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 J mol^{–1}.

#### Solution 1

It is given that the energy of an electron in the ground state of the hydrogen atom is –2.18 × 10^{–18 }J.

Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 × 10^{–18 }J.

∴ Ionization enthalpy of atomic hydrogen = 2.18 × 10^{–18 }J

Hence, ionization enthalpy of atomic hydrogen in terms of J mol^{–1} = 2.18 × 10^{–18} × 6.02 × 10^{23} J mol^{–1} = 1.31 × 10^{6} J mol^{–1}

#### Solution 2

The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of the , atoms may be expressed as E (ground state) = ( – 2.18 x 10^{-18} J) x(6.022 x 10^{23} mol^{-1})= -1.312 x 10^{6} J mol^{-1}

Ionisation enthalpy =E_{∞}_{–}E ground state

= 0-(-1.312 x 10^{6}mol^{-1})

= 1.312 x 10^{6} J mol^{-1}.