In Bohr’s model of the hydrogen atom, which of the following is true when the electron makes a transition from \( n=2 \) to \( n=1 \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The energy decreases.
In the Bohr model both the energy and the angular momentum depend on the orbit number, \( E_n=-13.6\,\mathrm{eV}/n^{2} \) and \( L=nh/2\pi \). So both change when n changes.
Option (1) The energy remains constant. — Incorrect. The two levels have different energies, \( E_2=-3.4\,\mathrm{eV} \) and \( E_1=-13.6\,\mathrm{eV} \). The difference leaves the atom as a photon.
Option (2) The angular momentum remains constant. — Incorrect. The angular momentum is \( 2h/2\pi \) in the orbit n = 2 and \( h/2\pi \) in the orbit n = 1, so it changes.
Option (3) The energy decreases. — Correct. The jump is to a level of lower energy. The energy falls from \( -3.4\,\mathrm{eV} \) to \( -13.6\,\mathrm{eV} \). The difference of 10.2 eV is emitted as a photon (Bohr’s third postulate).
Option (4) The angular momentum increases. — Incorrect. Since \( L \) is proportional to n, going from n = 2 to n = 1 halves the angular momentum. It does not increase.
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “(iii) Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by
\( h\nu=E_i-E_f \qquad (12.6) \)
where Ei and Ef are the energies of the initial and final states and Ei > Ef.”