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NMR Simple Question

  1. May 5, 2013 #1
    Ok, well I'm not really sure if this is the right place to post this, but I have a simple question regarding nuclear magnetic resonance.

    I know that the "spin" axis of a proton will align along the external magnetic field. That's pretty simple.

    What I don't understand is the direction they align.

    I'm under the impression from what I've read that approximately half will align with the field (low energy) and somewhat less will align against the field (high energy).

    Why do some align against the field in the first place? It seems to me that they would naturally assume the lowest energy potential.

    Sorry if I'm viewing this too simplistically.
  2. jcsd
  3. May 6, 2013 #2


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    The fraction of high to low energy states occupied is determined bny the Boltzmann statistics:
    ##n_h/n_l =\exp(-(E_h-E_l)/kT)## where ##E_h=-E_l## is the energy of a spin in the high/low energy state and k is the Boltzmann constant. Only at very low temperatures will all spins be in the low energy state. At room temperature it is entropically favourable for both energy states to be populated almost equally.
  4. Feb 25, 2014 #3
    Funny thing is that you can make a compass needle lay in oposite direction to the external field if you are precise enough. It is just a state where torque of the field B on the moment μ is zero: [itex]\vec{τ}[/itex] = [itex]\vec{B}[/itex] x [itex]\vec{μ}[/itex] ! The same is with protons, but you dont have to be precise cause this allignement is one of their two possible quantum states.
  5. Feb 25, 2014 #4
    So why would the spin of a single electron be equally likely to "line up" with a magnetic field as opposite with it? How does the entropy come into play?
  6. Feb 25, 2014 #5
    it is not equally likely. There is greater chance for spin to be aligned parallel to the field than antiparallel.
  7. Feb 25, 2014 #6
    Yes. But keep in mind that this difference is about few ppm. Insignificant on the long scale but still fair enough for NMR phenomenon.
  8. Feb 27, 2014 #7


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    The answer is that thermal agitation at room temperature results in almost equal spin up and spin down populations at equilibrium. This is because the energy difference between the states is small compared with the mean thermal excitation energy. Increasing the static (B0) magnetic field strength increases the energy difference, increasing the equilibrium population in the parallel (ground) state relative to the anti-parallel (excited) state. This, in turn, is why there is always pressure to increase the field strength at which NMR instruments and magnetic resonance imagers operate--the SNR of the NMR signal improves.
  9. Feb 27, 2014 #8
    The entropy is a function of the number of microscopic states available. There is only one state where all the atoms align one way. There are many more states where some of the atoms align the other way.
  10. Feb 27, 2014 #9
    Curious! If a Stern Gerlach experiment were performed close to zero degrees temperature with a strong enough field would you expect to get only one line?
  11. Feb 27, 2014 #10
    But what about a single lone electron?
  12. Feb 27, 2014 #11


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    I'm not sure that this applies directly to that. The Stern-Gerlach apparatus reads out the spin states of atoms passing through it but doesn't say anything about how those atoms were prepared, and it is for the latter that temperature would matter. If you somehow prepared all atoms to be in a single state, which certainly would be done at low temperature, then yes you would get one line.
    Last edited: Feb 27, 2014
  13. Feb 27, 2014 #12
    The entropy of a single particle isn't a particularly useful concept
  14. Feb 27, 2014 #13


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    You are derailing this thread a bit. Note that in a typical NMR experiment, you are dealing with a lot of spins and with energy difference between the two spin states that are of the order of the thermal energy. It is not a "single particle" environment.

    This thread is also almost a year old.

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