Consider two different hydrogen atoms. The electron in each atom is in first excited state. According to Bohr's model, electrons will have:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → same energies and the same orbital angular momentum
In Bohr's model, the state of the electron is fixed by the principal quantum number \( n \). Both the energy and the orbital angular momentum depend only on \( n \) (and on \( Z \), which is 1 for hydrogen). The first excited state is \( n = 2 \) in both atoms.
\( \displaystyle E_{n} = -\frac{13.6}{n^{2}}\,\mathrm{eV} \;\Rightarrow\; E_{2} = -\frac{13.6}{4}\,\mathrm{eV} = -3.4\,\mathrm{eV} \)
\( \displaystyle L_{n} = \frac{nh}{2\pi} \;\Rightarrow\; L_{2} = \frac{2h}{2\pi} = \frac{h}{\pi} \)
Both electrons have the same \( n \), so both quantities are the same.
Option (1) different energies but the same orbital angular momentum — Incorrect. The energy depends only on \( n \), and both electrons have \( n = 2 \), so their energies are equal.
Option (2) same energies and the same orbital angular momentum — Correct. Both electrons have \( E = -3.4\,\mathrm{eV} \) and \( L = h/\pi \), because both are in the same state \( n = 2 \).
Option (3) same energies but different orbital angular momentum — Incorrect. The angular momentum \( nh/2\pi \) is also fixed by \( n \), so it cannot differ between the two electrons.
Option (4) different energies and no orbital angular momentum — Incorrect. The energies are equal, and every Bohr orbit has a non-zero angular momentum \( nh/2\pi \).
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… (ii) Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π … That is
\( \displaystyle L = \frac{nh}{2\pi} \qquad (12.5) \)
… Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 10–19 J, Eq. (12.9) can be rewritten as
\( \displaystyle E_n = -\frac{13.6\ \text{eV}}{n^2} \qquad (12.10) \)”