# Basic question about NMR

1. Jul 5, 2010

### Moogie

Hi

My basic understanding of NMR is that the nucleus of an atom is like a little bar magnet and when placed in a magnetic field this magnet can align with the magnet (low energy state) or oppose the magnet (high energy). The nucleus can enter the opposed state when radio frequency radiation is applied. The specific frequency of radiation needed depends on the effective local magnetic field experienced by the nucleus.

I don't need an understanding any greater than the one give above.

One term I come across when reading is something like the following:

'bring the nucleus into resonance'
'when the nucleus is in resonance'

What does 'in resonance' mean at the level I am describing? Does it mean 'when the nucleus is in the high energy state opposed to the applied magnetic field?' It seems to from the context I read it in but I'd like to be sure

Thanks

2. Jul 5, 2010

### Dickfore

Apart from a static orienting magnetic field, there is also a perpendicular rotating magnetic field. It is this field which creates the resonance. The static field only serves to create a two-level system (spin up has lower energy than spin down, up meaning along the static field).

3. Jul 5, 2010

### Moogie

What is the resonance then? Perhaps i only understand the spin up/down

4. Jul 5, 2010

### Moogie

What does it mean (in nice simple terms) when the nucleus is in resonance then please?

5. Jul 5, 2010

### Dickfore

I think it has something to do with the frequency of the rotating magnetic field being equal to (or half that) of the difference in the energies of the two levels divided by h. If we use the fact that the potential energy of a magnetic dipole in a magnetic field is $U = - \mathbf{\mu} \cdot \mathbf{B}$, and that the magnetic moment due to the nuclear spin is $\mathbf{\mu} = g_{J} \, \mu_{N} \, \mathbf{J}$, then the energy in the state $|j, m \rangle$ is:

$$E_{m} = -g_{J} \, \mu_{N} \, B_{0} \, m$$

The energy difference between two neighboring energy levels is:

$$\Delta E = g_{J} \, \mu_{N} \, B_{0}$$

which leads to a transition frequency of:

$$f = \frac{\Delta E}{h} = \frac{g_{J} \, \mu_{N} \, B_{0}}{h}$$

but I am not completely sure about this. Someone else?

6. Jul 5, 2010

7. Jul 5, 2010

### Feldoh

Resonance is a term that is generally thrown around when discussing stuff that exhibits periodic motion. It's essentially a way to build up a lot of energy by somehow causing a system to oscillate at a particular frequency/frequencies.

8. Jul 6, 2010

### Bob S

The proton NMR resonance frequency is the proton dipole moment flipping direction in an external magnetic field. It is a very sharp and accurately known resonance proportional to the external magnetic field (42.5781 MHz per Tesla), and has been used as a precise method for measuring magnetic field. See

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nmr.html

Bob S

9. Jul 6, 2010

### alxm

You put the proton in a magnetic field, the spins align themselves with the field (and you split the spin levels). Then you hit it with an RF pulse at 90 degrees (you don't actually have to match the energy precisely here, one typically uses a broad band pulse)

The spins absorb the radiation, changes orientation and then re-emits the radiation, returning to their original state. This is resonance. Just like playing a certain note and making say, a crystal wine glass resonate and give off the same note.

Now, the nuclear spin does not immediately emit the radiation and return to its old position. Like a spinning top that's been nudged sideways, it http://www.youtube.com/watch?v=WUkUL3Hp67A" around the axis of the field as it "rights itself" and gives off radiation. The average time this takes is called the spin-lattice relaxation time (or simply T1, in NMR lingo). So I would interpret "being in resonance" as referring to the period of time between the pulse and T1.

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