Solution:
Second excited state corresponds to \(n = 3\). The energy of this state is \(E_3 = -\frac{13.6}{3^2} = -1.51 \text{eV}\). Therefore, the energy required to ionize the electron from this level is \(+1.51 \text{eV}\).
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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