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

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:

N_h/N_l = 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:

• N_βCH3/N_αCH3 = ratio for CH₃ at equilibrium
• N_βCH2/N_αCH2 = ratio for CH₂ at equilibrium
• N_βOH/N_αOH = 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_βCH390/N_αCH390 = ratio for CH₃ after 90 pulse
• N_βCH290/N_αCH290 = ratio for CH₂ after 90 pulse
• N_βOH90/N_αOH90 = 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_i and an excited ratio at time 0 = R_max 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 system will gradually approach the ratio at equilibrium again.

First order

For a first order reaction this rate = a constant times the ratio = k*R. An integration of this rate gives:

• R_t = R_i+R_max*e^-kt