Basic NMR physics – When a sample is placed in a strong external magnetic field (B0), how do nuclear spins behave at thermal equilibrium before excitation?

Introduction / Context:
NMR relies on the interaction between nuclear magnetic moments and an applied static magnetic field (B0). At equilibrium, populations of nuclear spin states are distributed according to Boltzmann statistics. A small excess occupies the lower-energy state aligned with B0, generating a net magnetization vector along the field. This macroscopic magnetization is what the spectrometer manipulates and detects after radio-frequency excitation.

Given Data / Assumptions:

• We consider spin-1/2 nuclei (e.g., 1H, 13C) at room temperature in a high-field magnet.
• Thermal equilibrium is established prior to any RF pulse.
• Net longitudinal magnetization (Mz) arises from a slight population difference.

Concept / Approach:
In B0, spin states split into two energy levels (Zeeman splitting). The lower-energy state corresponds to magnetic moments aligned with B0; the higher-energy state is anti-aligned. Because the energy gap is small compared to kT, the excess population is slight, but sufficient to create measurable net magnetization. No macroscopic “stopping” or instantaneous 90° rotation occurs until an RF pulse perturbs the equilibrium.

Step-by-Step Solution:

Verification / Alternative check:
Increasing field strength increases the population difference and thus sensitivity. Cryogenic temperatures further enhance polarization, consistent with the equilibrium alignment model.

Why Other Options Are Wrong:

• Spins do not stop; they precess at the Larmor frequency.
• They do not all reverse direction spontaneously; populations reflect Boltzmann statistics.
• Exact 90° orientation is an excitation outcome, not an equilibrium state.
• Liquid-helium cooling is not required for alignment; room temperature polarization is sufficient for NMR.

Common Pitfalls:
Confusing equilibrium magnetization with post-pulse transverse magnetization; only the latter produces the detectable FID immediately after excitation.

Final Answer:
They partially align with the magnetic field (with a slight excess in the lower-energy state)