What will be the energy required to remove the electron from the first excited state of $Li^{++}$, if the ground state energy of the electron in a hydrogen atom is -13.6 eV?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 30.6 eV
Energy levels of a hydrogen-like ion: $E_n = -Z^2 \frac{13.6\ \text{eV}}{n^2}$
For Li$^{++}$, $Z = 3$
Energy of first excited state ($n = 2$):
$E_2 = -3^2 \frac{13.6}{2^2} = -9 \cdot \frac{13.6}{4} = -30.6\ \text{eV}$
Energy required to remove the electron from this state (ionization energy from $n=2$):
$E_\text{required} = 0 - (-30.6) = 30.6\ \text{eV}$
Final Answer: $E_\text{required} = 30.6\ \text{eV}$