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NCERT Class XI Chemistry Classification of Elements and Periodicity in Properties Solutions

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Question : 15 of 40
Marks: +1, -0
Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10−1810^{-18} J. Calculate the ionization enthalpy of atomic hydrogen in terms of Jmol−1\mathrm{J} \mathrm{mol}^{-1}. [Hint : Apply the idea of mole concept to derive the answer]
Solution:  
Energy of the electron in the ground state of H-atom, E1E_1 = –2.18 × 10−1810^{-18} J
Ionisation energy = E∞−EnE_{\infty} - E_n
Ionisation enthalpy per mole of atomic hydrogen = (E∞−E1E_{\infty} - E_1) NA
= [0 – (– 2.18 × 10−1810^{-18})] × 6.023 × 102310^{23}
= 2.18 × 6.023 × 10510^5 J/mol = 13.13 × 10510^5 J/mol
= 1.313 × 10610^6 J/mol
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