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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 ${Li}^{2 +}$ ion will be

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122.4eV

–122.4 Ev

244.8 eV

–244.8 eV

answer is D.

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Detailed Solution

$E = - 13.6 \frac{Z^{2}}{n^{2}};$

$∴ \frac{E_{1}}{E_{2}} = {(\frac{Z_{1}}{Z_{2}})}^{2} \Rightarrow \frac{13.6}{E_{2}} = {(\frac{1}{3})}^{2}$

$E_{2} = 122.4 eV;$

$\to PE, U = - 2 KE = - 2 (122.4) = - 244.8 eV$

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