A dimensionless quantity is constructed in terms of electronic charge 𝑒, permittivity of free space 𝜀₀, Planck’s constant ℎ, and speed of light 𝑐. If the dimensionless quantity is written as 𝑒^𝛼𝜀₀ ^𝛽ℎ^𝛾𝑐^𝛿 and 𝑛 is a non-zero integer, then (𝛼, 𝛽, 𝛾, 𝛿) is given by

The correct option is (2n, -n, -n, -n).

To find the values of $\alpha$, $\mathrm{��}$, $\gamma$, and $\delta$ such that the quantity $e^{\alpha }{\epsilon }_0^{\beta }{h}^{\gamma }{c}^{\delta }$ is dimensionless, we can write down the dimensions of each fundamental quantity in terms of Mass (M), Length (L), Time (T), and Electric Current (A).

Let's find the dimensional formula for each parameter:
1. Electronic charge ($e$):
Charge is current multiplied by time.
$\left[e\right]=A^1{T}^1$

2. Permittivity of free space (${\epsilon }_0$):
From Coulomb's Law, the electrostatic force between two charges is $F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$, which gives ${\epsilon }_0=\frac{q^2}{4\pi F{r}^2}$.
Thus, the dimensions of ${\epsilon }_0$ are:
$\left[{\epsilon }_0\right]=\frac{A^2{T}^2}{\left(ML{T}^{-2}\right)\left(L^2\right)}=M^{-1}{L}^{-3}{T}^4{A}^2$

3. Planck’s constant ($h$):
From the relation $E=h\nu$, where $E$ is energy and $\nu$ is frequency:
$\left[h\right]=\frac{\left[E\right]}{\left[\nu \right]}=\frac{M{L}^2{T}^{-2}}{T^{-1}}=M{L}^2{T}^{-1}$

4. Speed of light ($c$):
$\left[c\right]=L{T}^{-1}$

Now, we write the dimension of the given quantity:
$\left[e^{\alpha }{\epsilon }_0^{\beta }{h}^{\gamma }{c}^{\delta }\right]=M^0{L}^0{T}^0{A}^0$

Substituting the dimensional formulas, we get:
${\left(AT\right)}^{\alpha }{\left(M^{-1}{L}^{-3}{T}^4{A}^2\right)}^{\beta }{\left(M{L}^2{T}^{-1}\right)}^{\gamma }{\left(L{T}^{-1}\right)}^{\delta }=M^0{L}^0{T}^0{A}^0$

Grouping the powers for each base unit:
Mass ($M$): $-\beta +\gamma =0\Rightarrow \beta =\gamma$
Length ($L$): $-3\beta +2\gamma +\delta =0$
Time ($T$): $\alpha +4\beta -\gamma -\delta =0$
Current ($A$): $\alpha +2\beta =0\Rightarrow \alpha =-2\beta$

From $\beta =\gamma$, let's substitute this into the equation for Length:
$-3\beta +2\beta +\delta =0\Rightarrow \mathrm{-\beta }+\delta =0\Rightarrow \delta =\beta$

Let's check the Time equation with these substitutions:
$(-2\beta )+4\beta -\beta -\beta =0$
This is consistent, meaning any non-zero value can be used as a scaling factor.

If we set $\beta =-n$ (where $n$ is a non-zero integer), we obtain:
$\alpha =-2\beta =2n$
$\beta =-n$
$\gamma =\beta =-n$
$\delta =\beta =-n$

Therefore, the coordinates $(\alpha ,\beta ,\gamma ,\delta )$ are given by $(2n,-n,-n,-n)$.