Question

energy of an electron in the ground state of hydrogen atom is - 2.8 ×10 ^-18j. calculate the ionization enthalpy of atomic hydrogen in terms of kj mol ^-1​

Answer 1

ionisation energy is the minimum amount of energy required to remove an electron from isolated gaseous atom.

i.e., I.E = -(energy of an electron in the ground state of atom)

= -(-2.8 × 10^-18) J

= 2.8 × 10^-18 J

so, ionisation energy of hydrogen atom = 2.8 × 10^-18 J/atom

we have to find ionisation energy of 1 mole of hydrogen atom.

we know, 1 mol = 6.023 × 10²³ atoms.

so, ionisation energy of hydrogen atom in term of KJ/mol = 2.8 × 10^-18 × 6.023 × 10²³/10³ KJ/mol

= 2.8 × 6.023 × 10² KJ/mol

= 1686.44 KJ/mol