A hydrogen‑like ion has nuclear charge $Z=3$ (e.g., Li${}^{2+}$). Using the Bohr model, the energy of the electron in the $n^{\text{th}}$ orbit is $E_n=-13.6Z^2/n^2\text{eV}$. What is the energy of the electron when it is in the $n=2$ orbit?

Step-by-step solution

1. Insert the given values into the Bohr energy formula: $E_2=-13.6\times Z^2/2^2$.
2. With $Z=3$, compute $Z^2=9$.
3. Thus $E_2=-13.6\times 9/4$ eV.
4. Calculate $13.6\times 9=122.4$, divide by 4 gives $30.6$.
5. Therefore $E_2=-30.6$ eV.

Concept tested: Energy levels of hydrogen-like ions in Bohr model

Published 5 September 2026 · How this solution was prepared and checked