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Chemistry Question on Electronic Configurations Of Elements And The Periodic Table

Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1 .

Answer

It is given that the energy of an electron in the ground state of the hydrogen atom is −2.18×10−18J.-2.18 × 10^{ -18} J.
Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18×10−18J.2.18 × 10^{ -18} J.
∴\therefore Ionization enthalpy of atomic hydrogen = 2.18×10−18J.2.18 × 10^{ -18} J.

Hence, ionization enthalpy of atomic hydrogen in terms of :
J mol−1J \,mol ^{-1} = 2.18×10−18×6.02×1023 J mol−12.18 × 10^{ -18} × 6.02 × 10^{ 23}\, J\, mol^{ -1}
= 1.31×106 J  mol−11.31 × 10^ 6 \,J \;mol ^{-1}