In the H–He⁺ mixture experiment, the ratio of potential energy of the n=2 electron for the H atom to that of He⁺ ion is

PE = 2 × TE. TE(H, n=2) = −3.4 eV. TE(He⁺, n=2) = −13.6 × 4 / 4 = −13.6 eV. Ratio = PE(H)/PE(He⁺) = TE(H)/TE(He⁺) = −3.4/(−13.6) = 1/4.