The ionization energy of hydrogen atom in the ground state is 2.17×10−11erg. The ionization energy of Li2+ in the ground state will be

ELi=Z2EH =322.17×10−11erg =1.953×10−10erg =1.953×10−17J

ELi=Z2EH =322.17×10−11erg =1.953×10−10erg =1.953×10−17J

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The ionization energy of hydrogen atom in the ground state is $2.17 \times 10^{- 11} erg$ . The ionization energy of ${Li}^{2 +}$ in the ground state will be

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$1.953 \times 10^{- 15} J$

$1.953 \times 10^{- 18} J$

$1.953 \times 10^{- 16} J$

$1.953 \times 10^{- 17} J$

answer is C.

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$E_{Li} = Z^{2} E_{H} = (3^{2}) (2.17 \times 10^{- 11} erg) = 1.953 \times 10^{- 10} erg = 1.953 \times 10^{- 17} J$

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