1. We need the dimensions of

$\frac{1}{4\pi\varepsilon_0}\cdot \frac{e^2}{hc}$

2. First, note that

$\frac{1}{4\pi\varepsilon_0}$

is Coulomb's constant, whose dimensions can be obtained from Coulomb's law:

$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$

So,

$\left[\frac{1}{4\pi\varepsilon_0}\right] = \frac{[F][r^2]}{[q]^2}$

Now,

$[F] = [MLT^{-2}], \quad [r^2] = [L^2], \quad [q] = [IT]$

Hence,

$\left[\frac{1}{4\pi\varepsilon_0}\right] = \frac{[MLT^{-2}][L^2]}{[I^2T^2]} = [ML^3T^{-4}I^{-2}]$

3. Dimensions of $e^2$:

Since $[e] = [IT]$,

$[e^2] = [I^2T^2]$

Therefore,

$\left[\frac{1}{4\pi\varepsilon_0}e^2\right] = [ML^3T^{-4}I^{-2}]\cdot [I^2T^2] = [ML^3T^{-2}]$

4. Dimensions of $hc$:

Planck's constant $h$ has dimensions of action:

$[h] = [\text{energy}]\cdot[\text{time}] = [ML^2T^{-2}]\cdot[T] = [ML^2T^{-1}]$

Speed of light:

$[c] = [LT^{-1}]$

So,

$[hc] = [ML^2T^{-1}]\cdot[LT^{-1}] = [ML^3T^{-2}]$

5. Now divide:

$\left[\frac{1}{4\pi\varepsilon_0}\frac{e^2}{hc}\right] = \frac{[ML^3T^{-2}]}{[ML^3T^{-2}]} = [M^0L^0T^0]$

So the quantity is dimensionless.

6. Checking options:

• A: $[MLT^{-1}]$ — incorrect
• B: $[MLT^0]$ — incorrect
• C: $[M^0L^0T^0]$ — correct
• D: $[LC^{-1}]$ — incorrect

Therefore, the correct answer is Option C.