Based on Boltzmann statistics, which molecule in question number one will have the greater signal? At which field strength will the signal be greater? Based on natural and biological abundance, which nucleus will have the greater signal?

The Bolztmann equation tells us that N-/N+ = e-ΔE/kT
Where N-/N+ is the ratio of spins in the upper to those in the lower spin states. The MRI signal is proportional to ( N+ - N- ), and

( N+ - N- ) = [1 - N-/N+ ] / [1 + N-/N+].
Taking only Boltzmann statistics into account and assuming that T=310K (body temperature):

 ( N+ - N- ) γ (MHz/T) Bo=1.5T B0=4.7T 1H 42.58 4.944024x10-6 1.549173x10-5 23Na 11.27 1.307501x10-6 4.09651x10-6 31P 17.25 2.002004x10-6 6.272539x10-6

In calculating the relative signals, we must take into account the natural abundances of the isotopes (NISO). In calculating the relative signals from the body we must also include the biological abundances (NBIO).

Signal = k NISO NBIO (N+ - N-)

In this euation, k is a proportionality constant.

 Relative Signals NISO NBIO Bo=1.5T B0=4.7T 1H 99.98 0.63 3.11411x10-6 9.75784x10-6 23Na 100 0.00041 5.36075x10-10 1.67957x10-9 31P 100 0.0024 4.80481x10-9 1.50541x10-8

Therefore, of the three nuclei, hydrogen will have the best signal. Of the two field strengths, 4.7 T will have the better signal. The signal will be proportional to the gyromagnetic ratio and the field strength.