Magnetism and Electrons — Question 4

Section 32-6 · Physics pages 5278–5279

Question 4 A spin resonance measures the energy gap between two projections An electron spin in field B=0.350 T has gs = 2. Use µB = 9.274 × 10−24 J/T and h = 6.626 × 10−34 J s. Ignore all other interactions. −µBB +µBB ∆E Schematic; labels, not drawing lengths, specify the data. (a) Find the energy gap and photon frequency for a spin transition. (b) Find the field required to double that frequency and identify which moment projection absorbs a photon.
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Question 4 - Solution A spin resonance measures the energy gap between two projections −µBB +µBB ∆E (a) Method and calculation ∆E = (+µBB) −(−µBB) = 2µBB = 6.4918 × 10−24 J . f = ∆E/h = 9.797 × 109 Hz . (b) Interpretation and further calculation f ∝B ⇒B2f = 0.700 T . µz = +µB −→−µB on absorption. Physical check The lower energy corresponds to moment parallel to B, which for an electron means spin angular momentum antiparallel to B.

Question 4

A spin resonance measures the energy gap between two projections

An electron spin in field B=0.350 T has gs=2g_s=2. Use μB=9.274×10−24J/T\mu_B=9.274\times10^{-24}\,\mathrm{J/T} and h=6.626×10−34Jsh=6.626\times10^{-34}\,\mathrm{J\,s}. Ignore all other interactions.

Diagram shown in the original SVG worksheet.

  1. Find the energy gap and photon frequency for a spin transition.

  2. Find the field required to double that frequency and identify which moment projection absorbs a photon.

Question 4 - Solution

A spin resonance measures the energy gap between two projections

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ΔE=(+μBB)−(−μBB)=2μBB=6.4918×10−24J.\Delta E=(+\mu_BB)-(-\mu_BB)=2\mu_BB=\boxed{6.4918\times10^{-24}\,\mathrm J}.f=ΔE/h=9.797×109Hz.\boxed{f=\Delta E/h=9.797\times10^9\,\mathrm{Hz}}.

(b) Interpretation and further calculation

f∝B⇒B2f=0.700T.f\propto B\Rightarrow\boxed{B_{2f}=0.700\,\mathrm T}.μz=+μB→−μBon absorption.\boxed{\mu_z=+\mu_B\ \longrightarrow\ -\mu_B}\quad\text{on absorption}.

Physical check

The lower energy corresponds to moment parallel to B, which for an electron means spin angular momentum antiparallel to B.