Magnetic Resonance — Question 1

Section 40-5 · Physics pages 6652–6653

Question 1 Electron and proton resonances probe very different frequency bands Use the magnitudes |γe|/(2π) = 28.0 GHz/T and γp/(2π) = 42.6 MHz/T. A spin-half Zeeman gap is ∆E = ℏ|γ|B. Both species experience B = 1.20 T. lower upper ∆E Schematic; labels, not drawing lengths, specify the data. (a) Find each resonance frequency. (b) Find the ratio of their photon energies and the proton field required to resonate at the electron frequency found in part (a).
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Question 1 - Solution Electron and proton resonances probe very different frequency bands lower upper ∆E (a) Method and calculation hf = ℏ|γ|B ⇒f = |γ|B/(2π). fe = 33.6 GHz, fp = 51.12 MHz . (b) Interpretation and further calculation Eγ,e/Eγ,p = fe/fp = 657.3 . Bp,required = 33600/42.6 = 788.7 T . Physical check The comparison uses the same field but different gyromagnetic ratios. The very large inferred proton field is an algebraic comparison, not an assumed laboratory operating condition.

Question 1

Electron and proton resonances probe very different frequency bands

Use the magnitudes |γe|/(2π)=28.0GHz/T|\gamma_e|/(2\pi)=28.0\,\mathrm{GHz/T} and γp/(2π)=42.6MHz/T\gamma_p/(2\pi)=42.6\,\mathrm{MHz/T}. A spin-half Zeeman gap is ΔE=ℏ|γ|B\Delta E=\hbar|\gamma|B. Both species experience B=1.20TB=1.20\,\mathrm T.

Diagram shown in the original SVG worksheet.

  1. Find each resonance frequency.

  2. Find the ratio of their photon energies and the proton field required to resonate at the electron frequency found in part (a).

Question 1 - Solution

Electron and proton resonances probe very different frequency bands

Diagram shown in the original SVG worksheet.

(a) Method and calculation

hf=ℏ|γ|B⇒f=|γ|B/(2π).hf=\hbar|\gamma|B\Rightarrow f=|\gamma|B/(2\pi).fe=33.6GHz,fp=51.12MHz.\boxed{f_e=33.6\,\mathrm{GHz},\quad f_p=51.12\,\mathrm{MHz}}.

(b) Interpretation and further calculation

Eγ,e/Eγ,p=fe/fp=657.3.\boxed{E_{\gamma,e}/E_{\gamma,p}=f_e/f_p=657.3}.Bp,required=33600/42.6=788.7T.\boxed{B_{p,\mathrm{required}}=33600/42.6=788.7\,\mathrm T}.

Physical check

The comparison uses the same field but different gyromagnetic ratios. The very large inferred proton field is an algebraic comparison, not an assumed laboratory operating condition.