expressed in terms of fundamental dimensions: mass ($M$), length ($L$), time ($T$), electric current ($I$), thermodynamic temperature ($\Theta$), amount of substance ($N$), and luminous intensity ($J$). The given expression involves:

• Electronic charge ($e$): Fundamental charge of an electron.
• Speed of light in free space ($c$): A universal constant with dimensions of velocity.
• Planck's constant ($h$): A fundamental constant in quantum mechanics, relating energy to frequency.
• Permittivity of free space ($\varepsilon_0$): A constant appearing in Coulomb's law, relating electric field to charge.

The quantity in question is: $\frac{1}{4\pi \varepsilon_0} \frac{|e|^2}{hc}$ Our goal is to determine its dimensions by expressing each term in terms of $M$, $L$, $T$, and $I$.

Step 1: Write down the dimensions of each constant.

We use the following well-known dimensional formulas:

• Electronic charge ($e$): $[e] = [IT]$ (since current × time = charge).
• Speed of light ($c$): $[c] = [LT^{-1}]$.
• Planck's constant ($h$): From $E = h\nu$, where $E$ is energy ($[ML^2T^{-2}]$) and $\nu$ is frequency ($[T^{-1}]$), we get: $[h] = \frac{[E]}{[\nu]} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$
• Permittivity of free space ($\varepsilon_0$): From Coulomb's law, $F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}$, where $F$ is force ($[MLT^{-2}]$), $q$ is charge ($[IT]$), and $r$ is distance ($[L]$). Rearranging: $[\varepsilon_0] = \frac{[q^2]}{[F][r^2]} = \frac{[I^2T^2]}{[MLT^{-2}][L^2]} = [M^{-1}L^{-3}T^4I^2]$

Step 2: Compute the dimensions of $\frac{1}{4\pi \varepsilon_0}$

Since $4\pi$ is dimensionless, we have: $\left[\frac{1}{4\pi \varepsilon_0}\right] = \frac{1}{[\varepsilon_0]} = [M^1L^3T^{-4}I^{-2}]$

Step 3: Compute the dimensions of $\frac{|e|^2}{hc}$

Substitute the dimensions of $e$, $h$, and $c$: $\left[\frac{|e|^2}{hc}\right] = \frac{[e]^2}{[h][c]} = \frac{[I^2T^2]}{[ML^2T^{-1}][LT^{-1}]} = \frac{[I^2T^2]}{[ML^3T^{-2}]} = [M^{-1}L^{-3}T^4I^2]$

Step 4: Multiply the two parts to get the dimensions of the full expression.

Now, multiply the dimensions from Step 2 and Step 3: $\left[\frac{1}{4\pi \varepsilon_0} \frac{|e|^2}{hc}\right] = \left[\frac{1}{4\pi \varepsilon_0}\right] \left[\frac{|e|^2}{hc}\right] = [M^1L^3T^{-4}I^{-2}] \cdot [M^{-1}L^{-3}T^4I^2]$ Simplify the exponents:

• Mass ($M$): $1 + (-1) = 0$
• Length ($L$): $3 + (-3) = 0$
• Time ($T$): $-4 + 4 = 0$
• Current ($I$): $-2 + 2 = 0$

Step 5: Match with the given options.

The dimensions $[M^0L^0T^0]$ correspond to option C.

Students often make the following mistakes:

1. Ignoring the dimensions of $\varepsilon_0$: Some assume $\varepsilon_0$ is dimensionless or forget to include its dimensions, leading to incorrect cancellation of terms.
2. Incorrect dimensions for $h$: Confusing $h$ with angular momentum (which has the same dimensions) is rare, but misremembering $[h] = [ML^2T^{-1}]$ can derail the entire solution.
3. Overcomplicating the problem: Some students try to relate the expression to known physical constants (e.g., fine-structure constant) instead of performing dimensional analysis. While the given expression is proportional to the fine-structure constant (which is dimensionless), this is not necessary to solve the problem.
4. Sign errors in exponents: When multiplying dimensions, a small sign error (e.g., $T^{-4} \cdot T^4 = T^0$ vs. $T^{-4} \cdot T^{-4} = T^{-8}$) can lead to incorrect conclusions.

Exam Tip: Always write down the dimensions of each term explicitly and verify each step. For dimensionless quantities, ensure all fundamental dimensions cancel out. In this case, the cancellation of $M$, $L$, $T$, and $I$ confirms the result.