In the expression for energy of an electron in the \( n^{\mathrm{th}} \) orbit in a hydrogen atom, \( E_{n} = \frac{-13.6}{n^{2}} \), what does the negative sign indicate?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Electron is bound with nucleus
Energy is measured from zero for an electron at rest far away from the nucleus (a free electron). A negative total energy means the electron has less energy than a free electron. Energy must be supplied (13.6 eV from the ground state) to free it, so the electron is bound to the nucleus.
In a Bohr orbit, the kinetic energy is \( K = +13.6/n^{2} \) eV and the potential energy is \( U = -2K \), so the total is \( E = K + U = -K \), which is negative.
Note: It seems that there is a mistake in the original NTA paper, which prints \( E_{n} = -13.6/n^{2} \) without the unit, when it should read \( E_{n} = -13.6/n^{2}\ \mathrm{eV} \). The question asks only about the negative sign, so it can still be solved.
NCERT (2026-27):
Electron Orbits – “… The total energy of the electron is negative. This implies the fact that the electron is bound to the nucleus. If \( E \) were positive, an electron will not follow a closed orbit around the nucleus.”