As higher values of n are considered for orbits in a hydrogen atom,...

The electric potential energy of an electron in the $n$-th orbit of a hydrogen atom is given by $U = -\dfrac{1}{4\pi\epsilon_0} \dfrac{e^2}{r_n}$.

The radius of the $n$-th orbit is directly proportional to the square of the principal quantum number, $r_n \propto n^2$.

Substituting this relation, we get $U \propto -\dfrac{1}{n^2}$.

As $n$ increases, the magnitude $\dfrac{1}{n^2}$ decreases. Due to the negative sign, the overall value of the potential energy becomes less negative, which means it increases.

Therefore, as higher values of $n$ are considered, the electric potential energy of the atom increases.