Spin physics – For a nucleus with spin quantum number I = 1/2 placed in an external magnetic field, how many distinct Zeeman energy orientations are possible?

Introduction / Context:
Nuclear spin states split in a magnetic field into discrete energy levels. The number of allowed orientations (and hence energy levels) is determined by the spin quantum number I. This quantization underlies the basic two-level system exploited by NMR for many common nuclei such as 1H and 13C.

Given Data / Assumptions:

• We consider a static magnetic field B0 and an isolated spin-1/2 nucleus.
• Magnetic quantum numbers mI take on values from +I to −I in integer steps.
• Selection rules allow transitions with ΔmI = ±1 under RF excitation.

Concept / Approach:
For I = 1/2, the allowed mI values are +1/2 and −1/2, yielding two Zeeman levels. The energy difference is proportional to the magnetic field and the gyromagnetic ratio (ΔE = ħ * γ * B0). This two-level system leads to a single resonance frequency at the Larmor frequency, broadened and shifted by chemical environment (chemical shift) and interactions (e.g., J-coupling).

Step-by-Step Solution:

Verification / Alternative check:
NMR of 1H and 13C shows single-quantum transitions characteristic of two-level systems; contrast with quadrupolar nuclei (e.g., 14N, I = 1) that exhibit additional relaxation behavior due to more levels and electric quadrupole moments.

Why Other Options Are Wrong:

• 3, 4, 6: correspond to spins I = 1 (three levels) or higher; not applicable to I = 1/2.
• 1: would imply no resonance; contradicts fundamental NMR behavior.

Common Pitfalls:
Assuming more lines automatically means more Zeeman levels; multiplets in 1H NMR arise from J-coupling, not additional intrinsic Zeeman states for I = 1/2 nuclei.

Final Answer:
2