For a hydrogen atom, this expression simplifies to ${\mathit{E}}_{\mathit{n}}=-\frac{13.6}{{\mathit{n}}^2}$ eV. From this relation, we observe that:
${\mathit{E}}_{\mathit{n}}\propto \frac{1}{{\mathit{n}}^2}$
As n increases, the energy becomes less negative (i.e., increases) and approaches zero as n → ∞.
The energy is proportional to $\frac{1}{{\mathit{n}}^2}$