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 × 1023 J mol–1 = 1.31 × 106 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 1023 mol-1)= -1.312 x 106 J mol-1
Ionisation enthalpy =E∞–E ground state
= 0-(-1.312 x 106mol-1)
= 1.312 x 106 J mol-1.
