Question

Energy of an electron in the ground state of the hydrogen atom is -$2.18\times {10}^{-18}J.$ The ionization enthalpy of atomic hydrogen is $a\times {10}^{5}J/mol$. Then $‘a^{\prime }$ is: (Take $N_A=6\times {10}^{23}$)

Answer 1

The ionization enthalpy is the amount of energy required to remove an electron from ground state to infinity. The energy of electron in the ground state is $-2.18\times {10}^{-18}J$
The energy of electron at infinity is zero.
The energy required to remove electron is $0-(-21.8\times {10}^{-18}J)=21.8\times {10}^{-18}J$
To remove 1 mole of electrons the amount of energy required is $2.18\times {10}^{-18}\times 6\times {10}^{23}=13.08\times {10}^{5}J$. This is the ionization enthalpy of hydrogen.