JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Atomic Structure

The energy of a one-electron species is $E_n=-2.18\times 10^{-18}\frac{Z^2}{n^2}\text{J}$, a quantity that is always negative. For the third orbit of ${\text{Li}}^{2+}$, with $Z=3$ and $n=3$, $E_3=-2.18\times 10^{-18}\times \frac{3^2}{3^2}=-2.18\times 10^{-18}\text{J}$. The remaining choices fail on sign or magnitude: the second orbit of hydrogen gives $-5.45\times 10^{-19}\text{J}$, and the He${}^{+}$ energies must be negative. The correct stationary-state energy is $-2.18\times 10^{-18}\text{J}$ for the third orbit of ${\text{Li}}^{2+}$.