How much energy is required to ionize a $H$ atom if the electron occupies the $n=5$ orbit? Compare your answer with the ionization enthalpy of the $H$ atom (energy required to remove the electron from the $n=1$ orbit).

(N/A) The energy of an electron in the $n^{\text{th}}$ orbit is given by:
$E_{n} = \frac{-2.18 \times 10^{-18} \times Z^{2}}{n^{2}} \ J$
For a $H$ atom,$Z = 1$,so $E_{n} = \frac{-2.18 \times 10^{-18}}{n^{2}} \ J$.
For ionization from $n=5$ to $n=\infty$:
$\Delta E = E_{\infty} - E_{5} = 0 - \left( \frac{-2.18 \times 10^{-18}}{5^{2}} \right) \ J$
$\Delta E = \frac{2.18 \times 10^{-18}}{25} \ J = 8.72 \times 10^{-20} \ J$.
For ionization from $n=1$ to $n=\infty$ (Ionization Enthalpy):
$\Delta E' = E_{\infty} - E_{1} = 0 - \left( \frac{-2.18 \times 10^{-18}}{1^{2}} \right) \ J$
$\Delta E' = 2.18 \times 10^{-18} \ J$.
Conclusion: The energy required to ionize an electron from the $n=5$ orbit $(8.72 \times 10^{-20} \ J)$ is significantly less than the ionization enthalpy of the $H$ atom $(2.18 \times 10^{-18} \ J)$.