JEE Main 2026 — Structure of Atom Question with Solution

The energy of an electron in the $n^{th}$ stationary state of a one-electron system is given by the Bohr's formula:
$E_n=-2.18\times 10^{-18}\times \frac{Z^2}{n^2}$ J
where $Z$ is the atomic number and $n$ is the principal quantum number.
For option (1): Second orbit of $H{e}^{+}$ ($Z=2,n=2$)
$E=-2.18\times 10^{-18}\times \frac{2^2}{2^2}=-2.18\times 10^{-18}$ J. (Incorrect sign in option)
For option (2): Third orbit of $L{i}^{2+}$ ($Z=3,n=3$)
$E=-2.18\times 10^{-18}\times \frac{3^2}{3^2}=-2.18\times 10^{-18}$ J. (Correct)
For option (3): Second orbit of $H$ atom ($Z=1,n=2$)
$E=-2.18\times 10^{-18}\times \frac{1^2}{2^2}=-0.545\times 10^{-18}$ J. (Incorrect value)
For option (4): First orbit of $H{e}^{+}$ ($Z=2,n=1$)
$E=-2.18\times 10^{-18}\times \frac{2^2}{1^2}=-8.72\times 10^{-18}$ J. (Incorrect sign in option)