Relaxation of spin 1/2 nuclei: two-state derivation

From apimba
Revision as of 03:51, 14 April 2020 by Milllo (talk | contribs)
Jump to navigation Jump to search

Introduction

When an NMR sample is placed in a static magnetic field and allowed to come to equilibrium the population of each group of spin 1/2 atoms is divided into 2, with the number in each state determined by the Boltzmann Distribution:

Nh/Nl = e-DE/kT

Let us call the two states α and β spin states. For a sample of ethanol (CH3CH2OH) there will be 3 sets of 2 states: the CH3 pair, the CH2 pair and the OH pair. Each ratio will be different since the transition energy is different:

  • NβCH3/NαCH3 = ratio for CH3 at equilibrium
  • NβCH2/NαCH2 = ratio for CH2 at equilibrium
  • NβOH/NαOH = ratio for OH at equilibrium

Once the sample is excited with a 90 degree Rf pulse the populations of the 3 sets of states will change to the maximum allowed by the different energies in the system:

  • NβCH390/NαCH390 = ratio for CH3 after 90 pulse
  • NβCH290/NαCH290 = ratio for CH2 after 90 pulse
  • NβOH90/NαOH90 = ratio for OH after 90 pulse

When the transmitter is turned off, the populations will transition back to their initial values, but some nuclei will reman in the excited state longer than others due to the local environment, thus there is an rate to the process = how long it takes for the ratios to revert back to equilibrium values.