Questions: Solving the Schrödinger Equation for Hydrogen Atom

Question 1

A hydrogen atom is in the state n=2, ℓ=1, m_ℓ=0. What is its energy?

• A. −3.4 eV, the same as all n=2 states regardless of ℓ or m_ℓ
• B. −1.51 eV, because ℓ=1 raises the energy compared to ℓ=0
• C. −13.6 eV, because a p orbital keeps the electron close to the nucleus
• D. −6.8 eV, determined by the product of n and ℓ

Answer: (A)

Energy in hydrogen depends only on the principal quantum number n: E_n = −13.6 eV/n². For n=2, E = −13.6/4 = −3.4 eV regardless of whether ℓ=0 (s orbital) or ℓ=1 (p orbital). This n²-fold degeneracy — all four n=2 states (2s, 2px, 2py, 2pz) having the same energy — is a special feature of the Coulomb 1/r potential. It breaks down in multi-electron atoms, but is exact for hydrogen.

Question 2

Which set of quantum numbers represents a valid hydrogen orbital?

• A. n=2, ℓ=2, m_ℓ=0
• B. n=3, ℓ=2, m_ℓ=−3
• C. n=3, ℓ=1, m_ℓ=−1
• D. n=1, ℓ=1, m_ℓ=0

Answer: (C)

The constraints are 0 ≤ ℓ ≤ n−1 and −ℓ ≤ m_ℓ ≤ ℓ. Option A fails because ℓ=2 requires n≥3. Option B fails because m_ℓ=−3 requires ℓ≥3. Option D fails because ℓ=1 requires n≥2. Only option C satisfies both rules: n=3, ℓ=1 satisfies ℓ ≤ n−1 = 2, and m_ℓ=−1 satisfies |m_ℓ| ≤ ℓ = 1.

Question 3

In hydrogen, the 2s and 2p orbitals have the same energy.

• T. True
• F. False

Answer: True

Because the hydrogen atom energy E_n = −13.6 eV/n² depends only on n, all orbitals with n=2 are exactly degenerate: 2s (ℓ=0) and all three 2p orbitals (ℓ=1, m_ℓ=−1,0,+1) all sit at −3.4 eV. This n²-fold degeneracy — four states for n=2, nine for n=3 — is a hidden symmetry specific to the 1/r potential. The degeneracy lifts in multi-electron atoms, which is why 2s and 2p electrons have different energies in carbon but not in hydrogen.

Question 4

The wavefunction ψ(r,θ,φ) of a hydrogen orbital describes the trajectory the electron follows as it orbits the nucleus.

• T. True
• F. False

Answer: False

The wavefunction is a probability amplitude, not a trajectory. |ψ(r,θ,φ)|² gives the probability density — the probability of finding the electron in a small volume element at that location. Electrons do not follow classical orbits; the orbital shape (s, p, d) represents a cloud of probability rather than a path. Treating orbitals as orbits was Bohr's model, which quantum mechanics replaced precisely because electrons do not have definite positions and momenta simultaneously.

Question 5

Why does the energy of a hydrogen atom depend only on the principal quantum number n, and what does this imply about the number of distinct quantum states at each energy level?

Model answer: Energy depends only on n because the 1/r Coulomb potential has a special hidden symmetry (beyond spherical symmetry) that makes all (ℓ, m_ℓ) combinations at a given n degenerate. For each n, ℓ runs from 0 to n−1, and for each ℓ, m_ℓ runs from −ℓ to +ℓ, giving n² distinct states all at the same energy E_n.

The n² degeneracy is not just a coincidence of the math — it reflects a conserved quantity (the Runge-Lenz vector) that is specific to inverse-square-law forces. For n=1: 1 state; for n=2: 4 states (1s + 3×2p); for n=3: 9 states (1s + 3×2p + 5×3d), all degenerate. This degeneracy has physical consequences: it is why hydrogen's spectral lines appear where they do and why atomic shells fill the way they do in the periodic table.