Difference between revisions of "Relaxation of spin 1/2 nuclei: two-state derivation"
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==Introduction== | ==Introduction== | ||
| − | When an NMR sample is placed in a static magnetic field and allowed to come to equilibrium | + | 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: |
| − | + | N<sub>h</sub>/N<sub>l</sub> = e<sup>-<span style="font-family:symbol;">D</span>E/kT</sup> | |
| − | + | 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. Each ratio will be different since the transition energy is different: | |
| + | *N<sub>βCH<sub>3</sub></sub>/N<sub>αCH<sub>3</sub></sub> = ratio for CH<sub>3</sub> at equilibrium | ||
| + | *N<sub>βCH<sub>2</sub></sub>/N<sub>αCH<sub>2</sub></sub> = ratio for CH<sub>2</sub> at equilibrium | ||
| + | *N<sub>βOH</sub>/N<sub>αOH</sub> = 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: | |
| + | *N<sub>βCH<sub>3</sub>90</sub>/N<sub>αCH<sub>3</sub>90</sub> = ratio for CH<sub>3</sub> after 90 pulse | ||
| + | *N<sub>βCH<sub>2</sub>90</sub>/N<sub>αCH<sub>2</sub>90</sub> = ratio for CH<sub>2</sub> after 90 pulse | ||
| + | *N<sub>βOH90</sub>/N<sub>αOH90</sub> = ratio for OH after 90 pulse | ||
| − | + | When the transmitter is turned off, the populations will transition back to their initial states, but some nuclei will remain in the excited state longer than others due to the local environment, thus there is a rate to the process = how long it takes for the ratios to revert back to equilibrium values. | |
| + | |||
| + | In general, there is an initial ratio = R<sub>i</sub>, an excited ratio at time 0 = R<sub>max</sub> and the rate of change of the ratios = dR/dt. Since β is the excited state, the ratio at equilibrium will be smaller than at any time t. So once excited the ratio will gradually decrease and approach the ratio at equilibrium again. | ||
| + | |||
| + | ==First order== | ||
| + | For a first order reaction this rate (dR/dt) = a constant times the ratio = k*R. An integration and rearrangement of this rate gives: | ||
| + | *R<sub>t</sub> = R<sub>i</sub>+(R<sub>max</sub>-R<sub>i</sub>)*e<sup>-kt</sup> | ||
| + | |||
| + | where 1/k is called the first order relaxation time, which is equal to the time it takes for 63% of the populations to return to equilibrium. This process is presumed to be due to a stabilization of the higher energy states by neighboring environments. Unfortunately, this delay time is due to multiple factors so it doesn't give much information about the system, other than the fact that long relaxation times means that there are stabilizing mechanisms in the system. | ||
Latest revision as of 16:37, 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. 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 states, but some nuclei will remain in the excited state longer than others due to the local environment, thus there is a rate to the process = how long it takes for the ratios to revert back to equilibrium values.
In general, there is an initial ratio = Ri, an excited ratio at time 0 = Rmax and the rate of change of the ratios = dR/dt. Since β is the excited state, the ratio at equilibrium will be smaller than at any time t. So once excited the ratio will gradually decrease and approach the ratio at equilibrium again.
First order
For a first order reaction this rate (dR/dt) = a constant times the ratio = k*R. An integration and rearrangement of this rate gives:
- Rt = Ri+(Rmax-Ri)*e-kt
where 1/k is called the first order relaxation time, which is equal to the time it takes for 63% of the populations to return to equilibrium. This process is presumed to be due to a stabilization of the higher energy states by neighboring environments. Unfortunately, this delay time is due to multiple factors so it doesn't give much information about the system, other than the fact that long relaxation times means that there are stabilizing mechanisms in the system.
