Energy of an electron in the ground state of the hydrogen atom is `-5.19 xx 10^(-19)Cal`. Calculate the ionization enthalpy of atomic hydrogen in terms of `Kcal//mol`.
Hint: Apply the idea of mole concept to derive the answer.

Answer

Ionization energy of atomic hydrogen is the minimum amount of energy required to remove the electron from the ground state to infinity. Now energy of the electron in the ground state `=- 5.19 xx 10^(-19)Cal`. Energy of the electron at infinity `= 0`
The energy required to remove an electron in the ground state of hydrogen atom `=0 -`(its energy in the ground state) `=- (-5.19 xx 10^(-19) Cal) = 5.19 xx 10^(-19) Cal`.
`:.` Ionization enthalpy per mole of hydrogen atoms `=(5.19 xx 10^(-19) xx 6.02 xx 10^(23))/(1000) Kcal`.
`= 312.4 KCal//mol`.