If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is
Answer & explanation
Correct answer: option 2
$E_n=\frac{13.6}{n^2} \times Z^2$. For first excited state n = 2 and for $L i^{++}, z=3 \Rightarrow E=\frac{13.6}{4} \times 9$ = 30.6 eV