Difference between revisions of "Relaxation of spin 1/2 nuclei: two-state derivation"

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==Introduction==
 
==Introduction==
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When an NMR sample is placed in a static magnetic field and allowed to come to equilibrium it is found that a net magnetization of the sample along the direction of the applied field (traditionally the z-axis) is developed.  Magnetization parallel to the applied field is termed longitudinal. This equilibrium magnetization arises from the unequal population of the two energy levels that correspond to the α and β spin states.  In fact, the z-magnetization, M<sub>z</sub>, is proportional to the population difference:
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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:
  
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M<sub>z</sub>z ∝ (n<sub>α</sub> - n<sub>β</sub>)
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N<sub>h</sub>/N<sub>l</sub> = e<sup>-<span style="font-family:symbol;">D</span>E/kT</sup>
  
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where n<sub>α</sub> and n<sub>β</sub> are the populations of the two corresponding energy levels. Ultimately, the constant of proportion just determines the absolute size of the signal we will observe. As we are generally interested in the relative size of magnetizations and signals we may just as well write:
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Let us call the two states α and β spin states. For a sample of ethanol (CH<sub>3</sub>CH<sub>2</sub>OH) there will be 3 sets of 2 states: the CH<sub>3</sub> pair, the CH<sub>2</sub> pair and the OH pair.  
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*N<sub>βCH<sub>3</sub></sub>/N<sub>α3CH<sub>3</sub></sub> = ratio for CH<sub>3</sub> at equilibrium
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*N<sub>β</sub>/N<sub>α2</sub> = ratio for CH<sub>2</sub> at equilibrium
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*N<sub>β</sub>/N<sub>α</sub> = ratio for OH at equilibrium
  
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M<sub>z</sub>z = (n<sub>α</sub> - n<sub>β</sub>)
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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:
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*N<sub>β90</sub>/N<sub>α90</sub> = ratio for CH<sub>3</sub> after 90 pulse
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Mathematically this system can be treated similar to chemical kinetics.
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*N<sub>β90</sub>/N<sub>α90</sub> = ratio for CH<sub>2</sub> after 90 pulse
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*N<sub>β90</sub>/N<sub>α90</sub> = ratio for OH after 90 pulse
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Let the populations of the α and β states at time t be n<sub>αt</sub> and n<sub>βt</sub> , respectively.  If these are not the equilibrium values, then for the system to reach equilibrium the population of one level must increase and that of the other must decrease.  This implies that there must be transitions between the two levels i.e. something must happen which causes a spin to move from the α state to the β state or vice versa.
 
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==First order rate assumption==
 
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First, assume that the rate of transitions is first order, so there will be a constant of proportionality (call it W) that relates the rate of transition to the populations of the states:
 
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rate = W*n<sub>α</sub>
 

Revision as of 03:39, 14 April 2020

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.
  • NβCH3/Nα3CH3 = ratio for CH3 at equilibrium
  • Nβ/Nα2 = ratio for CH2 at equilibrium
  • Nβ/Nα = 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:

  • Nβ90/Nα90 = ratio for CH3 after 90 pulse
  • Nβ90/Nα90 = ratio for CH2 after 90 pulse
  • Nβ90/Nα90 = ratio for OH after 90 pulse
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