Questions: Solution of the Hydrogen Atom

Question 1

A hydrogen electron is in a state with principal quantum number n = 3. Which of the following correctly lists all allowed values of the orbital quantum number l?

• A. l = 0, 1, 2, 3 — l can take any non-negative integer up to n
• B. l = 0, 1, 2 — l ranges from 0 to n − 1
• C. l = 1, 2, 3 — l starts at 1 for excited states
• D. l = 0 only — all n = 3 states are s-states

Answer: (B)

The constraint from solving the radial Schrödinger equation is l = 0, 1, 2, …, n − 1. For n = 3, this gives l = 0, 1, or 2 (the 3s, 3p, and 3d subshells). Option A is wrong by one — l cannot equal n. The constraint l ≤ n − 1 arises from requiring the radial wavefunction to be normalizable: trying to construct a solution with l ≥ n forces the power series in the associated Laguerre polynomial to diverge. The l values are not an independent postulate — they fall out of the mathematics of the radial equation.

Question 2

How many distinct quantum states does the n = 2 energy level of hydrogen contain, ignoring electron spin?

• A. 1 — there is only one n = 2 state
• B. 2 — one for each allowed value of l
• C. 4 — one 2s state and three 2p states (m_l = −1, 0, +1)
• D. 8 — including all spin states

Answer: (C)

For n = 2, l can be 0 or 1. For l = 0: only m_l = 0 (one state, the 2s orbital). For l = 1: m_l = −1, 0, +1 (three states, the 2p orbitals). Total ignoring spin: 1 + 3 = 4 states. This equals n² = 2² = 4, confirming the general formula. The degeneracy (all four states having the same energy) is a special feature of the 1/r Coulomb potential. Option D of 8 would be correct if spin were included (2 spin states per orbital).

Question 3

The principal quantum number n in hydrogen comes from the angular part of the wavefunction — specifically, from the spherical harmonics.

• T. True
• F. False

Answer: False

False. The principal quantum number n emerges from solving the *radial* part of the Schrödinger equation, not the angular part. The angular part gives the spherical harmonics, which are labeled by l (orbital angular momentum quantum number) and m_l (magnetic quantum number). When the Coulomb potential is substituted into the radial equation and normalizable solutions are required, the energy is forced to take discrete values — only specific values of n produce solutions that don't blow up at large r. So n is fundamentally a radial quantum number, determining both the energy and the overall spatial scale of the orbital.

Question 4

In hydrogen, two states with the same principal quantum number n but different orbital quantum numbers l have exactly the same energy (ignoring relativistic corrections and spin effects).

• T. True
• F. False

Answer: True

True. The energy of a hydrogen eigenstate depends only on n: E_n = −13.6 eV / n². This means all states within a given n — regardless of l and m_l — are degenerate (have the same energy). For example, the 2s (n=2, l=0) and 2p (n=2, l=1) states all have energy −3.4 eV. This 'accidental' degeneracy is a special consequence of the 1/r form of the Coulomb potential. In atoms with more than one electron, electron-electron repulsion breaks this degeneracy, making 2s and 2p have different energies. Fine structure (relativistic effects and spin-orbit coupling) partially lifts it even in hydrogen.

Question 5

The hydrogen atom wavefunction requires three quantum numbers n, l, and m_l. Where does each come from, and what physical quantity does each determine?

Model answer: The three quantum numbers arise from the three spatial degrees of freedom in the Schrödinger equation after separation of variables. The φ (azimuthal) equation yields m_l, the magnetic quantum number, which determines the z-component of angular momentum: L_z = m_l ℏ, with m_l = −l, …, +l. The θ (polar) equation yields l, the orbital quantum number, which determines the total angular momentum magnitude: |L| = √(l(l+1)) ℏ, with l = 0, 1, 2, …. The radial equation yields n, the principal quantum number, which determines the energy: E_n = −13.6 eV / n², with n = 1, 2, 3, … and the constraint l ≤ n − 1.

This question tests whether students understand where the quantum numbers come from rather than just knowing their names and ranges. The key insight is that three quantum numbers emerge naturally because the wavefunction depends on three coordinates (r, θ, φ), and separation of variables produces an independent equation for each. Each equation, when solved with physical boundary conditions (normalizability, single-valuedness), produces one quantum number. The quantum numbers are not postulated — they are forced by the mathematics.