What is the formla for the energy difference between the allowed nuclear spin states for $""^1 H$ nuclei?

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Answer 1 · 2016-01-05T02:31:38

The formula is $ΔE = hν = γh(B_0-B_e)$ .

In the presence of an applied external magnetic field, $""_1"H"$ nuclei exist in two nuclear spin states of different energy.

A large magnet creates a homogeneous external magnetic field, $B_0$ .

The energy difference $ΔE$ between the spin states is proportional to the strength of $B_0$ .

NMR spectroscopy records transitions between these spin states induced by a radio frequency electromagnetic field called the $B_1$ field.

In practice, the electrons surrounding the nucleus produce a small magnetic field $B_e$ that opposes the $B_0$ field.

The field experienced by the proton is then $(B_0-B_e)$ .

Finally, there is a proportionality constant $γ$ (the gyromagnetic ratio) that is characteristic of a proton.

The final equation is

$ΔE = hν = γh(B_0-B_e)$ , where

$ν$ is the frequency of the $B_1$ field.

The graph below shows how the energy difference between the two spin states varies with the applied field.

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What is the formla for the energy difference between the allowed nuclear spin states for ""^1 H""^1 H nuclei?