10.3 NMR Fundamentals

Transcription

1 10.3 NMR Fundamentals nuclear spin calculations and examples NMR properties of selected nuclei the nuclear magnetic moment and precession around a magnetic field the spin quantum number and the NMR transition absorption strength and the number density of states NMR sensitivity saturation and absorption with incoherent detection 10.3 : 1/12

2 Nuclear Spin Both protons and neutrons have a spin of 1/2. Protons and neutrons have separate sets of "orbits" which can hold two particles with their spins paired. protons neutrons To estimate the spin use the nuclear shell model, which arranges the "orbits" according to the above graph. if the number of protons is even, they are all paired with a net spin of zero if the number of neutrons is even, they are all paired with a net spin of zero if the number of protons is odd, place them into the orbits one at a time using Hund's rule repeat for an odd number of neutrons the nuclear spin is the sum of the spins for the protons and neutrons 10.3 : 2/12

3 Example Nuclear Spins 1 H - the proton is odd, the nuclear spin is 1/2 2 H - both the proton and neutron numbers are odd, no spin is paired, the nuclear spin is 2 1/2 = 1 10 B - both the proton and neutron numbers are odd, the proton configuration is ( ) (,, ) the neutron configuration is ( ) (,, ), the nuclear spin is 6 1/2 = 3 11 B - the proton number is odd, the neutron number is even, the proton configuration is ( ) (,, ), the nuclear spin is 3 1/2 = 3/2 12 C - both the proton and neutron numbers are even, all spins are paired, the nuclear spin is 0 13 C - the proton number is even, the neutron number is odd, the neutron configuration is ( ) (,, ), the nuclear spin is 1/2 14 N - both the proton and neutron numbers are odd, the proton configuration is ( ) (,, ), the neutron configuration is ( ) (,, ), the nuclear spin is 2 1/2 = 1 15 N - the proton number is odd the neutron number even, the proton configuration is ( ) (,, ), the nuclear spin is 1/ : 3/12

4 Properties of Selected Nuclei isotope abundance (%) nuclear spin nuclear magnetic moment, µ N (magnetons)* magnetogyric ratio, γ (Hz T -1 ) 1 H / H C 1.1 1/ N P 100 1/ F 100 1/ * one nuclear magneton is J T -1. One Tesla is the magnetic field force necessary to move one coulomb of charge one meter per second along a path perpendicular to the field : 4/12

5 Nuclear Magnetic Moment A spinning charge, I, generates a magnetic moment. The nuclear magnetic moment, µ N, is related to the nuclear spin by the following expression, µ N = γhi where h is Plank's constant ( ), γ is the magnetogyric ratio, and I is unitless. The magnetogyric ratio is characteristic of the nucleus and is due to the summation of the magnetic moments for all the neutrons and protons. Example: 1 H µ N = ( Hz T -1 )( J Hz -1 ) 0.5 = J T -1 = 2.79 magneton 10.3 : 5/12

6 Precession or Larmor Frequency A nucleus with a magnetic moment can be treated as if it were a bar magnetic. When a bar magnetic is placed in an external magnetic field it precesses clockwise in a circle about the applied field. It is this motion that couples to electromagnetic radiation. When the radiation frequency equals the precession frequency, energy is coupled into the spinning magnetic moment and changes its orientation with respect to the field. B µ N 10.3 : 6/12

7 Spin Quantum Number The magnetic moment can align itself with an external magnetic field in certain allowed ways given by the magnetic quantum number. m I = I, (I - 1),..., -(I -1), -I Where positive m I are aligned with the external field while negative values are aligned against the field. A value of zero is perpendicular to the external field. The energy of each magnetic moment orientation is given by E = -γ h m I B where a negative energy is lower than a positive energy. The energy of the orientation depends upon the nucleus, spin quantum number, and the strength of the external field : 7/12

8 The NMR Transition For a spin 1/2 nucleus the transition energy, E, is given by the following. E = -γh(-1/2)b -(-γh(+1/2)b) = γhb E 0 E m I = -½ The frequency of the transition can be determined by using Planck's Law. ν = E/h = γb 0 B m I = +½ Thus the frequency of an NMR transition depends upon the magnetogyric ratio (nuclear identity) and the applied field. To obtain an NMR spectrum one could fix the frequency and sweep the magnetic field. This is the way cw NMR is done. Alternatively, one could fix the magnetic field and sweep the frequencies. This is the way FT-NMR is done : 8/12

9 Absorption Consider a two level system. Absorption of radiation at the resonance frequency is given by, A = σl(n α - N β ) where σ is the transition cross section, l is the sample path length, and N is the number density of each state. With electronic transitions in the UV visible, N β is approximately zero, yielding Beer's Law. E β > α > A = σln α Note that Beer's Law is more familiar when written with molar absorptivity and molar concentrations. A = εlc 10.3 : 9/12

10 NMR Sensitivity In the NMR spectral region ( MHz) kt is far greater than E. at 300 K, kt = J at 300 MHz, E = J The thermal population of N β is given by the Boltzmann distribution. Since N = N α + N β, N α - N β = N, or, the technique is about 10,000 times less sensitive than UV visible absorption, based on N α - N β alone. This makes sense since UV visible requires ~10-6 molar solutions to obtain a spectrum while NMR requires millimolar solutions. Sensitivity and field strength: as the field strength increases, E increases as E increases, the N α /N β ratio increases as the ratio increases, the sensitivity increases 10.3 : 10/12

11 Saturation and Absorption (1) Absorption of radiation converts nuclei in α levels into nuclei in β levels - one photon is lost from the field and added to the nuclear magnetic energy. Stimulated emission of radiation converts nuclei in β levels into nuclei in α levels - one photon is added to the field and lost from the nuclear magnetic energy. In the UV visible, stimulated emission is difficult to observe because N β is ordinarily very small. That is, the number of photons added back into the radiation field is much smaller than the number taken from the field. In the NMR region, N β is virtually the same value as N α. Therefore stimulated emission and absorption occur to about the same extent. If we start with a thermal N α and apply incoherent radiation, N α will decrease until N α = N β. At this point the rate of incoherent absorption equals the rate of stimulated emission, and the sample appears to be transparent. This is called optical saturation : 11/12

12 Saturation and Absorption (2) Operationally, applying incoherent electromagnetic radiation will produce absorption, which will decrease in strength as the nuclei in the α levels are converted to β levels. Ultimately, the absorption will go to zero. Some method must exist to return nuclei in the β levels back to the α levels. In the UV visible, a β state will return to the α state in nanoseconds. This process is called spontaneous emission. The rate of spontaneous emission is proportional to the inverse of the transition frequency squared. An NMR frequency at 10 8 Hz is 10 6 smaller than an optical frequency at Hz. This means that the spontaneous emission rate will be times smaller, or around 10 3 seconds. As a result, incoherent absorption will quickly disappear and does not produce a usable NMR signal : 12/12