In Bohr’s model of the hydrogen atom, which of the following is true…

The correct answer is Option (4) → It is different for each orbit.

In the Bohr model, the energy of the electron in the nth orbit of the hydrogen atom is \( E_n=-13.6\,\mathrm{eV}/n^2 \). So it depends on the principal quantum number n.

Option (1) It goes on decreasing as the orbit moves outwards. — Incorrect. As n increases, the magnitude \( 13.6/n^2 \) eV becomes smaller, so the energy becomes less negative. The energy rises in the outer orbits, from \( -13.6\,\mathrm{eV} \) for n = 1 to \( -3.40\,\mathrm{eV} \) for n = 2.

Option (2) It is the same for every orbit. — Incorrect. The energy changes with n. The electron in the first orbit has \( -13.6\,\mathrm{eV} \), but in the third orbit it has \( -1.51\,\mathrm{eV} \).

Option (3) It is positive in every orbit. — Incorrect. The energy is negative in every orbit because the electron is bound to the nucleus. It reaches 0 eV only when the electron is completely removed (\( n=\infty \)).

Option (4) It is different for each orbit. — Correct. Each stationary orbit has its own definite total energy, fixed by \( 1/n^2 \). These definite values are the energy levels of the atom.

NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom.”

Energy Levels – “The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. For n = 2, 3, ... the absolute value of the energy E is smaller, hence the energy is progressively larger in the outer orbits.”