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

Question

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 ${J} {mol}^{-1}$.

Answer 1

SOLUTION : Energy of the electron in the ground state of H-atom,

$E_1=-2.18 \times 10^{-18} {~J}$

Ionisation energy $=E_{\infty}-E_n$

Ionisation enthalpy per mole of atomic hydrogen $=\left(E_{\infty}-E_1\right) N_A$

$=\left[0-\left(-2.18 \times 10^{-18}\right)\right] \times 6.023 \times 10^{23}$

$=2.18 \times 6.023 \times 10^5 {~J} / \mathrm{mol}=13.13 \times 10^5 {~J} / {mol} $

$=1.313 \times 10^6 {~J} / {mol}$