H3 Chemistry 9813 · Study focus: H3 Chemistry: Explain energy absorption in NMR
H3 Chemistry: Explain energy absorption in NMR
Start from the governing chemical model, test it against evidence, then transfer the reasoning to an unfamiliar case.
Your success criteria
- Explain energy absorption in NMR
- Use named chemical evidence.
- Transfer the governing reason to an unfamiliar case.
Match radiofrequency to ΔE
For a supplied proton gap 2.00×10⁻²⁶ J, absorption occurs when hf equals that gap.
With h=6.63×10⁻³⁴ J s, the matching frequency is 3.02×10⁷ Hz; quantitative transition calculations are illustrative, not required by the outcome.
Resonance
Resonance is absorption when radiofrequency photon energy matches the separation between field-split nuclear-spin states.
The transition takes α to β; relaxation subsequently returns population toward equilibrium.
Increasing effective field increases ΔE and the matching frequency. Electrons shield a nucleus so chemically different protons experience slightly different effective fields.
A signal requires a population difference; equal upward and downward transition rates with equal populations would give no net absorption.
| B₀ condition | Effective field | ΔE / J | Matching f / Hz | Transition |
|---|---|---|---|---|
| B₀ = a | a | 2.00×10⁻²⁶ | 3.02×10⁷ | α→β |
| B₀ = b | b > a | >2.00×10⁻²⁶ | >3.02×10⁷ | α→β |
Text alternative: Text alternative states state order, arrow direction, all values/units and the qualitative effect of increasing field.
Field, shielding and frequency
Increasing effective field increases ΔE and the matching frequency. Electrons shield a nucleus so chemically different protons experience slightly different effective fields.
A signal requires a population difference; equal upward and downward transition rates with equal populations would give no net absorption.
Calculate a matching frequency
Use f=ΔE/h=(2.00×10⁻²⁶)/(6.63×10⁻³⁴)=3.02×10⁷ Hz.
Label the photon radiofrequency and the transition nuclear α→β.
- Find f for ΔE=2.00×10⁻²⁶ J using the supplied h.
Open the feedback checkpoint after attempting
- Credit 3.02×10⁷ Hz with units and nuclear-state interpretation.
Predict a field change
A stronger B0 produces a larger split for the same nucleus.
Because ΔE=hf, the resonance frequency rises rather than falls.
- State the frequency response when B0 increases.
Open the feedback checkpoint after attempting
- Frequency increases; explicitly link larger ΔE to f through Planck's relation.
Use shielding qualitatively
A more shielded proton experiences a lower effective field at fixed B0.
Its resonance differs slightly from a deshielded proton, creating chemical-shift information.
- Compare effective field and resonance for two differently shielded protons.
Open the feedback checkpoint after attempting
- State lower effective field for the more shielded environment and avoid claiming a different isotope.
Intensity cannot replace energy matching
Increasing radiofrequency power supplies more photons but does not make a photon of the wrong frequency match ΔE.
Resonance remains frequency-selective.
- Repair: “Any radiofrequency is absorbed if the pulse is intense enough.”
Open the feedback checkpoint after attempting
- Require hf=ΔE; intensity alone cannot repair a frequency mismatch.
From resonance to chemical shift
Complete twelve energy-matching checks, then unseen field-scaling assessment and different shielding re-test.
Next objective: interpret chemical shift at /learning/h3-nmr-chemical-shift-lesson.html.
- Annotate a resonance energy diagram with B0, ΔE, hf and α→β.
Open the feedback checkpoint after attempting
- Credit exact matching, upward absorption and the nuclear-state boundary.