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Question about the Boltzmann distribution in combination with NMR

  1. Apr 22, 2006 #1
    I need to answer the following question:

    A simple energy level system has two energy levels. These are the energy levels matching the spin of a proton in a magnetic field. This is important for NMR. In that case the energy difference depends on the used magnetic field, but for a typical NMR machine of 400 MHz (that's the frequency in which the transitions take place for the proton) that energy difference is 0.013 cm-1 (or 2,5823616 * 10-25 J)

    a) Calculate the chance that the system is in the ground state at room temperature (293 K).

    I know the Boltzmann distribution is:

    P(Ei) = (e-EiB) / ([tex]\Sigma[/tex]n e-EnB)

    with [tex]B= \frac{1}{k_B T}[/tex] (at room temperature I've calculated that B= 2,470736379 * 10 20)

    The only question is: how do I calculate the chance for the proton to be in the ground state if I don't know the energy of the ground state (only the energy difference between the two states) ????

    b) While doing an NMR experiment all the spins are reversed. Why do people sometimes describe the result of this action with a negative temperature?

    Well as for this question I know it has something to do with the Boltzmann distribution and the fact that the chances can't be bigger than 1, but could somebody please help me to figure out the EXACT reason?

    I'll appreciate that a hell of a lot!!!
     
  2. jcsd
  3. Apr 22, 2006 #2
    UPDATE: I've managed to figure out the first question!!!
    I'd appreciate a bit of help with the second one very much!.
     
  4. Apr 22, 2006 #3

    siddharth

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