The hydrogen atom has states labeled n = 2, l = 0 (the 2s orbital) and n = 2, l = 1 (the 2p orbital). How do their energies compare?
AE(2p) > E(2s), because higher angular momentum corresponds to higher energy
BE(2s) > E(2p), because s orbitals have electrons more concentrated near the nucleus
CE(2s) = E(2p), because energy in hydrogen depends only on the principal quantum number n
DE(2s) = E(2p) only in the ground state; for n > 1 they differ
In hydrogen, E_n = −13.6 eV/n² depends only on n, not on l or m_l. The 2s and 2p states are both at n = 2 and have exactly the same energy. This l-degeneracy is a special property of the Coulomb potential (related to a hidden SO(4) symmetry) and does not hold for other central potentials. In multi-electron atoms, where the effective potential deviates from pure Coulomb, this degeneracy is broken and 2s lies lower than 2p. Option A applies the classical intuition that angular momentum raises energy, which fails for the quantum hydrogen atom.
Question 2 Multiple Choice
How many distinct quantum states (including spin) are at the n = 3 energy level of hydrogen?
A3, because there are 3 allowed values of l (0, 1, 2)
B9, because there are 9 combinations of l and m_l
C18, because there are 9 spatial states and 2 spin states each
D6, because only l = 0 and l = 1 are physically relevant for n = 3
The degeneracy is 2n². For n = 3: l can be 0, 1, or 2. The m_l count for each l is 2l+1: 1 + 3 + 5 = 9 spatial states. With 2 spin states each, total = 18 = 2(3²). Option B counts only spatial states (n² = 9) and forgets spin. Option A counts only the number of l values, not the m_l sublevels within each. Option D incorrectly excludes l = 2 (the 3d orbitals), which are fully accessible at n = 3.
Question 3 True / False
The fact that hydrogen's energy depends mainly on n, and not on l, holds for most central potentials — it is a universal quantum mechanical result.
TTrue
FFalse
Answer: False
The l-degeneracy is specific to the Coulomb potential V(r) ∝ 1/r, which has a hidden symmetry (SO(4)) beyond the rotational symmetry (SO(3)) that all central potentials share. For any other central potential — a harmonic oscillator, a finite square well, the screened potential of multi-electron atoms — states with different l at the same n generally have different energies. This is precisely why s, p, d orbitals split in energy in atoms beyond hydrogen.
Question 4 True / False
The negative sign in E_n = −13.6 eV / n² indicates that higher n states have lower energy than the ground state.
TTrue
FFalse
Answer: False
The negative sign indicates the electron is bound (bound below the ionization energy of 0 eV). Higher n states have energies closer to zero — less negative, so higher in energy. E₁ = −13.6 eV, E₂ = −3.4 eV, E₃ ≈ −1.5 eV: as n increases, energy increases toward 0. The ground state (n = 1) is the most tightly bound, with the most negative energy. The misconception reverses the ordering by confusing 'more negative' with 'lower.'
Question 5 Short Answer
Why does energy in hydrogen depend only on n and not on l, and what physical symmetry explains this?
Think about your answer, then reveal below.
Model answer: The Coulomb potential V(r) = −e²/r has a hidden dynamical symmetry beyond ordinary rotational symmetry (SO(3)): the SO(4) symmetry corresponding to conservation of the Laplace-Runge-Lenz vector (which constrains classical Kepler orbits to be closed ellipses). Quantum mechanically, this symmetry means states with different l but the same n are related by symmetry operations and must have the same energy. For any other central potential, the LRL vector is not conserved, the SO(4) symmetry is broken, and different-l states at the same n have different energies. The hydrogen atom's l-degeneracy is a fingerprint of the 1/r potential, not a general quantum mechanical fact.
The classical analogue is illuminating: Kepler orbits are the only bound orbits (besides the harmonic oscillator) that are closed ellipses rather than precessing rosettes. Both the classical orbit closure and the quantum l-degeneracy stem from the same underlying SO(4) symmetry of the Coulomb potential.