Dam ...not one comment after almost 2 years
why lol
The picture indicates that the two alignments are equivalent to two quantum energy states separated by an energy difference, ΔE. More nuclei occupy the lower energy than the higher energy state, but their excess is incredibly small - less than 0.001%.
This energy difference is directly related to the precessional frequency through the following equation:
ΔE = h f
where h is Planck's Constant and f is the precessional frequency, related to the Larmor frequency, ω, through:
ω = 2 π f
The important point here is that:
transitions from the parallel to the anti-parallel state can be induced in the sample when it is excited with electromagnetic radiation of energy, ΔE.
This energy is about 1.75x10-7 eV for a proton in a 1 T field, i.e. a tiny amount of energy compared with electron binding energies, for instance. Compared to a 140 keV gamma-ray emitted by 99mTc, its energy is over a million, million times smaller.
The energy difference is equivalent to the energy of electromagnetic radiation in the radio frequency (RF) region of the electromagnetic spectrum.
Does this sounds ok?