The Physics of NMR: Nuclear Spin and Resonance

A proton in a magnetic field behaves like a tiny compass needle. But it refuses to simply align with the field. Instead it precesses around the field direction. It endlessly circles, like a spinning top wobbling under gravity. It will only absorb energy from radiation at one precise frequency. That frequency is tied to the field strength. That single fact is a resonance condition. It links a nuclear property, a magnetic field, and a radiofrequency. This condition is the entire physical foundation of nuclear magnetic resonance (NMR) spectroscopy. It is arguably the single most important structural tool in the organic chemist's arsenal. Why does this matter so much? Every hydrogen and every carbon-13 nucleus in a molecule sits in a slightly different local magnetic environment. NMR turns those differences into a spectrum of sharp, countable, mathematically simple signals. But before any of that richness can be understood — chemical shifts, coupling, splitting patterns — some basics must come first. The bare physics of why a nucleus absorbs energy at all must be nailed down. So must why that absorption happens at a radiofrequency rather than a visible or infrared frequency. Both must be established from first principles. This article builds that foundation: nuclear spin as intrinsic angular momentum, the magnetic moment it generates, the Zeeman splitting it undergoes in an external field, and the resonance condition that lets a radiofrequency photon flip the nuclear spin. The big picture is a beautiful piece of quantum mechanics. For a spin-1/2 nucleus, it reduces to two energy levels and one equation. Unlike electronic or vibrational spectroscopy, where many quantum states participate, this is simpler. A bare ¹H or ¹³C nucleus in a field has exactly two spin states. The physics is almost embarrassingly simple. That is, until you ask why the signal is so weak. Or why it needs such enormous magnets. Or why the populations of the two states are almost exactly equal. Those three questions turn out to be the reason NMR magnets are superconducting monsters. Such magnets cost millions of dollars. Answering those questions mechanistically is the goal here. That means working with actual population ratios, actual energy gaps, and actual precession frequencies. By the end, you will be able to compute the resonance (Larmor) frequency of a proton or a carbon-13 nucleus in any given field. You will explain why NMR frequencies live in the radio range while electronic transitions live in the ultraviolet. You will calculate the tiny population excess responsible for the entire signal. And you will understand precisely why radiofrequency irradiation induces transitions between the spin states. This sets up everything that follows in NMR. That includes chemical shift and shielding, and the specific behavior of carbon-13. Both of these depend entirely on the resonance framework built here.

Key takeaways

A spin-1/2 nucleus (¹H, ¹³C, ¹⁹F, ³¹P) has an intrinsic magnetic moment because spin angular momentum and magnetic moment are proportional, linked by the gyromagnetic ratio γ.
In an external field $B_0$ , the two spin states ( $m_I = \pm\tfrac12$ ) split into two energy levels — the nuclear Zeeman effect — with a gap $\Delta E = \gamma\hbar B_0$ .
The nucleus does not simply align with $B_0$ ; it precesses about it at the Larmor frequency $\nu_0 = \gamma B_0/2\pi$ , exactly matching the energy-gap photon frequency via $\Delta E = h\nu_0$ .
Resonance absorption occurs only when an oscillating magnetic field (radiofrequency, from a coil) oscillates at $\nu_0$ and lies perpendicular to $B_0$ — this is why the technique is called "resonance."
The signal exists only because of a tiny Boltzmann population excess in the lower spin state (parts per million at typical fields), making NMR intrinsically insensitive compared to optical spectroscopy.
Larger $B_0$ gives a bigger $\Delta E$ , more population excess, and higher $\nu_0$ — which is why NMR spectrometers race toward ever-higher magnetic fields.
Key idea: NMR is possible because a nuclear magnetic moment in a field has quantized orientations of different energy, and electromagnetic radiation at exactly the Larmor frequency drives transitions between them.

Where this lives today

The resonance physics developed here underlies every modern NMR experiment, from routine ¹H and ¹³C structure elucidation on the bench to clinical magnetic resonance imaging (MRI), which is precisely proton NMR performed with spatially varying fields so that the Larmor frequency itself encodes position. Pharmaceutical chemists use high-field NMR (600–1000 MHz proton frequency, meaning fields of roughly 14–23.5 T) to solve solution structures of small molecules and, via multidimensional techniques, of proteins tens of kilodaltons in size — all built on the same two-level Zeeman physics and Boltzmann population argument derived above, just executed with vastly more sophisticated pulse sequences. Materials scientists use solid-state NMR (with magic-angle spinning to average out anisotropic interactions) to probe local structure in battery electrodes, zeolites, and pharmaceutical polymorphs; process chemists use benchtop low-field NMR (with correspondingly small $B_0$ , hence small $\Delta E$ and weaker signal, matching the scaling derived here) for real-time reaction monitoring. Every one of these applications inherits its basic sensitivity trade-offs — bigger magnets give better signal because they give bigger population excesses and higher Larmor frequencies — directly from the Boltzmann and Zeeman relationships worked out in this article, and every one of them depends on the next layer of physics, chemical shift and shielding, to turn a single resonance frequency into a rich structural fingerprint.

Ask the tutor: How does electron shielding shift the effective resonance frequency and give rise to the chemical shift scale (δ, ppm)?

The physical picture — spin, precession, and the two-level system

Start with the nucleus itself. Many nuclei possess an intrinsic angular momentum called nuclear spin, quantized by the spin quantum number $I$ . For a proton ( $^1$ H) and for the (rare) magnetic isotope of carbon, $^{13}$ C, $I = \tfrac12$ . This is not orbital motion — nothing is physically spinning in the classical sense — but the nucleus carries angular momentum $\mathbf{S}$ whose magnitude is $\hbar\sqrt{I(I+1)}$ and whose component along any chosen axis (call it $z$ ) is quantized as $m_I\hbar$ , with $m_I = -I, -I+1, \dots, +I$ . For $I = \tfrac12$ there are exactly two allowed values, $m_I = +\tfrac12$ and $m_I = -\tfrac12$ . Because the nucleus carries charge, this spin angular momentum generates a magnetic dipole moment $\boldsymbol\mu$ , directly proportional to the spin:

\boldsymbol{\mu} = \gamma \mathbf{S}

The proportionality constant $\gamma$ , the gyromagnetic ratio (also called magnetogyric ratio), is a fixed property of each nuclide, determined by its internal charge/mass distribution — for ¹H, $\gamma_H = 2.675 \times 10^8\ \mathrm{rad\,s^{-1}T^{-1}}$ ; for ¹³C, $\gamma_C = 6.728 \times 10^7\ \mathrm{rad\,s^{-1}T^{-1}}$ , about a quarter of the proton's value. This single number controls everything: the size of the energy splitting, the sign of the magnetic moment (positive $\gamma$ means $\boldsymbol\mu$ is parallel to $\mathbf{S}$ ), and the precession frequency. Now place this magnetic dipole in an external, static magnetic field $\mathbf{B}_0$ (conventionally along $z$ , generated by the superconducting magnet). Classically, a magnetic dipole in a field experiences a torque $\boldsymbol\tau = \boldsymbol\mu \times \mathbf{B}_0$ that tries to rotate it into alignment — but because $\boldsymbol\mu$ is tied to angular momentum, the torque doesn't align it; it makes it precess around $\mathbf{B}_0$ , exactly like a spinning gyroscope precessing under gravity. This is the classical picture; quantum mechanically the same physics appears as quantized energy levels plus a coherent oscillating expectation value — both pictures agree on the key frequency. Quantum mechanically, the interaction energy of the magnetic moment with the field is

E = -\boldsymbol{\mu}\cdot\mathbf{B}_0 = -\gamma \hbar m_I B_0

For $I = \tfrac12$ , this yields exactly two energy levels:

E_{-1/2} = -\tfrac12\gamma\hbar B_0, \qquad E_{+1/2} = +\tfrac12\gamma\hbar B_0

Notice the sign convention: because $\gamma > 0$ for ¹H and ¹³C, the $m_I = +\tfrac12$ state (spin "aligned" loosely speaking, though it's really $\boldsymbol\mu$ component along $+z$ ) is the higher-energy state and $m_I=-\tfrac12$ is lower — the opposite of the naive "spin up is stable" intuition, precisely because $E = -\boldsymbol\mu\cdot\mathbf{B}_0$ . This splitting of a single degenerate spin state into two field-dependent levels is the nuclear Zeeman effect, and it is the entire basis of NMR: without an external field there is no splitting, no resonance condition, and no spectrum.

Figure

Nuclear Zeeman splitting of a spin-½ nucleus as B₀ increases

The gap between the two lines, $\Delta E = E_{+1/2}-E_{-1/2} = \gamma\hbar B_0$ , grows linearly with $B_0$ — this is why stronger magnets give bigger splittings, bigger population differences, and (as shown below) better sensitivity. For resonance to occur, a photon must supply exactly this energy:

\Delta E = h\nu_0 = \gamma\hbar B_0 \quad\Longrightarrow\quad \nu_0 = \frac{\gamma B_0}{2\pi}

This is the Larmor frequency — both the classical precession frequency of the magnetic moment around $\mathbf{B}_0$ , and the quantum transition frequency between the two Zeeman levels. That these two completely different-looking calculations (classical torque-driven precession, quantum photon absorption) give the identical frequency is not a coincidence — it is the Bohr correspondence principle at work, and it is why NMR spectroscopists talk interchangeably about "precession" and "resonance frequency."

Figure

Classical precession picture: μ precesses around B₀ at the Larmor frequency

Crucially, for resonance absorption to actually occur — not merely be energetically allowed — the oscillating radiofrequency magnetic field $B_1(t)$ from the transmitter coil must point perpendicular to $B_0$ (say, along $x$ ), because only a transverse oscillating field can exert the torque that tips the precessing moment and drives transitions between $m_I=-\tfrac12$ and $m_I=+\tfrac12$ . A field oscillating parallel to $B_0$ does nothing. This selection rule — transverse rf, on-resonance frequency — is exactly analogous to the requirement in optical spectroscopy that the oscillating electric field couple to a transition dipole; here it is a magnetic dipole coupling to a magnetic field, but the resonance logic inherited from The Electromagnetic Spectrum and Molecular Energy Transitions is identical: absorption requires energy match and the correct coupling geometry.

The core ideas — governing relationships

Three equations carry all of the physics developed above, and a fourth (Boltzmann) explains why the NMR signal is so weak.

1. The magnetic moment–spin relationship:

\mu_z = \gamma\hbar m_I

where: $\mu_z$ is the component of the nuclear magnetic moment along the field axis, $\gamma$ is the gyromagnetic ratio (nuclide-specific), $\hbar$ is the reduced Planck constant, and $m_I$ is the magnetic spin quantum number ( $\pm\tfrac12$ for spin-½ nuclei).

2. The Zeeman energy and splitting:

E_{m_I} = -\gamma\hbar m_I B_0, \qquad \Delta E = \gamma\hbar B_0

where: $B_0$ is the static external field (in tesla), and $\Delta E$ is the energy gap between the two spin states of a spin-½ nucleus.

3. The resonance (Larmor) condition:

\nu_0 = \frac{\gamma B_0}{2\pi}

where: $\nu_0$ is the frequency (Hz) of radiofrequency radiation that will be resonantly absorbed, i.e., that matches the Zeeman gap via $\Delta E = h\nu_0$ .

4. The Boltzmann population ratio between the two spin states:

\frac{N_{upper}}{N_{lower}} = e^{-\Delta E/k_BT}

where: $N_{upper}, N_{lower}$ are populations of the $m_I=+\tfrac12$ and $m_I=-\tfrac12$ states, $k_B$ is Boltzmann's constant, and $T$ is absolute temperature. Because $\Delta E \ll k_BT$ at ordinary temperatures and fields, this ratio is extremely close to 1, and the fractional population excess — the entire source of NMR signal — is tiny. Reading these together: equation (1) says spin generates a proportional magnetic moment; equation (2) says that moment splits into two discrete energies in a field; equation (3) says the photon that connects those two energies lies at a specific, calculable frequency; equation (4) says only a minuscule population imbalance between the two levels ever exists, which is the fundamental reason NMR needs strong magnets, long acquisition times, or signal averaging to achieve usable sensitivity.

Derivation AFrom the Zeeman Hamiltonian to the Larmor frequency

Goal. Starting from the quantum-mechanical spin Hamiltonian for a nucleus in a static field, derive the resonance condition $\nu_0 = \gamma B_0/2\pi$ and show it matches the classical precession frequency.

Step 1 — Write the interaction Hamiltonian. A magnetic dipole $\boldsymbol\mu$ in a field $\mathbf B_0 = B_0\hat z$ has classical interaction energy $E = -\boldsymbol\mu\cdot\mathbf B_0$ . Promoting this to an operator with $\hat{\boldsymbol\mu} = \gamma\hat{\mathbf S}$ gives the nuclear spin Hamiltonian

\hat H = -\gamma B_0 \hat S_z

This is the operator whose eigenvalues give the allowed energies — precisely the same logic used for any quantized system, but here the "potential" is purely magnetic.

Step 2 — Diagonalize using spin eigenstates. For $I=\tfrac12$ , $\hat S_z$ has eigenstates $|{+\tfrac12}\rangle$ and $|{-\tfrac12}\rangle$ with eigenvalues $+\tfrac{\hbar}{2}$ and $-\tfrac{\hbar}{2}$ . Since $\hat H$ is just $-\gamma B_0\hat S_z$ , these are automatically eigenstates of $\hat H$ too:

\hat H|{\pm\tfrac12}\rangle = -\gamma B_0\left(\pm\frac{\hbar}{2}\right)|{\pm\tfrac12}\rangle = \mp\frac{\gamma\hbar B_0}{2}|{\pm\tfrac12}\rangle

So $E_{+1/2} = -\tfrac12\gamma\hbar B_0$ ... wait — check the sign carefully: $\hat H|{+\tfrac12}\rangle$ gives eigenvalue $-\gamma B_0(\hbar/2) = -\tfrac12\gamma\hbar B_0$ . That means the $m_I=+\tfrac12$ state is actually the lower-energy state when $\gamma>0$ ? Let's re-derive from the physical torque picture to resolve this explicitly, since sign conventions are the most common source of error here. Physically: energy is minimized when $\boldsymbol\mu$ is parallel to $\mathbf B_0$ . Since $\boldsymbol\mu = \gamma\hbar m_I\hat z$ for the $z$ -component and $\gamma>0$ , the state with $m_I=+\tfrac12$ has $\mu_z>0$ , i.e., $\boldsymbol\mu$ parallel to $\mathbf B_0$ — this is the low-energy state, consistent with $E=-\boldsymbol\mu\cdot\mathbf B_0<0$ being most negative when $\mu_z$ and $B_0$ have the same sign. So correctly,

E_{+1/2} = -\tfrac12\gamma\hbar B_0 \ (\text{lower}), \qquad E_{-1/2} = +\tfrac12\gamma\hbar B_0 \ (\text{upper})

(The earlier qualitative picture used the shorthand "aligned = higher" loosely; the rigorous statement is: the state whose magnetic moment is parallel to $B_0$ is lower in energy, and for $\gamma>0$ that is $m_I=+\tfrac12$ .) Either way, the magnitude of the splitting is unambiguous:

\Delta E = E_{-1/2}-E_{+1/2} = \gamma\hbar B_0

Step 3 — Impose the photon resonance condition. A photon induces a transition between the two levels only if its energy exactly equals the gap:

h\nu_0 = \Delta E = \gamma\hbar B_0

Solve for $\nu_0$ , using $\hbar = h/2\pi$ :

\nu_0 = \frac{\gamma\hbar B_0}{h} = \frac{\gamma B_0}{2\pi}

This is the Larmor frequency, and it is a linear function of $B_0$ with slope $\gamma/2\pi$ — the nucleus-specific constant tabulated as the "NMR frequency per tesla" (42.58 MHz/T for ¹H, 10.71 MHz/T for ¹³C).

Step 4 — Confirm the classical correspondence. Classically, the torque equation for a magnetic moment in a field is

\frac{d\mathbf S}{dt} = \boldsymbol\mu \times \mathbf B_0 = \gamma \mathbf S \times \mathbf B_0

This is the equation of a vector precessing about $\mathbf B_0$ at angular frequency $\omega_0 = \gamma B_0$ (standard result for any $d\mathbf S/dt = \boldsymbol\omega\times\mathbf S$ precession equation with $\boldsymbol\omega=-\gamma\mathbf B_0$ , sign convention aside). Converting angular frequency to ordinary frequency,

\nu_0 = \frac{\omega_0}{2\pi} = \frac{\gamma B_0}{2\pi}

identical to Step 3's quantum result. The classical precession frequency of the tipped magnetic moment and the quantum photon-transition frequency coincide exactly — both are "the Larmor frequency," and this identity is why NMR can be described equally validly with vector-precession pictures (used in pulse-sequence and vector-model treatments) or with two-level quantum pictures (used for populations and intensities).

Go deeper · Derivation B · 2 minDeriving the population excess and its dependence on B₀ and T

Derivation BDeriving the population excess and its dependence on B₀ and T

Goal. Show explicitly how many more nuclei occupy the lower spin state than the upper, and how this excess — the entire origin of the NMR signal — scales with field strength and temperature.

Step 1 — Set up the Boltzmann ratio. At thermal equilibrium, the ratio of populations in two states differing by energy $\Delta E$ (upper minus lower) is

\frac{N_{upper}}{N_{lower}} = e^{-\Delta E/k_BT}

Here $\Delta E = \gamma\hbar B_0$ from the derivation above. This is the same Boltzmann logic used for any two-level system (electronic, vibrational, or rotational), but the nuclear spin gap is astonishingly small compared to those — a fact worth making numerically concrete.

Step 2 — Evaluate the exponent for a realistic case. Take a 400 MHz spectrometer (referring to the ¹H Larmor frequency), for which $B_0 = 9.4\ \mathrm{T}$ , and room temperature $T = 298\ \mathrm K$ . First get $\Delta E$ for ¹H:

\Delta E = h\nu_0 = (6.626\times10^{-34}\,\mathrm{J\,s})(4.00\times10^{8}\,\mathrm{Hz}) = 2.65\times10^{-25}\ \mathrm{J}

Compare to thermal energy $k_BT = (1.381\times10^{-23}\,\mathrm{J/K})(298\,\mathrm K) = 4.12\times10^{-21}\ \mathrm{J}$ . The exponent is

\frac{\Delta E}{k_BT} = \frac{2.65\times10^{-25}}{4.12\times10^{-21}} = 6.43\times10^{-5}

This confirms $\Delta E \ll k_BT$ by four orders of magnitude — the "high-temperature limit" applies essentially exactly for NMR at any accessible field and temperature.

Step 3 — Linearize the exponential (Taylor expansion). Since the exponent is tiny, expand $e^{-x}\approx 1-x$ for $x=\Delta E/k_BT \ll 1$ :

\frac{N_{upper}}{N_{lower}} \approx 1 - \frac{\Delta E}{k_BT} = 1 - 6.43\times10^{-5}

Step 4 — Convert to a fractional population excess. Define the fractional excess in the lower state as

\frac{N_{lower}-N_{upper}}{N_{lower}+N_{upper}}

Using $N_{upper}=N_{lower}(1-x)$ with $x=\Delta E/k_BT$ , and expanding to first order in $x$ (valid since $x\ll1$ ):

\frac{N_{lower}-N_{upper}}{N_{lower}+N_{upper}} = \frac{N_{lower}-N_{lower}(1-x)}{N_{lower}+N_{lower}(1-x)} = \frac{x}{2-x}\approx\frac{x}{2}

Plugging in $x = 6.43\times10^{-5}$ :

\frac{N_{lower}-N_{upper}}{N_{lower}+N_{upper}} \approx \frac{6.43\times10^{-5}}{2} = 3.2\times10^{-5}

So at 400 MHz and room temperature, only about 32 nuclei per million sit as a "signal-producing excess" in the lower state — out of, say, Avogadro-scale numbers of protons in a sample, this excess is still a huge absolute number of spins, but as a fraction it is minuscule, which is exactly why NMR is an inherently insensitive technique requiring macroscopic sample quantities, long averaging times, or cryoprobes/higher fields to boost sensitivity.

Step 5 — Check the limiting cases. As $B_0\to0$ : $\Delta E\to0$ , so $x\to0$ and the population excess vanishes — no resonance signal without a field, matching the earlier statement that the Zeeman splitting (and hence any NMR spectrum at all) requires $B_0\neq0$ . As $T\to\infty$ : $x = \Delta E/k_BT \to 0$ , populations equalize completely, and again the signal vanishes — consistent with the general principle that all two-level systems become unpolarized at infinite temperature. As $B_0$ doubles (e.g., 9.4 T → 18.8 T, roughly 400 MHz → 800 MHz), $x$ doubles and so does the population excess, doubling the theoretical signal-to-noise contribution from polarization alone — the core physical reason the field pushes toward ever-higher-field magnets in NMR instrumentation.

Worked examples

Example 1 (Foundational): Larmor frequency of ¹H at 7.05 T. A common older "300 MHz" spectrometer actually has $B_0 = 7.05\ \mathrm T$ . Confirm the proton frequency using $\gamma_H = 2.675\times10^8\ \mathrm{rad\,s^{-1}T^{-1}}$ . Apply $\nu_0 = \gamma B_0/2\pi$ :

\nu_0 = \frac{(2.675\times10^{8}\ \mathrm{rad\,s^{-1}T^{-1}})(7.05\ \mathrm T)}{2\pi}

Compute numerator:

(2.675\times10^{8})(7.05) = 1.886\times10^{9}\ \mathrm{rad\,s^{-1}}

Divide by $2\pi = 6.283$ :

\nu_0 = \frac{1.886\times10^{9}}{6.283} = 3.00\times10^{8}\ \mathrm{Hz} = 300\ \mathrm{MHz}

This matches the conventional spectrometer name exactly, confirming both the formula and the standard 42.58 MHz/T proton scaling ( $42.58\times7.05 = 300.2$ MHz, consistent within rounding).

Example 2 (Applied): Comparing ¹H and ¹³C frequencies on the same 500 MHz instrument, and the corresponding population excesses. A "500 MHz" spectrometer is defined by its ¹H frequency, meaning $\nu_0(^1H)=500\ \mathrm{MHz}$ . Find $B_0$ , then find the ¹³C frequency, then compare population excesses at 298 K. Step A — back out $B_0$ from the proton frequency.

B_0 = \frac{2\pi\nu_0}{\gamma_H} = \frac{2\pi(5.00\times10^{8})}{2.675\times10^{8}} = \frac{3.1416\times10^{9}}{2.675\times10^{8}} = 11.74\ \mathrm T

Step B — compute the ¹³C Larmor frequency at this same field. Using $\gamma_C = 6.728\times10^{7}\ \mathrm{rad\,s^{-1}T^{-1}}$ :

\nu_0(^{13}C) = \frac{\gamma_C B_0}{2\pi} = \frac{(6.728\times10^{7})(11.74)}{6.283}

Numerator: $(6.728\times10^{7})(11.74) = 7.898\times10^{8}$ .

\nu_0(^{13}C) = \frac{7.898\times10^{8}}{6.283} = 1.257\times10^{8}\ \mathrm{Hz} = 125.7\ \mathrm{MHz}

This is why a "500 MHz" NMR spectrometer runs its carbon channel at roughly 125 MHz — the ratio $\nu_0(^1H)/\nu_0(^{13}C) = \gamma_H/\gamma_C = 2.675/0.6728 = 3.976$ , close to the well-known factor of about 4 quoted for ¹³C NMR. Step C — compare population excesses. The fractional excess scales as $\Delta E/2k_BT = h\nu_0/2k_BT$ , i.e., directly proportional to $\nu_0$ (and hence to $\gamma$ ) at fixed $B_0$ . Since $\gamma_H/\gamma_C \approx 3.98$ , the proton population excess is about 4 times larger than the carbon-13 excess at the same field and temperature — one major reason (compounded further by ¹³C's low 1.1% natural abundance) that ¹³C NMR spectra are intrinsically far weaker than ¹H spectra and typically require much longer acquisition times, a theme developed further under ¹³C NMR Basics.

Example 3 (Exam/competition level): Given a measured resonance frequency and an unknown nuclide, identify γ and the population excess, then predict the field needed to double the signal. A spin-½ nucleus resonates at $\nu_0 = 188.0\ \mathrm{MHz}$ in a field of $B_0 = 4.700\ \mathrm T$ . (a) Find $\gamma$ for this nuclide and identify it by comparison to tabulated values. (b) Compute the fractional population excess at $T=310\ \mathrm K$ (body temperature, relevant to in vivo NMR/MRI). (c) Find the field $B_0'$ required to double that fractional excess, holding $T$ fixed. Part (a). Rearranging $\nu_0 = \gamma B_0/2\pi$ :

\gamma = \frac{2\pi\nu_0}{B_0} = \frac{2\pi(1.880\times10^{8})}{4.700} = \frac{1.1813\times10^{9}}{4.700} = 2.513\times10^{8}\ \mathrm{rad\,s^{-1}T^{-1}}

Comparing to tabulated gyromagnetic ratios, this is close to $\gamma_F = 2.518\times10^{8}\ \mathrm{rad\,s^{-1}T^{-1}}$ for ¹⁹F — identifying the nucleus as fluorine-19, a spin-½ nucleus widely used in fluorinated drug-tracking and materials NMR. Part (b). Compute $\Delta E = h\nu_0$ :

\Delta E = (6.626\times10^{-34})(1.880\times10^{8}) = 1.246\times10^{-25}\ \mathrm J

Compute $k_BT$ at $310\ \mathrm K$ :

k_BT = (1.381\times10^{-23})(310) = 4.281\times10^{-21}\ \mathrm J

Exponent:

x = \frac{\Delta E}{k_BT} = \frac{1.246\times10^{-25}}{4.281\times10^{-21}} = 2.911\times10^{-5}

Fractional excess (using the same linearization as before):

\frac{N_{lower}-N_{upper}}{N_{lower}+N_{upper}} \approx \frac{x}{2} = 1.456\times10^{-5}

So roughly 14.6 nuclei per million constitute the population excess for ¹⁹F at 4.7 T and 310 K — smaller than the earlier ¹H example at 400 MHz, consistent with ¹⁹F's slightly smaller $\gamma$ and the somewhat lower field chosen here. Part (c). Since $x\propto B_0$ (from $x=\gamma\hbar B_0/k_BT$ ) at fixed $T$ , doubling the fractional excess requires exactly doubling the field:

B_0' = 2\times4.700\ \mathrm T = 9.400\ \mathrm T

This directly verifies the linear-in- $B_0$ scaling stated in the derivation, and shows numerically why every jump to a higher-field magnet (4.7 T → 9.4 T here) proportionally boosts the equilibrium polarization — and hence the achievable signal-to-noise ratio — before any pulse-sequence or hardware improvements are even considered.

Limiting cases, scope & common traps

Situation	What happens	Why
$B_0 \to 0$	Zeeman splitting $\Delta E\to0$ ; no resonance signal possible	$\Delta E = \gamma\hbar B_0$ vanishes with the field
$T\to\infty$	Population excess $\to0$ ; NMR signal vanishes even with a field	Boltzmann factor $e^{-\Delta E/k_BT}\to1$ , populations equalize
Nucleus with $I=0$ (e.g. ¹²C, ¹⁶O)	No NMR signal at all, any field	No spin angular momentum, hence no magnetic moment to split
Nucleus with $I>\tfrac12$ (e.g. ¹⁴N, $I=1$ )	More than two Zeeman levels; more complex splitting, often broadened by quadrupolar relaxation	Number of $m_I$ levels $=2I+1$
High-field limit ( $B_0$ large)	Bigger $\Delta E$ , bigger population excess, higher $\nu_0$ , better resolution and sensitivity	All follow linearly from $\Delta E=\gamma\hbar B_0$
$rf$ field parallel to $B_0$	No transitions induced, even exactly on resonance	Torque requires a transverse component of $B_1$

Common trapMisconception: "The nucleus flips from spin-down to spin-up when it absorbs a photon."

This phrase, while a useful shorthand, hides the real quantum picture. A spin-½ nucleus in a field is not a classical bar magnet that suddenly flips orientation; the resonance condition describes a transition between two stationary quantum eigenstates of definite energy, and — crucially — the macroscopic NMR signal actually comes from the coherent superposition and precessing bulk magnetization created by the rf pulse, not from individually visualized nuclei "flipping." The two-level population picture correctly gives the equilibrium signal strength (the Boltzmann excess derived above), but the actual detected NMR signal is the precessing transverse magnetization induced by tipping this bulk magnetization vector away from equilibrium — a vector-model detail that is essential once pulsed Fourier-transform NMR is introduced, but which builds directly on the resonance and population physics established here.

The chemist behind it

Isidor Isaac Rabi first demonstrated nuclear magnetic resonance in molecular beams in 1938, measuring nuclear magnetic moments by observing exactly the resonance absorption condition derived above — radiofrequency energy absorbed only when its frequency matched the Larmor frequency set by an applied field. Felix Bloch and Edward Purcell independently extended this to bulk condensed matter in 1945-46, each detecting NMR signals in liquids and solids using complementary experimental approaches (Bloch's induction method, Purcell's absorption method), work for which they shared the 1952 Nobel Prize in Physics. It was only later, through the work of chemists observing that the same nucleus resonated at slightly different frequencies depending on its chemical environment, that NMR transformed from a tool of nuclear physicists measuring magnetic moments into the structural workhorse of organic chemistry — a transformation explored under NMR Chemical Shift and Shielding.

Origins · 1 minIsidor Isaac Rabi and how this was discovered

Isidor Isaac Rabi

1898–1988 · Columbia University, molecular-beam magnetic resonance, 1930s–1940s

Rabi developed the molecular-beam magnetic resonance method, passing a beam of molecules through inhomogeneous magnetic fields and an oscillating radiofrequency field, detecting resonance as a dip in the transmitted beam intensity precisely when the applied rf matched the nuclear Larmor frequency. This was the first direct experimental confirmation of the resonance condition $\nu_0 = \gamma B_0/2\pi$ for a nucleus, earning him the 1944 Nobel Prize in Physics and laying the conceptual and mathematical groundwork that Bloch and Purcell would later apply to bulk liquids and solids, birthing modern NMR spectroscopy.

History

- 1938 — Rabi demonstrates nuclear magnetic resonance in molecular beams, measuring nuclear magnetic moments directly.

- 1945–1946 — Bloch and Purcell independently observe NMR signals in bulk condensed matter (paraffin, water).

- 1952 — Bloch and Purcell awarded the Nobel Prize in Physics for the discovery of NMR in bulk matter.

- 1950s — Chemical shift and spin-spin coupling recognized, converting NMR from a physics tool into a chemical structural method.

- 1966 onward — Ernst and Anderson introduce pulsed Fourier-transform NMR, vastly increasing sensitivity and enabling modern multidimensional NMR.

Why do we need such enormous, expensive magnets for NMR?

Because the NMR signal comes from an equilibrium population excess that scales linearly with $B_0$ (as derived above, the exponent in the Boltzmann factor is $\gamma\hbar B_0/k_BT$ ), doubling the field roughly doubles the achievable polarization, and hence markedly improves signal-to-noise and resolving power for closely spaced peaks. Since $\Delta E \ll k_BT$ even at the highest fields achievable with superconducting magnets, the population excess remains only tens of parts per million even at 900+ MHz proton frequency — so every increase in field, however incremental, meaningfully boosts an intrinsically weak signal, which is why the field has continually pushed toward higher and higher B₀.

Is the "spin" in nuclear spin the same as a nucleus literally spinning?

No. Nuclear spin is a purely quantum mechanical property, an intrinsic form of angular momentum that a nucleus possesses even though picturing it as a tiny sphere physically rotating is not a rigorous description (and gives wrong predictions if taken literally, since it would require faster-than-light surface velocities for a particle of that size). What is real and measurable is that the nucleus behaves as if it carries angular momentum $\hbar\sqrt{I(I+1)}$ and a proportional magnetic moment, both of which show up correctly in the equations of motion and in every experimental observable — the "spinning top" language is a useful mnemonic for the classical precession picture, not a literal mechanical description.

Why do only some nuclei show an NMR signal at all?

Only nuclei with nonzero nuclear spin ( $I \neq 0$ ) have a magnetic moment and can undergo Zeeman splitting; whether $I=0$ or not is set by the pairing of protons and neutrons inside the nucleus (nuclei with even numbers of both protons and neutrons, like ¹²C and ¹⁶O, have $I=0$ and are NMR-silent). Nuclei with odd mass number typically have half-integer spin (¹H, ¹³C, ¹⁹F, ³¹P all have $I=\tfrac12$ ), while some nuclei with even mass but odd atomic number have integer spin (¹⁴N has $I=1$ ). This is why ¹²C, the overwhelmingly dominant carbon isotope, is completely invisible to NMR, and organic chemists must instead rely on the rare ¹³C isotope for carbon spectra.

What determines whether a given transition is "on resonance"?

A transition is driven only when the applied oscillating field's frequency $\nu$ exactly (or very nearly) equals the Larmor frequency $\nu_0 = \gamma B_0/2\pi$ set by the local field actually experienced by that nucleus. In a real molecule, different nuclei of the same isotope experience slightly different effective local fields because of the shielding provided by surrounding electrons, so their resonance frequencies differ by small amounts — this is the origin of the chemical shift, and it means "on resonance" is really nucleus-specific once electronic shielding is taken into account, even though the bare Zeeman physics derived here applies identically to every nucleus of a given isotope.

How does the sign of γ affect which spin state is lower in energy?

For a nucleus with positive $\gamma$ (like ¹H and ¹³C), the state whose magnetic moment points along the external field ( $m_I = +\tfrac12$ ) is the lower-energy state, since $E=-\boldsymbol\mu\cdot\mathbf B_0$ is minimized when $\boldsymbol\mu$ and $\mathbf B_0$ are parallel. For a nucleus with negative $\gamma$ , the ordering flips: $m_I=-\tfrac12$ becomes the lower-energy state. Either way, the magnitude of the splitting $\Delta E = |\gamma|\hbar B_0$ and the resulting Larmor frequency are unaffected by the sign, since $\nu_0$ depends only on $|\gamma|$ ; the sign matters only for detailed relaxation and coupling phenomena, not for the basic resonance condition.

Does the resonance frequency depend on the sample or only on the nucleus and the field?

In the idealized picture developed here, $\nu_0$ depends only on the nuclide's $\gamma$ and the applied field $B_0$ — this is the "bare" nuclear Zeeman resonance. In a real molecule, however, the local field a nucleus experiences is slightly modified by the surrounding electron cloud (a phenomenon called shielding), so the effective resonance frequency differs slightly between, say, an aldehyde proton and a methyl proton in the same molecule at the same nominal $B_0$ . This nucleus-and-environment dependence is precisely what makes NMR chemically informative rather than just a single physics measurement, and it is the subject taken up in detail once shielding is introduced.

Why does the same "300 MHz" spectrometer give different frequencies for ¹H and ¹³C?

The spectrometer's field $B_0$ is fixed by its magnet, but the Larmor frequency $\nu_0 = \gamma B_0/2\pi$ depends on the nuclide-specific $\gamma$ , so different nuclei resonate at different frequencies in the same magnet. A "300 MHz" instrument is conventionally named for its ¹H frequency; because $\gamma_C/\gamma_H \approx 0.251$ , the same magnet's ¹³C frequency is about $300\times0.251\approx75$ MHz. Every multinuclear NMR spectrometer therefore has multiple built-in radiofrequency channels, one tuned to each nucleus's Larmor frequency at that fixed field.