Bohr's atom model

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Question

If the binding energy (in eV) of the electron in a hydrogen atom is 13.6 eV, the energy (in eV ) required to remove the electron from the first excited state of $L i^{+ +}$ is

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Solution

$For lithium, E_{2} = - 13.6 \frac{z^{2}}{n^{2}} = - \frac{13.6 \times 3^{2}}{2^{2}} = - 30.6 eV So energy needed to remove the electron = 30.6 eV$

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