The total energy of an electron in the ground state of hydrogen atom is –13.6 eV. The potential energy of an electron is the ground state of Li²⁺ ion will be

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–122.4 eV

122.4eV

244.8 eV

–244.8 eV

answer is D.

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Detailed Solution

$\begin{array}{l} E = - 13 .6 \frac{Z^{2}}{n^{2}} ∴ \frac{E_{1}}{E_{2}} = (\frac{Z_{1}}{Z_{2}}) ∴ \frac{13 .6}{E_{2}} = {(\frac{1}{3})}^{2} \\ ∴ E_{2} = 122 .4 eV; \to PE, U = - 2 KE \end{array}$
= –2(122.4) = –244.8eV.

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