1. We need the combination $e^{\alpha}\,\varepsilon_0^{\beta}\,h^{\gamma}\,c^{\delta}$ to be dimensionless.

So its overall dimensions must be $M^0L^0T^0I^0.$

2. Write dimensions of each quantity:

• Charge: $[e]=[Q]=IT$

• Permittivity of free space: Using Coulomb's law, $\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2} \sim F$ So, $[\varepsilon_0]=\frac{Q^2}{F\,L^2}$ Now, $[F]=MLT^{-2}$ Hence, $[\varepsilon_0]=\frac{(IT)^2}{(MLT^{-2})L^2}=M^{-1}L^{-3}T^4I^2$

• Planck's constant: $[h]=[\text{energy}][\text{time}]=ML^2T^{-1}$

• Speed of light: $[c]=LT^{-1}$

3. Therefore, $[e^{\alpha}\varepsilon_0^{\beta}h^{\gamma}c^{\delta}]$ becomes $(IT)^\alpha\,(M^{-1}L^{-3}T^4I^2)^\beta\,(ML^2T^{-1})^\gamma\,(LT^{-1})^\delta.$

Collecting powers of each fundamental dimension:

• Mass $M$: $-\beta+\gamma=0$

• Length $L$: $-3\beta+2\gamma+\delta=0$

• Time $T$: $\alpha+4\beta-\gamma-\delta=0$

• Current $I$: $\alpha+2\beta=0$

4. Solve these equations.

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

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

Now use length equation: $-3\beta+2\gamma+\delta=0$ Substitute $\gamma=\beta$: $-3\beta+2\beta+\delta=0$ $\delta=\beta$

Now check time equation: $\alpha+4\beta-\gamma-\delta=0$ Substitute $\alpha=-2\beta$, $\gamma=\beta$, $\delta=\beta$: $-2\beta+4\beta-\beta-\beta=0$ $0=0$ So it is consistent.

Thus, $\alpha=-2\beta,\qquad \gamma=\beta,\qquad \delta=\beta.$

Let $\beta=-n$ where $n$ is a non-zero integer. Then $\alpha=2n,\qquad \gamma=-n,\qquad \delta=-n.$

Hence, $(\alpha,\beta,\gamma,\delta)=(2n,-n,-n,-n).$

5. Compare with options:

• A: $(2n,-n,-n,-n)$ ✅
• B, C, D do not match.

So the correct option is A.