Questions: Quantum Mechanical Treatment of Hydrogen

Question 1

The hydrogen atom has states labeled by quantum numbers n, l, and m. A student claims that the n=2, l=1, m=0 state has higher energy than the n=2, l=0, m=0 state because it has greater angular momentum. Is this correct?

• A. Yes — higher angular momentum means higher rotational kinetic energy
• B. No — energy depends only on n; both states are degenerate with energy E₂ = −13.6/4 eV
• C. Yes — the l=1 state has a node structure that pushes the electron further from the nucleus, raising energy
• D. No — the l=0 state actually has higher energy because the electron spends more time near the nucleus

Answer: (B)

The energy eigenvalues E_n = −13.6 eV / n² depend only on the principal quantum number n, not on l or m. This is a special feature of the 1/r Coulomb potential. Both the (n=2, l=0, m=0) and (n=2, l=1, m=0) states have exactly the same energy −3.4 eV. For a given n, all states with l ranging from 0 to n−1 and m ranging from −l to l are degenerate — giving n² degenerate states per energy level. Small corrections (spin-orbit coupling, relativistic effects) lift this degeneracy.

Question 2

A student describes the hydrogen atom's ground state by saying: 'The electron orbits the proton at a fixed distance of one Bohr radius, a₀.' What is the fundamental error in this description?

• A. The ground state electron orbits at twice the Bohr radius, not one
• B. The electron does not follow a definite orbital path; the wavefunction gives a probability distribution, and the Bohr radius marks the most probable radial distance
• C. The description is correct — the Bohr model gives the exact ground state behavior
• D. The error is that the electron is stationary in the ground state, not orbiting

Answer: (B)

The ground state wavefunction ψ₁₀₀ is a probability amplitude, not a trajectory. The electron has no definite position between measurements — |ψ₁₀₀|² gives the probability density for finding it at any point in space. The Bohr radius a₀ ≈ 0.053 nm marks the peak of the radial probability distribution (the most probable distance), not a fixed orbital radius. The electron can be found anywhere from r=0 to r→∞ with varying probability. This replaces the Bohr model's definite circular orbit with a genuine quantum probability cloud.

Question 3

In the hydrogen atom, changing the orbital quantum number l while keeping n fixed does not change the electron's energy.

• T. True
• F. False

Answer: True

The energy E_n = −13.6 eV / n² depends only on n. For n=3, for example, the states l=0, l=1, and l=2 all have energy −13.6/9 ≈ −1.51 eV. This degeneracy in l (and m) is a special property of the pure Coulomb potential. In multielectron atoms, electron-electron repulsion and screening break this degeneracy, making different l values at the same n have different energies — which is why s, p, d, f orbitals fill in the order seen in the periodic table.

Question 4

The quantization of hydrogen's energy levels (E_n = −13.6 eV/n²) is an additional postulate imposed on the Schrödinger equation by hand, not derived from it.

• T. True
• F. False

Answer: False

This is false — quantization emerges from the mathematics of the Schrödinger equation itself. The requirement that the wavefunction be normalizable (square-integrable, finite at r=0, decaying to zero as r→∞) forces a quantization condition on the radial solutions. Only specific discrete values of energy allow normalizable solutions; all other energies lead to wavefunctions that diverge at infinity. Quantization is a consequence of boundary conditions, not an assumption — which is why this result was such a triumph over the ad hoc quantization rules of the old Bohr model.

Question 5

Why does the hydrogen atom have exactly n² degenerate energy eigenstates for each principal quantum number n?

Model answer: For a given n, the orbital quantum number l can range from 0 to n−1 (giving n possible values). For each l, the magnetic quantum number m ranges from −l to l, giving 2l+1 states. Summing over all l: Σ(l=0 to n−1) (2l+1) = n². This n²-fold degeneracy arises because the Coulomb potential energy depends only on r, not on angular direction, so states with different angular momentum quantum numbers l and m have the same energy.

The degeneracy in m (−l to l) follows directly from the spherical symmetry of the potential — no spatial direction is preferred. The additional degeneracy in l (all l values at a given n have the same energy) is a deeper, accidental symmetry of the 1/r Coulomb potential specifically, sometimes called the SO(4) symmetry. It is lifted by any correction that breaks this special symmetry, such as fine-structure effects. The total n² degeneracy per level is what produces the structure of the periodic table through Aufbau filling.