Difficulty: Easy
Correct Answer: n must be any positive integer (1, 2, 3, …)
Explanation:
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:
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, …)
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