Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18} \ J$. Calculate the ionization enthalpy of atomic hydrogen in terms of $J \ mol^{-1}$.

(N/A) Ionization enthalpy is the amount of energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$.
Energy of electron in ground state $(E_1)$ $= -2.18 \times 10^{-18} \ J$
Energy of electron at infinity $(E_{\infty})$ $= 0 \ J$
Energy required to remove one electron $= E_{\infty} - E_1 = 0 - (-2.18 \times 10^{-18} \ J) = 2.18 \times 10^{-18} \ J$
To calculate the ionization enthalpy for $1 \ mol$ of hydrogen atoms,we multiply by Avogadro's constant $(N_A = 6.022 \times 10^{23} \ mol^{-1})$:
Ionization enthalpy $= 2.18 \times 10^{-18} \ J \times 6.022 \times 10^{23} \ mol^{-1}$
$= 13.12796 \times 10^{5} \ J \ mol^{-1}$
$\approx 1.313 \times 10^{6} \ J \ mol^{-1}$ or $13.13 \times 10^{5} \ J \ mol^{-1}$.