Allowed principal quantum numbers for the hydrogen atom When solving the time-independent Schrödinger equation for an electron in the Coulomb field of a proton (hydrogen atom), which statement about the principal quantum number n yields physically acceptable bound-state wavefunctions?

Introduction / Context:Quantization in the hydrogen atom arises from boundary conditions on the Schrödinger equation. The principal quantum number n labels discrete bound states with distinct energies and radial wavefunctions.

Given Data / Assumptions:

• Nonrelativistic Coulomb potential V(r) = −e^2/(4π ε0 r).
• Bound states (E < 0) are sought.
• Wavefunction must be normalizable and finite everywhere.

Concept / Approach:Solving the radial equation yields associated Laguerre polynomials with quantization from regularity and normalizability. These conditions require the principal quantum number to be a positive integer n = 1, 2, 3, … . The energy levels are E_n ∝ −1/n^2, and each n admits angular quantum numbers l = 0, 1, …, n−1.

Step-by-Step Solution:Impose boundary conditions at r = 0 and r → ∞.Series termination (polynomial condition) enforces integer n.Acceptable bound solutions exist for n = 1, 2, 3, …Thus, the correct general statement is “n must be any positive integer”.

Verification / Alternative check:Spectroscopic series (Lyman, Balmer, etc.) correspond to transitions between integer-n levels with energies scaling as 1/n^2.

Why Other Options Are Wrong:Restricting to n = 1 ignores excited bound states; “≥ 2 only” excludes ground state; non-integer or even-only values contradict the mathematical solution.

Common Pitfalls:Confusing n with l or m; assuming continuum (E > 0) states also quantized in n—they are not bound and use different labeling.

Final Answer:n must be any positive integer (1, 2, 3, …)