Solution:
Second excited state corresponds to \(n = 3\). The energy is \(E_3 = -\frac{13.6}{3^2} = -1.51\text{ eV}\). The ionization energy required is \(E_{\text{ion}} = 0 - E_3 = 1.51\text{ eV}\).
The ground state energy of hydrogen atom is \(-13.6\text{ eV}\). The energy needed to ionize hydrogen atom from its second excited state will be
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