Nuclear Spin and Energy Absorption | A-Level H3 Chemistry

NMR spectroscopy can eventually tell us an extraordinary amount about molecular structure, but the experiment begins with a much smaller idea: an atomic nucleus can behave like a tiny magnetic system. Place suitable nuclei in a strong magnetic field, supply radiofrequency energy, and they can move between quantised nuclear spin energy states.

The aim of this chapter is to understand where those energy states come from and what the word resonance actually means. Once this foundation is secure, ideas such as shielding, chemical shift, integration and spin-spin splitting become much easier to understand.

Diagnostic Quiz

What Do You Already Know About Nuclear Spin?

Try these five questions before reading the chapter. Focus on the relationship between nuclear spin, magnetic fields, energy levels and radiofrequency absorption.

5 diagnostic questions
Energy-level reasoning where useful
Instant explanation for every answer option

Question 1 of 5

Which statement best describes an NMR-active nucleus?

Every nucleus is NMR-active because every nucleus contains protons.

Not all nuclei possess a non-zero nuclear spin. For example, ¹²C and ¹⁶O both have nuclear spin quantum number I = 0 and are not observed by conventional NMR.

It has a nuclear spin quantum number I that is not zero.

Correct. A non-zero nuclear spin gives the nucleus a magnetic moment that can interact with an external magnetic field.

It must contain an even number of both protons and neutrons.

Nuclei with even numbers of both protons and neutrons are typically precisely the nuclei that have I = 0.

It must have exactly one proton.

¹H is important in NMR, but many other nuclei are NMR-active, including ¹³C, ¹⁹F and ³¹P.

NMR activity is fundamentally determined by nuclear spin. A nucleus with I = 0 has no nuclear magnetic moment associated with spin and therefore does not undergo the nuclear spin transitions used in NMR.

What happens to the two spin states of a ¹H nucleus when it is placed in a strong external magnetic field B₀?

Both states disappear.

The states remain available. The magnetic field changes their energies relative to one another.

They remain at exactly the same energy.

In the absence of the field they are degenerate, but B₀ removes this degeneracy.

They separate into a lower-energy state and a higher-energy state.

Correct. This separation is called Zeeman splitting. For ¹H, the α state is lower in energy and the β state is higher.

They become electronic excited states.

NMR concerns nuclear spin energy levels, not electronic excitation of the atom or molecule.

An external magnetic field removes the equal-energy degeneracy of the nuclear spin states. This creates the energy gap that makes an NMR transition possible.

Which equation gives the resonance condition for absorption between two nuclear spin energy levels separated by ΔE?

ΔE = hν₀

Correct. The energy supplied by the radiofrequency radiation must exactly match the separation between the spin states.

ΔE = h / ν₀

Photon energy is proportional to frequency, not inversely proportional to frequency.

ΔE = ν₀ / h

This does not have the correct relationship between energy and frequency. Planck's relation is E = hν.

ΔE = hcν₀

The factor c is not part of the energy-frequency relation. E = hc/λ is the corresponding wavelength form.

Resonance requires an exact energy match. Since electromagnetic radiation of frequency ν has energy hν per photon, the resonance condition is ΔE = hν₀.

What happens to the resonance frequency of ¹H when the external magnetic field B₀ is increased?

It decreases.

Increasing B₀ increases the energy separation between the spin states, so a larger rather than smaller frequency is required.

It increases.

Correct. For a given nucleus, ν₀ is directly proportional to B₀.

It remains constant.

The resonance frequency is field-dependent because the energy separation is field-dependent.

It becomes zero.

A stronger field increases rather than eliminates the energy gap.

For a nucleus with gyromagnetic ratio γ, ν₀ = γB₀/(2π). A stronger B₀ therefore gives a proportionally larger resonance frequency.

When does resonance occur in an NMR experiment?

Whenever any electromagnetic radiation reaches the sample.

The frequency must satisfy the nuclear spin transition energy requirement. Arbitrary frequencies do not produce resonance.

Only when visible light is absorbed.

NMR uses radiofrequency radiation rather than visible light.

When all nuclei have entered the higher-energy state.

Resonance concerns transitions between states; the entire population does not have to occupy the higher state.

When the applied radiofrequency matches the energy separation between the nuclear spin states.

Correct. At the resonance frequency, ΔE = hν₀ and energy can be transferred efficiently between the RF field and the nuclear spin system.

Resonance is an energy-matching condition. It is the same general idea seen throughout spectroscopy: a transition occurs when the supplied electromagnetic energy matches an allowed energy gap.

1. What is nuclear spin?

Protons and neutrons possess an intrinsic quantum-mechanical property called spin. The combined spins of all the protons and neutrons in a nucleus give the nucleus a total nuclear spin quantum number, I.

The word spin is historical. It is tempting to imagine the nucleus as a tiny charged ball physically rotating about its own axis, but this picture should not be taken literally. Nuclear spin is an intrinsic quantum property.

What matters for NMR is that a nucleus with I ≠ 0 also possesses a nuclear magnetic moment. It can therefore interact with a magnetic field.

NMR activity requires a non-zero nuclear spin quantum number: I ≠ 0.

Which nuclei are NMR-active?

The nuclear spin depends on the numbers of protons and neutrons in the nucleus. A particularly useful pattern is that nuclei containing even numbers of both protons and neutrons normally have I = 0.

In contrast, nuclei with an odd mass number have half-integer nuclear spin values, while nuclei containing odd numbers of both protons and neutrons can have integer nuclear spin values.

Nucleus	Nuclear spin I	NMR-active?
¹H	½	Yes
¹³C	½	Yes
¹²C	0	No
¹⁶O	0	No
¹⁴N	1	Yes

Carbon gives a useful comparison. ¹²C contains 6 protons and 6 neutrons, so it has I = 0 and is NMR-inactive. ¹³C still has 6 protons but now has 7 neutrons, giving I = ½. This is why ¹³C can be observed in carbon-13 NMR spectroscopy.

Allowed spin orientations

A nucleus with spin quantum number I has 2I + 1 possible magnetic spin states, described by the magnetic quantum number m_I.

Number of allowed spin states = 2I + 1

For a spin-½ nucleus such as ¹H:

m_I = +½ or −½

Thus a proton has exactly two allowed nuclear spin states when placed in a magnetic field.

A schematic magnetic-moment picture. Without an applied field there is no preferred direction. In B₀, a spin-½ nucleus has two allowed magnetic states. The arrows should not be interpreted as tiny nuclei literally rotating like classical objects.

2. What does an external magnetic field do?

In the absence of an external magnetic field, the magnetic spin states of a spin-½ nucleus have the same energy. They are said to be degenerate.

When the sample is placed in a strong external magnetic field B₀, the magnetic moment of the nucleus interacts with the field. The two spin states no longer have the same energy.

Applying B₀ removes the degeneracy of the nuclear spin states. This is called Zeeman splitting.

The α and β states of a proton

For nuclei such as ¹H and ¹³C, which have positive gyromagnetic ratios, the two spin-½ states are commonly described as:

α state, m_I = +½: the nuclear magnetic moment is aligned with B₀. This is the lower-energy state.

β state, m_I = −½: the nuclear magnetic moment is opposed to B₀. This is the higher-energy state.

The language of “parallel” and “anti-parallel” is a useful way to visualise this particular spin-½ system, but the underlying behaviour is quantum mechanical.

Zeeman splitting for a spin-½ nucleus such as ¹H. The applied magnetic field separates the α and β nuclear spin states.

The energy separation depends on B₀

The strength of the interaction between a nucleus and the magnetic field depends on a property of that particular nucleus called its gyromagnetic ratio, γ.

For a spin-½ nucleus, the energy separation between the two states is:

ΔE = γħB₀

Since ħ = h/(2π), this can also be written in terms of the resonance frequency:

ΔE = hν₀

ν₀ = γB₀ / 2π

The frequency ν₀ is called the Larmor frequency. Each type of nucleus has its own γ, so different nuclei have different Larmor frequencies in the same magnetic field.

The proton has γ/(2π) ≈ 42.58 MHz T⁻¹. In a field of about 9.40 T, its resonance frequency is therefore about 400 MHz. This is why an NMR spectrometer may be described as a “400 MHz” instrument.

3. Why is the lower-energy state slightly more populated?

Once B₀ separates the two spin states, the nuclei are distributed between them according to the Boltzmann distribution. At thermal equilibrium, slightly more spin-½ nuclei occupy the lower-energy α state than the higher-energy β state.

The difference is actually very small because the nuclear spin energy separation is tiny compared with ordinary thermal energies.

N_α / N_β = e^ΔE/kT

Here, N_α and N_β are the populations of the lower and higher states respectively, k is the Boltzmann constant, and T is the absolute temperature.

NMR produces a net signal because there is a small excess population in the lower-energy state.

If the two states contained exactly equal populations, transitions in opposite directions would cancel one another and there would be no net absorption associated with the population difference.

Why stronger magnets improve NMR sensitivity

Increasing B₀ increases ΔE. This slightly increases the population difference between the two states, so there is more net nuclear magnetisation available to produce an NMR signal.

Stronger magnetic fields therefore provide two major advantages: a larger resonance frequency and, in general, better sensitivity.

4. Energy absorption and resonance

The nucleus is now in a magnetic field and has quantised spin energy levels. The next question is how we make a nucleus move from the lower-energy state to the higher-energy state.

Electromagnetic radiation carries energy according to Planck's relation:

E = hν

The energy separation between nuclear spin states is very small, so the required frequencies lie in the radiofrequency region of the electromagnetic spectrum.

For a transition to occur efficiently, the RF frequency must satisfy:

hν = ΔE

Therefore:

ν = ν₀

When this condition is satisfied, the nuclear spin system is said to be in resonance.

At resonance, the frequency of the applied radiofrequency field matches the Larmor frequency, so the energy supplied matches the nuclear spin energy gap: ΔE = hν₀.

What does the RF field actually do?

An NMR spectrometer applies an oscillating radiofrequency magnetic field, conventionally called B₁, perpendicular to B₀.

When the frequency of B₁ matches the Larmor frequency of the nucleus, the RF field can drive transitions between the nuclear spin states. In a pulsed NMR experiment, a short RF pulse causes the overall nuclear magnetisation to move away from its equilibrium direction.

Off resonance: the RF frequency does not match ν₀, so energy transfer is inefficient.

At resonance: the RF frequency matches ν₀, allowing efficient interaction with the nuclear spin system.

5. The NMR experiment step by step

It is useful to connect the energy-level picture to what an NMR spectrometer actually does.

Step 1: place the sample in B₀

The sample is placed inside a very strong, highly uniform external magnetic field. The nuclear spin states become separated in energy, and a small excess of nuclei occupies the lower-energy state.

Step 2: apply a radiofrequency pulse

The sample is exposed to an RF magnetic field B₁. Frequencies capable of interacting with the nuclei are supplied during the pulse.

Step 3: excite the nuclear spin system

When the RF field is resonant with the nuclei, it changes the orientation of the net nuclear magnetisation. In the simple energy-level picture, this corresponds to inducing transitions between nuclear spin states.

Step 4: allow the nuclei to evolve after the pulse

Once the RF pulse ends, the nuclear magnetisation continues to precess around B₀. At the same time, the system gradually returns towards thermal equilibrium in processes collectively described as relaxation.

Step 5: detect the free induction decay

The precessing transverse nuclear magnetisation produces a changing magnetic field in the receiver coil. This induces a small alternating voltage that is recorded as a time-domain signal called the free induction decay, or FID.

Step 6: Fourier-transform the signal

The FID contains many frequency components superimposed on one another. A mathematical operation called a Fourier transform converts this time-domain signal into the familiar frequency-domain NMR spectrum containing peaks.

Modern pulsed NMR does not normally create a spectrum by simply measuring an absorption peak directly. It records an FID after the RF pulse and Fourier-transforms that signal into the spectrum.

6. Why nuclei in different chemical environments resonate differently

So far, we have treated every nucleus of one isotope as if it experienced exactly the same magnetic field. If that were completely true, all ¹H nuclei in a sample would resonate at essentially the same frequency and NMR would tell us very little about molecular structure.

The electrons surrounding a nucleus respond to B₀ and generate small local magnetic fields of their own. As a result, the nucleus experiences an effective magnetic field that is slightly different from the applied field.

B_effective = B₀(1 − σ)

Here, σ represents the degree of electronic shielding. Because the resonance frequency depends on the effective magnetic field, nuclei in different electronic environments can resonate at slightly different frequencies.

Nuclear spin and Zeeman splitting explain why NMR is possible. Electronic shielding explains why NMR can distinguish different chemical environments.

This leads directly to the idea of chemical shift, which allows an NMR spectrum to reveal information about molecular structure.

7. Connecting field strength, energy and frequency

Three quantities are therefore directly linked in NMR:

Quantity increased	Immediate effect	Result
B₀	Greater Zeeman splitting	ΔE increases
ΔE	More energy required for transition	ν₀ increases
B₀	Larger Boltzmann population difference	Usually improved NMR sensitivity

B₀ ↑ ⇒ ΔE ↑ ⇒ ν₀ ↑

Suppose the energy separation between two proton spin states is 2.65 × 10⁻²⁵ J. Using h = 6.63 × 10⁻³⁴ J s, ν = ΔE/h ≈ 4.00 × 10⁸ Hz = 400 MHz.

Common Mistakes

Common mistake	Why it is wrong	Better approach
Saying that nuclear spin means the nucleus is literally a tiny ball spinning about its axis.	Spin is an intrinsic quantum-mechanical property and cannot be interpreted completely using a classical rotating-ball model.	Use the rotating-magnet picture only as an analogy for the magnetic behaviour, not as a literal description.
Assuming every atomic nucleus can be observed by NMR.	Nuclei with I = 0 are NMR-inactive.	First determine whether the isotope has non-zero nuclear spin.
Saying that all proton spins become parallel to B₀.	A spin-½ nucleus has two allowed states. Both are populated, but slightly more nuclei occupy the lower-energy state.	Describe α and β states together with the small Boltzmann population excess in α.
Treating radiofrequency absorption as possible at any frequency.	Efficient energy transfer requires the resonance condition ΔE = hν.	Match the RF frequency to the Larmor frequency ν₀.
Saying that increasing B₀ decreases the resonance frequency.	ν₀ = γB₀/(2π), so ν₀ is directly proportional to B₀ for a given nucleus.	Remember the chain B₀ ↑ → ΔE ↑ → ν₀ ↑.
Saying that a modern pulsed NMR spectrum is obtained by directly measuring absorbed RF energy as a peak.	The receiver normally measures the time-dependent FID after the RF pulse.	Use the sequence RF pulse → FID → Fourier transform → spectrum.

What you should be able to do

Identify NMR-active nuclei

Explain why nuclei with I ≠ 0 can interact with an external magnetic field and distinguish common active and inactive isotopes.

Describe spin states

Use 2I + 1 to determine the number of allowed magnetic spin states and describe the two states of a spin-½ nucleus.

Explain Zeeman splitting

Explain how applying B₀ separates previously degenerate nuclear spin states into different energies.

Apply the resonance equations

Use ΔE = hν₀ and ν₀ = γB₀/(2π) to connect magnetic field strength, energy separation and frequency.

Explain the population difference

Recognise that slightly more nuclei occupy the lower-energy state at thermal equilibrium and explain why this matters for NMR sensitivity.

Describe how an NMR signal is produced

Follow the sequence B₀ → RF pulse → nuclear response → FID → Fourier transform → NMR spectrum.

Mastery Quiz

Can You Apply Nuclear Spin and Resonance?

These ten questions test whether you can connect nuclear spin, magnetic fields, quantised energy levels, resonance frequency and NMR signal detection.

10 mastery questions
Quantitative and conceptual reasoning
Instant explanation for every answer option

Question 1 of 10

Which pair contains two nuclei that are both NMR-active?

¹²C and ¹⁶O

Both are even-even nuclei with I = 0, so neither is NMR-active.

¹H and ¹³C

Correct. Both ¹H and ¹³C have I = ½ and are widely used in NMR spectroscopy.

¹²C and ¹³C

¹³C is active, but ¹²C has I = 0.

¹H and ¹²C

¹H is active, but ¹²C is not.

NMR activity depends on non-zero nuclear spin. The important common nuclei ¹H and ¹³C both have I = ½.

A nucleus has nuclear spin quantum number I = 1. How many magnetic spin states are possible?

Two states apply to I = ½, not I = 1.

A spin-1 nucleus has more than one allowed mI value.

Correct. 2I + 1 = 2(1) + 1 = 3, corresponding to mI = +1, 0 and −1.

Four states would correspond to I = 3/2.

The number of allowed magnetic states is 2I + 1. A spin-1 nucleus therefore has three allowed states.

For ¹H in an external magnetic field, which statement about the α and β states is correct?

The α state is lower in energy than the β state.

Correct. For ¹H, the magnetic moment aligned with B₀ corresponds to the lower-energy α state.

The α and β states always remain degenerate.

They are degenerate only in the absence of the applied magnetic field.

The β state is lower in energy for a proton.

For ¹H, the β state is the higher-energy state.

The states are electronic orbitals.

They are nuclear spin states, not electronic orbitals.

Applying B₀ separates the proton's two spin states. The α state is lower and the β state is higher in energy.

The energy separation between two nuclear spin states is 2.65 × 10⁻²⁵ J. Taking h = 6.63 × 10⁻³⁴ J s, what radiofrequency is required for resonance?

ΔE = hν

2.50 × 10⁻⁹ Hz

This corresponds to reversing the required division. The frequency is ΔE/h.

1.76 × 10⁻⁵⁸ Hz

Multiplying ΔE by h does not give the resonance frequency.

4.00 × 10⁶ Hz

This is two orders of magnitude too small.

4.00 × 10⁸ Hz

Correct. ν = ΔE/h = (2.65 × 10⁻²⁵)/(6.63 × 10⁻³⁴) ≈ 4.00 × 10⁸ Hz.

NMR energy gaps correspond to radiofrequencies. Here the required frequency is 400 MHz.

For the same type of nucleus, the external field B₀ is doubled. What happens to its Larmor frequency?

It becomes half as large.

ν₀ is directly rather than inversely proportional to B₀.

It doubles.

Correct. ν₀ = γB₀/(2π), and γ is constant for a particular isotope.

It becomes four times as large.

The relationship is linear, not proportional to B₀².

It remains unchanged.

The Larmor frequency depends directly on field strength.

Doubling B₀ doubles both the energy separation and the resonance frequency for a given nucleus.

At thermal equilibrium, how are the α and β populations of ¹H related?

Every proton is in the α state.

Both states are thermally populated.

Every proton is in the β state.

The higher-energy state is populated, but it does not contain all the nuclei.

Slightly more protons are in the α state than in the β state.

Correct. The Boltzmann distribution gives a small excess in the lower-energy α state.

The β population is always twice the α population.

The lower-energy state, not the higher-energy state, has the larger equilibrium population.

The population difference is small but crucial. It produces the net nuclear magnetisation that ultimately allows an NMR signal to be observed.

A proton has Larmor frequency ν₀. What is required for efficient resonance?

Apply visible light of any frequency.

Nuclear spin transitions involve radiofrequency rather than visible radiation.

Set the RF frequency to zero.

A non-zero energy separation requires a corresponding non-zero resonant frequency.

Make the RF frequency much greater than ν₀.

Resonance requires frequency matching rather than merely using a very high frequency.

Apply an RF field at frequency ν₀.

Correct. When ν = ν₀, the applied RF field is resonant with the nuclear spin system.

Resonance occurs when the frequency of the RF field matches the Larmor frequency associated with the nuclear spin energy gap.

What signal is recorded directly by the receiver immediately after an RF pulse in a modern pulsed NMR experiment?

A mass spectrum

Mass spectra are produced by mass spectrometry, not NMR.

A free induction decay

Correct. The receiver coil records the time-dependent FID produced by the precessing transverse magnetisation.

A chromatogram

Chromatograms are produced by chromatographic separation techniques.

A visible-light absorption spectrum

NMR detects an RF signal from nuclear magnetisation rather than visible-light absorption.

The FID is recorded in the time domain. A Fourier transform then converts it into the familiar frequency-domain NMR spectrum.

At the same applied field B₀, a proton becomes more strongly electronically shielded. What happens to the effective magnetic field experienced by that proton?

It increases because shielding strengthens B₀.

The induced electronic field opposes part of the applied field at the nucleus.

It is unaffected by the electrons.

Electronic shielding exists precisely because the surrounding electrons modify the local field at the nucleus.

It decreases slightly.

Correct. Greater shielding means a larger opposing local field, so the nucleus experiences a slightly smaller effective field.

It becomes exactly zero.

Shielding changes the field only slightly; it does not normally cancel the external magnetic field completely.

Differences in electronic shielding give nuclei slightly different effective fields and therefore slightly different resonance frequencies. This is the origin of chemical shift.

Which sequence best represents how a modern pulsed NMR spectrum is obtained?

Visible light → ionisation → detector → NMR spectrum

This does not describe an NMR experiment.

Magnetic field → molecular fragmentation → FID → spectrum

Molecular fragmentation is associated with techniques such as mass spectrometry, not normal NMR detection.

B₀ → RF pulse → FID → Fourier transform → NMR spectrum

Correct. This is the essential sequence in pulsed Fourier-transform NMR.

RF pulse → chromatography → magnetic field → spectrum

Chromatography is not a necessary step in producing the NMR signal itself.

The strong magnetic field establishes the nuclear spin energy levels. The RF pulse perturbs the nuclear magnetisation, the receiver detects the FID, and a Fourier transform converts the FID into the NMR spectrum.

Frequently asked questions

Questions students often ask about nuclear spin and resonance

The basic equations are short, but several ideas behind them can be easy to mix up. These questions connect the energy-level model to what physically happens in an NMR experiment.

Does a nucleus literally spin like a rotating ball?

No. Nuclear spin is an intrinsic quantum-mechanical property. Imagining the nucleus as a tiny spinning magnet can help build intuition about its magnetic behaviour, but the model should not be interpreted literally.

Why are ¹H and ¹³C useful for NMR but ¹²C is not?

¹H and ¹³C both have nuclear spin I = ½, so they possess nuclear magnetic moments and are NMR-active. ¹²C has six protons and six neutrons and has I = 0, so it does not give the ordinary nuclear-spin resonance signal used in NMR spectroscopy.

Why are there exactly two energy states for a proton?

A proton has I = ½. The number of allowed magnetic spin states is 2I + 1, so a proton has two states. Their magnetic quantum numbers are m_I = +½ and m_I = −½.

Are the α and β states different even without a magnetic field?

The states exist as possible spin states, but in the absence of an external magnetic field their magnetic energies are equal. Applying B₀ removes this degeneracy and produces the energy separation ΔE.

Why does NMR use radiofrequency radiation rather than infrared or visible light?

Nuclear spin energy separations are extremely small. Since ΔE = hν, a small energy gap corresponds to a relatively low electromagnetic frequency, which lies in the radiofrequency region. Infrared and visible photons carry much larger energies.

What exactly does resonance mean?

Resonance occurs when the frequency of the applied RF field matches the Larmor frequency of the nuclear spin system. In the simple energy-level picture, the condition is ΔE = hν₀.

Why does increasing B₀ increase the resonance frequency?

Increasing B₀ increases the Zeeman energy separation because ΔE = γħB₀. A larger energy gap requires higher-frequency radiation because ΔE = hν₀. Therefore ν₀ is directly proportional to B₀.

Why are stronger NMR magnets useful?

A stronger magnetic field increases the nuclear spin energy separation and slightly increases the population difference between the lower- and higher-energy states. This generally improves sensitivity. Higher-field instruments also provide greater separation of resonances when frequencies are converted into the chemical-shift scale.

Does the NMR instrument simply detect the energy that the nuclei absorb?

That picture is useful for introducing resonance, but modern pulsed NMR normally works differently at the detection stage. After an RF pulse, the precessing transverse nuclear magnetisation induces an electrical signal in the receiver coil. The resulting free induction decay is recorded and then Fourier-transformed to produce the spectrum.

How does nuclear spin lead to chemical shift?

Nuclear spin makes magnetic resonance possible in the first place. The surrounding electrons then slightly shield each nucleus from B₀. Nuclei in different electronic environments therefore experience slightly different effective magnetic fields and resonate at slightly different frequencies. Those differences are expressed as chemical shifts.

The essential chain is now complete: an NMR-active nucleus possesses nuclear spin, B₀ separates its magnetic spin states in energy, and an RF field can interact with the nucleus when its frequency matches the corresponding energy gap. The next step is to ask why apparently identical nuclei of the same isotope can resonate at slightly different frequencies — the idea that leads directly to shielding and chemical shift.